判断一个数组是否是另一个数组的子集|添加方法5

给定两个数组：arr1[0..m-1] 和 arr2[0..n-1]。查找 arr2[] 是否是 arr1[] 的子集。两个数组都没有排序。可以假设两个数组中的元素是不同的。

例子：

Input: arr1[] = {11, 1, 13, 21, 3, 7}, arr2[] = {11, 3, 7, 1} 
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {1, 2, 3, 4, 5, 6}, arr2[] = {1, 2, 4} 
Output: arr2[] is a subset of arr1[]

Input: arr1[] = {10, 5, 2, 23, 19}, arr2[] = {19, 5, 3} 
Output: arr2[] is not a subset of arr1[] 

方法一（简单） ：
使用两个循环：外层循环一一选取 arr2[] 的所有元素。内循环线性搜索外循环选取的元素。如果找到所有元素，则返回 1，否则返回 0。

C++

// C++ program to find whether an array
// is subset of another array
#include 
 
/* Return 1 if arr2[] is a subset of
arr1[] */
bool isSubset(int arr1[], int arr2[],
              int m, int n)
{
    int i = 0;
    int j = 0;
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            if (arr2[i] == arr1[j])
                break;
        }
 
        /* If the above inner loop was
        not broken at all then arr2[i]
        is not present in arr1[] */
        if (j == m)
            return 0;
    }
 
    /* If we reach here then all
    elements of arr2[] are present
    in arr1[] */
    return 1;
}
 
// Driver code
int main()
{
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
 
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
 
    if (isSubset(arr1, arr2, m, n))
        printf("arr2[] is subset of arr1[] ");
    else
        printf("arr2[] is not a subset of arr1[]");
 
    getchar();
    return 0;
}

Java

// Java program to find whether an array
// is subset of another array
 
class GFG {
 
    /* Return true if arr2[] is a subset
    of arr1[] */
    static boolean isSubset(int arr1[],
                            int arr2[],
                            int m, int n)
    {
        int i = 0;
        int j = 0;
        for (i = 0; i < n; i++) {
            for (j = 0; j < m; j++)
                if (arr2[i] == arr1[j])
                    break;
 
            /* If the above inner loop
            was not broken at all then
            arr2[i] is not present in
            arr1[] */
            if (j == m)
                return false;
        }
 
        /* If we reach here then all
        elements of arr2[] are present
        in arr1[] */
        return true;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr1[] = { 11, 1, 13, 21, 3, 7 };
        int arr2[] = { 11, 3, 7, 1 };
 
        int m = arr1.length;
        int n = arr2.length;
 
        if (isSubset(arr1, arr2, m, n))
            System.out.print("arr2[] is "
                             + "subset of arr1[] ");
        else
            System.out.print("arr2[] is "
                             + "not a subset of arr1[]");
    }
}

Python3

# Python 3 program to find whether an array
# is subset of another array
 
# Return 1 if arr2[] is a subset of
# arr1[]
def isSubset(arr1, arr2, m, n):
    i = 0
    j = 0
    for i in range(n):
        for j in range(m):
            if(arr2[i] == arr1[j]):
                break
         
        # If the above inner loop was
        # not broken at all then arr2[i]
        # is not present in arr1[]
        if (j == m):
            return 0
     
    # If we reach here then all
    # elements of arr2[] are present
    # in arr1[]
    return 1
 
# Driver code
if __name__ == "__main__":
     
    arr1 = [11, 1, 13, 21, 3, 7]
    arr2 = [11, 3, 7, 1]
 
    m = len(arr1)
    n = len(arr2)
 
    if(isSubset(arr1, arr2, m, n)):
        print("arr2[] is subset of arr1[] ")
    else:
        print("arr2[] is not a subset of arr1[]")
 
# This code is contributed by ita_c

C#

// C# program to find whether an array
// is subset of another array
using System;
 
class GFG {
 
    /* Return true if arr2[] is a
    subset of arr1[] */
    static bool isSubset(int[] arr1,
                         int[] arr2,
                         int m, int n)
    {
        int i = 0;
        int j = 0;
        for (i = 0; i < n; i++) {
            for (j = 0; j < m; j++)
                if (arr2[i] == arr1[j])
                    break;
 
            /* If the above inner loop
            was not broken at all then
            arr2[i] is not present in
            arr1[] */
            if (j == m)
                return false;
        }
 
        /* If we reach here then all
        elements of arr2[] are present
        in arr1[] */
        return true;
    }
 
    // Driver function
    public static void Main()
    {
        int[] arr1 = { 11, 1, 13, 21, 3, 7 };
        int[] arr2 = { 11, 3, 7, 1 };
 
        int m = arr1.Length;
        int n = arr2.Length;
 
        if (isSubset(arr1, arr2, m, n))
            Console.WriteLine("arr2[] is subset"
                              + " of arr1[] ");
        else
            Console.WriteLine("arr2[] is not a "
                              + "subset of arr1[]");
    }
}
 
// This code is contributed by Sam007

Javascript

 
// JavaScript program to find whether an array
// is subset of another array
 
    /* Return true if arr2[] is a subset
    of arr1[] */
    function isSubset(arr1, arr2, m, n)
    {
        let i = 0;
        let j = 0;
        for (i = 0; i < n; i++) {
            for (j = 0; j < m; j++)
                if (arr2[i] == arr1[j])
                    break;
 
            /* If the above inner loop
            was not broken at all then
            arr2[i] is not present in
            arr1[] */
            if (j == m)
                return false;
        }
 
        /* If we reach here then all
        elements of arr2[] are present
        in arr1[] */
        return true;
    }
 
// Driver Code
    let arr1 = [ 11, 1, 13, 21, 3, 7 ];
    let arr2 = [ 11, 3, 7, 1 ];
 
    let m = arr1.length;
    let n = arr2.length;
 
    if (isSubset(arr1, arr2, m, n))
      document.write("arr2[] is "
                     + "subset of arr1[] ");
    else
      document.write("arr2[] is "
                    + "not a subset of arr1[]");
 

输出

arr2[] is subset of arr1[] 

时间复杂度： O(m*n)

方法 2（使用排序和二分搜索） ：

• 排序 arr1[] 需要 O(mLogm)
• 对于 arr2[] 的每个元素，在已排序的 arr1[] 中对其进行二分搜索。
• 如果未找到该元素，则返回 0。
• 如果所有元素都存在，则返回 1。

C++

// C++ program to find whether an array
// is subset of another array
#include 
using namespace std;
 
/* Function prototypes */
void quickSort(int* arr, int si, int ei);
int binarySearch(int arr[], int low,
                 int high, int x);
 
/* Return 1 if arr2[] is a subset of arr1[] */
bool isSubset(int arr1[], int arr2[],
              int m, int n)
{
    int i = 0;
 
    quickSort(arr1, 0, m - 1);
    for (i = 0; i < n; i++) {
        if (binarySearch(arr1, 0, m - 1,
                         arr2[i])
            == -1)
            return 0;
    }
 
    /* If we reach here then all elements
     of arr2[] are present in arr1[] */
    return 1;
}
 
/* FOLLOWING FUNCTIONS ARE ONLY FOR
    SEARCHING AND SORTING PURPOSE */
/* Standard Binary Search function*/
int binarySearch(int arr[], int low,
                 int high, int x)
{
    if (high >= low)
    {
        /*low + (high - low)/2;*/
        int mid = (low + high) / 2;
 
        /* Check if arr[mid] is the first
        occurrence of x. arr[mid] is first
        occurrence if x is one of the following
        is true:
        (i) mid == 0 and arr[mid] == x
        (ii) arr[mid-1] < x and arr[mid] == x    */
        if ((mid == 0 || x > arr[mid - 1]) && (arr[mid] == x))
            return mid;
        else if (x > arr[mid])
            return binarySearch(arr, (mid + 1), high, x);
        else
            return binarySearch(arr, low, (mid - 1), x);
    }
    return -1;
}
 
void exchange(int* a, int* b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}
 
int partition(int A[], int si, int ei)
{
    int x = A[ei];
    int i = (si - 1);
    int j;
 
    for (j = si; j <= ei - 1; j++) {
        if (A[j] <= x) {
            i++;
            exchange(&A[i], &A[j]);
        }
    }
    exchange(&A[i + 1], &A[ei]);
    return (i + 1);
}
 
/* Implementation of Quick Sort
A[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
void quickSort(int A[], int si, int ei)
{
    int pi; /* Partitioning index */
    if (si < ei) {
        pi = partition(A, si, ei);
        quickSort(A, si, pi - 1);
        quickSort(A, pi + 1, ei);
    }
}
 
/*Driver code */
int main()
{
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
 
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
 
    if (isSubset(arr1, arr2, m, n))
        cout << "arr2[] is subset of arr1[] ";
    else
        cout << "arr2[] is not a subset of arr1[] ";
 
    return 0;
}
 
// This code is contributed by Shivi_Aggarwal

C

// C program to find whether an array
// is subset of another array
#include 
#include 
/* Function prototypes */
void quickSort(int* arr, int si, int ei);
int binarySearch(int arr[], int low,
                 int high, int x);
 
/* Return 1 if arr2[] is a subset of arr1[] */
bool isSubset(int arr1[], int arr2[],
              int m, int n)
{
    int i = 0;
 
    quickSort(arr1, 0, m - 1);
    for (i = 0; i < n; i++) {
        if (binarySearch(arr1, 0, m - 1,
                         arr2[i]) == -1)
            return 0;
    }
 
    /* If we reach here then all elements of arr2[]
      are present in arr1[] */
    return 1;
}
 
/* FOLLOWING FUNCTIONS ARE ONLY FOR SEARCHING
AND SORTING PURPOSE */
/* Standard Binary Search function*/
int binarySearch(int arr[], int low, int high, int x)
{
    if (high >= low)
    {
        /*low + (high - low)/2;*/
        int mid = (low + high) / 2;
 
        /* Check if arr[mid] is the first
        occurrence of x.
        arr[mid] is first occurrence if x is
        one of the following
        is true:
        (i)  mid == 0 and arr[mid] == x
        (ii) arr[mid-1] < x and arr[mid] == x
     */
        if ((mid == 0 || x > arr[mid - 1]) && (arr[mid] == x))
            return mid;
        else if (x > arr[mid])
            return binarySearch(arr, (mid + 1), high, x);
        else
            return binarySearch(arr, low, (mid - 1), x);
    }
    return -1;
}
 
void exchange(int* a, int* b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}
 
int partition(int A[], int si, int ei)
{
    int x = A[ei];
    int i = (si - 1);
    int j;
 
    for (j = si; j <= ei - 1; j++) {
        if (A[j] <= x) {
            i++;
            exchange(&A[i], &A[j]);
        }
    }
    exchange(&A[i + 1], &A[ei]);
    return (i + 1);
}
 
/* Implementation of Quick Sort
A[] --> Array to be sorted
si  --> Starting index
ei  --> Ending index
*/
void quickSort(int A[], int si, int ei)
{
    int pi; /* Partitioning index */
    if (si < ei) {
        pi = partition(A, si, ei);
        quickSort(A, si, pi - 1);
        quickSort(A, pi + 1, ei);
    }
}
 
/*Driver code */
int main()
{
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
 
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
 
    if (isSubset(arr1, arr2, m, n))
        printf("arr2[] is subset of arr1[] ");
    else
        printf("arr2[] is not a subset of arr1[] ");
 
    return 0;
}

Java

// Java program to find whether an array
// is subset of another array
class Main {
    /* Return true if arr2[] is a subset of arr1[] */
    static boolean isSubset(int arr1[],
                            int arr2[], int m,
                            int n)
    {
        int i = 0;
 
        sort(arr1, 0, m - 1);
        for (i = 0; i < n; i++) {
            if (binarySearch(arr1,
                             0, m - 1,
                             arr2[i]) == -1)
                return false;
        }
 
        /* If we reach here then all elements of arr2[]
          are present in arr1[] */
        return true;
    }
 
    /* FOLLOWING FUNCTIONS ARE ONLY
    FOR SEARCHING AND
     * SORTING PURPOSE */
    /* Standard Binary Search function*/
    static int binarySearch(int arr[],
                            int low, int high,
                            int x)
    {
        if (high >= low)
        {
            /*low + (high - low)/2;*/
            int mid = (low + high)
                      / 2;
 
            /* Check if arr[mid] is the first occurrence of
            x. arr[mid] is first occurrence if x is one of
            the following is true: (i)  mid == 0 and
            arr[mid] == x (ii) arr[mid-1] < x and arr[mid]
            == x
         */
            if ((mid == 0 || x > arr[mid - 1])
                && (arr[mid] == x))
                return mid;
            else if (x > arr[mid])
                return binarySearch(arr,
                                    (mid + 1), high,
                                    x);
            else
                return binarySearch(arr, low,
                                    (mid - 1), x);
        }
        return -1;
    }
 
    /* This function takes last element as pivot,
       places the pivot element at its correct
       position in sorted array, and places all
       smaller (smaller than pivot) to left of
       pivot and all greater elements to right
       of pivot */
    static int partition(int arr[], int low, int high)
    {
        int pivot = arr[high];
        int i = (low - 1);
       
        for (int j = low; j < high; j++)
        {
            // If current element is smaller than or
            // equal to pivot
            if (arr[j] <= pivot)
            {
                i++;
 
                // swap arr[i] and arr[j]
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
 
        // swap arr[i+1] and arr[high] (or pivot)
        int temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;
 
        return i + 1;
    }
 
    /* The main function that implements QuickSort()
      arr[] --> Array to be sorted,
      low  --> Starting index,
      high  --> Ending index */
    static void sort(int arr[], int low, int high)
    {
        if (low < high) {
            /* pi is partitioning index, arr[pi] is
              now at right place */
            int pi = partition(arr, low, high);
 
            // Recursively sort elements before
            // partition and after partition
            sort(arr, low, pi - 1);
            sort(arr, pi + 1, high);
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int arr1[] = { 11, 1, 13, 21, 3, 7 };
        int arr2[] = { 11, 3, 7, 1 };
 
        int m = arr1.length;
        int n = arr2.length;
 
        if (isSubset(arr1, arr2, m, n))
            System.out.print("arr2[] is subset of arr1[] ");
        else
            System.out.print(
                "arr2[] is not a subset of arr1[]");
    }
}

蟒蛇3

# Python3 program to find whether an array
# is subset of another array
 
# Return 1 if arr2[] is a subset of arr1[]
 
 
def isSubset(arr1, arr2, m, n):
    i = 0
 
    quickSort(arr1, 0, m-1)
    for i in range(n):
        if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1):
            return 0
 
    # If we reach here then all elements
    # of arr2[] are present in arr1[]
    return 1
 
# FOLLOWING FUNCTIONS ARE ONLY FOR
# SEARCHING AND SORTING PURPOSE
# Standard Binary Search function
 
 
def binarySearch(arr, low, high, x):
    if(high >= low):
        mid = (low + high)//2
 
        # Check if arr[mid] is the first
        # occurrence of x.
        # arr[mid] is first occurrence if x is
        # one of the following
        # is true:
        # (i) mid == 0 and arr[mid] == x
        # (ii) arr[mid-1] < x and arr[mid] == x
        if((mid == 0 or x > arr[mid-1]) and (arr[mid] == x)):
            return mid
        elif(x > arr[mid]):
            return binarySearch(arr, (mid + 1), high, x)
        else:
            return binarySearch(arr, low, (mid - 1), x)
 
    return -1
 
 
def partition(A, si, ei):
    x = A[ei]
    i = (si - 1)
 
    for j in range(si, ei):
        if(A[j] <= x):
            i += 1
            A[i], A[j] = A[j], A[i]
    A[i + 1], A[ei] = A[ei], A[i + 1]
    return (i + 1)
 
# Implementation of Quick Sort
# A[] --> Array to be sorted
# si --> Starting index
# ei --> Ending index
 
 
def quickSort(A, si, ei):
    # Partitioning index
    if(si < ei):
        pi = partition(A, si, ei)
        quickSort(A, si, pi - 1)
        quickSort(A, pi + 1, ei)
 
 
# Driver code
arr1 = [11, 1, 13, 21, 3, 7]
arr2 = [11, 3, 7, 1]
 
m = len(arr1)
n = len(arr2)
 
if(isSubset(arr1, arr2, m, n)):
    print("arr2[] is subset of arr1[] ")
else:
    print("arr2[] is not a subset of arr1[] ")
 
 
# This code is contributed by chandan_jnu

C＃

// C# program to find whether an array
// is subset of another array
using System;
 
public class GFG {
    /* Return true if arr2[] is a subset of arr1[] */
    static bool isSubset(int[] arr1,
                         int[] arr2,
                         int m, int n)
    {
        int i = 0;
 
        sort(arr1, 0, m - 1);
        for (i = 0; i < n; i++)
        {
            if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1)
                return false;
        }
 
        /* If we reach here then all elements of arr2[]
          are present in arr1[] */
        return true;
    }
 
    /* FOLLOWING FUNCTIONS ARE ONLY FOR SEARCHING AND SORTING PURPOSE */
    /* Standard Binary Search function*/
    static int binarySearch(int[] arr,
                            int low,
                            int high, int x)
    {
        if (high >= low)
        {
            int mid = (low + high) / 2;
 
            /* Check if arr[mid] is the first
            occurrence of x.
            arr[mid] is first occurrence if x
            is one of the following
            is true:
            (i)  mid == 0 and arr[mid] == x
            (ii) arr[mid-1] < x and arr[mid] == x
         */
            if ((mid == 0 || x > arr[mid - 1])
                && (arr[mid] == x))
                return mid;
            else if (x > arr[mid])
                return binarySearch(arr,
                                    (mid + 1), high, x);
            else
                return binarySearch(arr,
                                    low, (mid - 1), x);
        }
        return -1;
    }
 
    /* This function takes last element as pivot,
       places the pivot element at its correct
       position in sorted array, and places all
       smaller (smaller than pivot) to left of
       pivot and all greater elements to right
       of pivot */
    static int partition(int[] arr, int low, int high)
    {
        int pivot = arr[high];
        int i = (low - 1);
        int temp = 0;
        for (int j = low; j < high; j++)
        {
            // If current element is smaller than or
            // equal to pivot
            if (arr[j] <= pivot)
            {
                i++;
                // swap arr[i] and arr[j]
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
 
        // swap arr[i+1] and arr[high] (or pivot)
        temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;
 
        return i + 1;
    }
 
    /* The main function that implements QuickSort()
      arr[] --> Array to be sorted,
      low  --> Starting index,
      high  --> Ending index */
    static void sort(int[] arr, int low, int high)
    {
        if (low < high) {
            /* pi is partitioning index, arr[pi] is
              now at right place */
            int pi = partition(arr, low, high);
 
            // Recursively sort elements before
            // partition and after partition
            sort(arr, low, pi - 1);
            sort(arr, pi + 1, high);
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr1 = { 11, 1, 13, 21, 3, 7 };
        int[] arr2 = { 11, 3, 7, 1 };
 
        int m = arr1.Length;
        int n = arr2.Length;
 
        if (isSubset(arr1, arr2, m, n))
            Console.Write("arr2[] is subset of arr1[] ");
        else
            Console.Write("arr2[] is not a subset of arr1[]");
    }
}
// This code is contributed by 29AjayKumar

PHP

= $low)
    {
        $mid = (int)(($low + $high)/2);
 
        /* Check if arr[mid] is the first
        occurrence of x.
        arr[mid] is first occurrence if
        x is one of the following
        is true:
        (i) mid == 0 and arr[mid] == x
        (ii) arr[mid-1] < x and arr[mid] == x */
        if(( $mid == 0 || $x > $arr[$mid-1])
           && ($arr[$mid] == $x))
            return $mid;
        else if($x > $arr[$mid])
            return binarySearch($arr,
                                ($mid + 1), $high, $x);
        else
            return binarySearch($arr,
                                $low, ($mid -1), $x);
    }
    return -1;
}
 
function exchange(&$a, &$b)
{
 
    $temp = $a;
    $a = $b;
    $b = $temp;
}
 
function partition(&$A, $si, $ei)
{
    $x = $A[$ei];
    $i = ($si - 1);
 
    for ($j = $si; $j <= $ei - 1; $j++)
    {
        if($A[$j] <= $x)
        {
            $i++;
            exchange($A[$i], $A[$j]);
        }
    }
    exchange ($A[$i + 1], $A[$ei]);
    return ($i + 1);
}
 
/* Implementation of Quick Sort
A[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
function quickSort(&$A, $si, $ei)
{
    /* Partitioning index */
    if($si < $ei)
    {
        $pi = partition($A, $si, $ei);
        quickSort($A, $si, $pi - 1);
        quickSort($A, $pi + 1, $ei);
    }
}
 
    /*Driver code */
    $arr1 = array(11, 1, 13, 21, 3, 7);
    $arr2 = array(11, 3, 7, 1);
 
    $m = count($arr1);
    $n = count($arr2);
 
    if(isSubset($arr1, $arr2, $m, $n))
        echo "arr2[] is subset of arr1[] ";
    else
        echo "arr2[] is not a subset of arr1[] ";
 
 
// This code is contributed by chandan_jnu
?>

Javascript

 
// Javascript program to find whether an array
// is subset of another array
 
// Return true if arr2[] is a subset of arr1[]
function isSubset(arr1, arr2, m, n)
{
    let i = 0;
     
    sort(arr1, 0, m - 1);
     
    for(i = 0; i < n; i++)
    {
        if (binarySearch(arr1, 0, m - 1,
                         arr2[i]) == -1)
            return false;
    }
 
    // If we reach here then all elements
    // of arr2[] are present in arr1[]
    return true;
}
     
/* FOLLOWING FUNCTIONS ARE ONLY
   FOR SEARCHING AND
 * SORTING PURPOSE */
/* Standard Binary Search function*/
function binarySearch(arr, low, high, x)
{
    if (high >= low)
    {
         
        // low + (high - low)/2;
        let mid = Math.floor((low + high) / 2);
 
        // Check if arr[mid] is the first occurrence of
        // x. arr[mid] is first occurrence if x is one of
        // the following is true: (i)  mid == 0 and
        // arr[mid] == x (ii) arr[mid-1] < x and arr[mid] == x
        if ((mid == 0 || x > arr[mid - 1]) &&
            (arr[mid] == x))
            return mid;
        else if (x > arr[mid])
            return binarySearch(arr, (mid + 1),
                                high, x);
        else
            return binarySearch(arr, low,
                               (mid - 1), x);
    }
    return -1;
}
 
/* This function takes last element as pivot,
   places the pivot element at its correct
   position in sorted array, and places all
   smaller (smaller than pivot) to left of
   pivot and all greater elements to right
   of pivot */
function partition(arr, low, high)
{
    let pivot = arr[high];
    let i = (low - 1);
    
    for(let j = low; j < high; j++)
    {
         
        // If current element is smaller than or
        // equal to pivot
        if (arr[j] <= pivot)
        {
            i++;
 
            // Swap arr[i] and arr[j]
            let temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }
 
    // Swap arr[i+1] and arr[high] (or pivot)
    let temp = arr[i + 1];
    arr[i + 1] = arr[high];
    arr[high] = temp;
 
    return i + 1;
}
 
/* The main function that implements QuickSort()
   arr[] --> Array to be sorted,
   low  --> Starting index,
   high  --> Ending index */
function sort(arr,low,high)
{
    if (low < high)
    {
         
        // pi is partitioning index, arr[pi]
        // is now at right place
        let pi = partition(arr, low, high);
 
        // Recursively sort elements before
        // partition and after partition
        sort(arr, low, pi - 1);
        sort(arr, pi + 1, high);
    }
}
 
// Driver Code
let arr1 = [ 11, 1, 13, 21, 3, 7 ];
let arr2 = [ 11, 3, 7, 1 ];
let m = arr1.length;
let n = arr2.length;
 
if (isSubset(arr1, arr2, m, n))
    document.write("arr2[] is subset of arr1[] ");
else
    document.write("arr2[] is not a subset of arr1[]");
 
// This code is contributed by patel2127
 

输出

arr2[] is subset of arr1[] 

时间复杂度： O(mLogm + nLogm)。请注意，如果使用 mLogm 算法进行排序，这将是复杂性，而在上述代码中则不然。在上面的代码中使用了快速排序，快速排序的最坏情况时间复杂度为 O(m^2)

方法3（使用排序和合并）

• 对两个数组进行排序：arr1[] 和 arr2[] 需要 O(mLogm + nLogn)
• 使用 Merge 类型的进程查看已排序的 arr2[] 的所有元素是否都存在于已排序的 arr1[] 中。

感谢Parthsarthi提出这种方法。
下图是上述方法的试运行：

下面是上述方法的实现：

C++

// C++ program to find whether an array
// is subset of another array
#include 
using namespace std;
 
/* Return 1 if arr2[] is a subset of arr1[] */
bool isSubset(int arr1[], int arr2[],
              int m, int n)
{
    int i = 0, j = 0;
 
    if (m < n)
        return 0;
 
    // Sort both the arrays
    sort(arr1, arr1 + m);
    sort(arr2, arr2 + n);
 
    // Iterate till they donot exceed their sizes
    while (i < n && j < m)
    {
        // If the element is smaller than
        // Move aheaad in the first array
        if (arr1[j] < arr2[i])
            j++;
        // If both are equal, then move
        // both of them forward
        else if (arr1[j] == arr2[i])
        {
            j++;
            i++;
        }
 
        // If we donot have a element smaller
        // or equal to the second array then break
        else if (arr1[j] > arr2[i])
            return 0;
    }
 
    return (i < n) ? false : true;
}
 
// Driver Code
int main()
{
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
 
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
 
    if (isSubset(arr1, arr2, m, n))
        printf("arr2[] is subset of arr1[] ");
    else
        printf("arr2[] is not a subset of arr1[] ");
 
    return 0;
}

Java

// Java code to find whether an array is subset of
// another array
import java.util.Arrays;
class GFG
{
    /* Return true if arr2[] is a subset of arr1[] */
    static boolean isSubset(int arr1[],
                            int arr2[], int m,
                            int n)
    {
        int i = 0, j = 0;
 
        if (m < n)
            return false;
 
        Arrays.sort(arr1); // sorts arr1
        Arrays.sort(arr2); // sorts arr2
 
        while (i < n && j < m) {
            if (arr1[j] < arr2[i])
                j++;
            else if (arr1[j] == arr2[i]) {
                j++;
                i++;
            }
            else if (arr1[j] > arr2[i])
                return false;
        }
 
        if (i < n)
            return false;
        else
            return true;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr1[] = { 11, 1, 13, 21, 3, 7 };
        int arr2[] = { 11, 3, 7, 1 };
 
        int m = arr1.length;
        int n = arr2.length;
 
        if (isSubset(arr1, arr2, m, n))
            System.out.println("arr2 is a subset of arr1");
        else
            System.out.println("arr2 is not a subset of arr1");
    }
}
// This code is contributed by Kamal Rawal

蟒蛇3

# Python3 program to find whether an array
# is subset of another array
 
# Return 1 if arr2[] is a subset of arr1[] */
 
 
def isSubset(arr1, arr2, m, n):
    i = 0
    j = 0
    if m < n:
        return 0
 
    arr1.sort()
    arr2.sort()
 
    while i < n and j < m:
        if arr1[j] < arr2[i]:
            j += 1
        elif arr1[j] == arr2[i]:
            j += 1
            i += 1
        elif arr1[j] > arr2[i]:
            return 0
    return False if i < n else True
 
 
# Driver code
arr1 = [11, 1, 13, 21, 3, 7]
arr2 = [11, 3, 7, 1]
 
m = len(arr1)
n = len(arr2)
if isSubset(arr1, arr2, m, n) == True:
    print("arr2 is subset of arr1 ")
else:
    printf("arr2 is not a subset of arr1 ")
 
# This code is contributed by Shrikant13

C＃

// C# code to find whether an array
// is subset of another array
using System;
class GFG {
 
    // Return true if arr2[] is
    // a subset of arr1[] */
    static bool isSubset(int[] arr1,
                         int[] arr2, int m,
                         int n)
    {
        int i = 0, j = 0;
 
        if (m < n)
            return false;
 
        // sorts arr1
        Array.Sort(arr1);
 
        // sorts arr2
        Array.Sort(arr2);
 
        while (i < n && j < m)
        {
            if (arr1[j] < arr2[i])
                j++;
            else if (arr1[j] == arr2[i])
            {
                j++;
                i++;
            }
            else if (arr1[j] > arr2[i])
                return false;
        }
 
        if (i < n)
            return false;
        else
            return true;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr1 = { 11, 1, 13, 21, 3, 7 };
        int[] arr2 = { 11, 3, 7, 1 };
 
        int m = arr1.Length;
        int n = arr2.Length;
 
        if (isSubset(arr1, arr2, m, n))
            Console.Write("arr2 is a subset of arr1");
        else
            Console.Write("arr2 is not a subset of arr1");
    }
}
 
// This code is contributed by nitin mittal.

PHP

 $arr2[$i] )
            return 0;
    }
 
    return ($i < $n) ? false : true;
}
 
/*Driver code */
 
    $arr1 = array(11, 1, 13, 21, 3, 7);
    $arr2 = array(11, 3, 7, 1);
 
    $m = count($arr1);
    $n = count($arr2);
 
    if(isSubset($arr1, $arr2, $m, $n))
        echo "arr2[] is subset of arr1[] ";
    else
        echo "arr2[] is not a subset of arr1[] ";
 
// This code is contributed by anuj_67.
?>

Javascript

 
// JavaScript program to find whether an array
// is subset of another array
 
// Return 1 if arr2[] is a subset of arr1[]
function isSubset(arr1, arr2, m, n)
{
    let i = 0, j = 0;
 
    if (m < n)
        return 0;
 
    // Sort both the arrays
    arr1.sort((a, b) => a - b);
    arr2.sort((a, b) => a - b);
 
    // Iterate till they donot exceed their sizes
    while (i < n && j < m)
    {
         
        // If the element is smaller than
        // Move aheaad in the first array
        if (arr1[j] < arr2[i])
            j++;
             
        // If both are equal, then move
        // both of them forward
        else if (arr1[j] == arr2[i])
        {
            j++;
            i++;
        }
 
        // If we donot have a element smaller
        // or equal to the second array then break
        else if (arr1[j] > arr2[i])
            return 0;
    }
    return (i < n) ? false : true;
}
 
// Driver Code
let arr1 = [ 11, 1, 13, 21, 3, 7 ];
let arr2 = [ 11, 3, 7, 1 ];
 
let m = arr1.length;
let n = arr2.length;
 
if (isSubset(arr1, arr2, m, n))
    document.write("arr2[] is subset of arr1[] ");
else
    document.write("arr2[] is not a subset of arr1[] ");
 
// This code is contributed by Manoj.
 

输出

arr2[] is subset of arr1[] 

时间复杂度： O(mLogm + nLogn)，比方法 2 好。请注意，如果使用 nLogn 算法对两个数组进行排序，这将是复杂度，而上面的代码并非如此。在上面的代码中使用了快速排序，快速排序的最坏情况时间复杂度为 O(n^2)

方法四（使用哈希）

• 为 arr1[] 的所有元素创建一个哈希表。
• 遍历 arr2[] 并在哈希表中搜索 arr2[] 的每个元素。如果未找到元素，则返回 0。
• 如果找到所有元素，则返回 1。

下面是上述方法的实现：

C++

// C++ code to find whether an array is subset of
// another array
#include 
using namespace std;
 
/* Return true if arr2[] is a subset of arr1[] */
bool isSubset(int arr1[], int m,
              int arr2[], int n)
{
 
    // Using STL set for hashing
    set hashset;
 
    // hset stores all the values of arr1
    for (int i = 0; i < m; i++)
    {
        hashset.insert(arr1[i]);
    }
 
    // loop to check if all elements of arr2 also
    // lies in arr1
    for (int i = 0; i < n; i++) {
        if (hashset.find(arr2[i]) == hashset.end())
            return false;
    }
    return true;
}
 
// Driver Code
int main()
{
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
 
    if (isSubset(arr1, m, arr2, n))
        cout << "arr2[] is subset of arr1[] "
             << "\n";
    else
        cout << "arr2[] is not a subset of arr1[] "
             << "\n";
    return 0;
}
// This code is contributed by Satvik Shrivas

Java

// Java code to find whether an array is subset of
// another array
import java.util.HashSet;
class GFG
{
    /* Return true if arr2[] is a subset of arr1[] */
    static boolean isSubset(int arr1[],
                            int arr2[], int m,
                            int n)
    {
        HashSet hset = new HashSet<>();
 
        // hset stores all the values of arr1
        for (int i = 0; i < m; i++) {
            if (!hset.contains(arr1[i]))
                hset.add(arr1[i]);
        }
 
        // loop to check if all elements
        //  of arr2 also lies in arr1
        for (int i = 0; i < n; i++)
        {
            if (!hset.contains(arr2[i]))
                return false;
        }
        return true;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr1[] = { 11, 1, 13, 21, 3, 7 };
        int arr2[] = { 11, 3, 7, 1 };
 
        int m = arr1.length;
        int n = arr2.length;
 
        if (isSubset(arr1, arr2, m, n))
            System.out.println("arr2 is a subset of arr1");
        else
            System.out.println(
                "arr2 is not a subset of arr1");
    }
}
// This code is contributed by Kamal Rawal

蟒蛇3

# Python3 program to find whether an array
# is subset of another array
 
# Return true if arr2[] is a subset
# of arr1[]
def isSubset(arr1, m, arr2, n):
     
    # Using STL set for hashing
    hashset = set()
 
    # hset stores all the values of arr1
    for i in range(0, m):
        hashset.add(arr1[i])
 
    # Loop to check if all elements
    # of arr2 also lies in arr1
    for i in range(0, n):
        if arr2[i] in hashset:
            continue
        else:
            return False
 
    return True
 
# Driver Code
if __name__ == '__main__':
     
    arr1 = [ 11, 1, 13, 21, 3, 7 ]
    arr2 = [ 11, 3, 7, 1 ]
     
    m = len(arr1)
    n = len(arr2)
     
    if (isSubset(arr1, m, arr2, n)):
        print("arr2[] is subset of arr1[] ")
    else:
        print("arr2[] is not a subset of arr1[] ")
 
# This code is contributed by akhilsaini

C＃

// C# code to find whether an array is
// subset of another array
using System;
using System.Collections.Generic;
 
class GFG {
    /* Return true if arr2[] is a
   subset of arr1[] */
    public static bool isSubset(int[] arr1,
                                int[] arr2,
                                int m, int n)
    {
        HashSet hset = new HashSet();
 
        // hset stores all the values of arr1
        for (int i = 0; i < m; i++)
        {
            if (!hset.Contains(arr1[i]))
            {
                hset.Add(arr1[i]);
            }
        }
 
        // loop to check if all elements
        // of arr2 also lies in arr1
        for (int i = 0; i < n; i++)
        {
            if (!hset.Contains(arr2[i]))
            {
                return false;
            }
        }
        return true;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr1 = new int[] { 11, 1, 13, 21, 3, 7 };
        int[] arr2 = new int[] { 11, 3, 7, 1 };
 
        int m = arr1.Length;
        int n = arr2.Length;
 
        if (isSubset(arr1, arr2, m, n)) {
            Console.WriteLine("arr2 is a subset of arr1");
        }
        else {
            Console.WriteLine(
                "arr2 is not a subset of arr1");
        }
    }
}
 
// This code is contributed by Shrikant13

Javascript

 
// Javascript code to find whether an
// array is subset of another array
 
// Return true if arr2[] is a
// subset of arr1[]
function isSubset(arr1, arr2, m, n)
{
    let hset = new Set();
  
    // hset stores all the values of arr1
    for(let i = 0; i < m; i++)
    {
        if (!hset.has(arr1[i]))
            hset.add(arr1[i]);
    }
 
    // Loop to check if all elements
    // of arr2 also lies in arr1
    for(let i = 0; i < n; i++)
    {
        if (!hset.has(arr2[i]))
            return false;
    }
    return true;
}
 
// Driver Code
let arr1 = [ 11, 1, 13, 21, 3, 7 ];
let arr2 = [ 11, 3, 7, 1 ];
let m = arr1.length;
let n = arr2.length;
 
if (isSubset(arr1, arr2, m, n))
    document.write("arr2 is a subset of arr1");
else
    document.write("arr2 is not a subset of arr1");
 
// This code is contributed by unknown2108
 

输出

arr2[] is subset of arr1[] 

方法五（使用集合）

• 插入到第一个数组的集合中；这就是我们如何知道数组中的元素。
• 插入第一个数组元素后保存数组的大小。
• 为第二个数组插入相同的集合。
• 检查集合的大小是否仍然相同，如果是，则为真，否则为假。

下面是上述方法的实现：

C++

#include 
using namespace std;
 
int main()
{
    // code
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    unordered_set s;
    for (int i = 0; i < m; i++) {
        s.insert(arr1[i]);
    }
    int p = s.size();
    for (int i = 0; i < n; i++) {
        s.insert(arr2[i]);
    }
    if (s.size() == p) {
       cout << "arr2[] is subset of arr1[] "
             << "\n";
    }
    else {
        cout << "arr2[] is not subset of arr1[] "
             << "\n";
    }
    return 0;
}

Java

import java.io.*;
import java.util.*;
 
class GFG
{
  public static void main (String[] args)
  {
 
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
    int m=arr1.length;
    int n=arr2.length;
 
    Set s = new HashSet();
    for (int i = 0; i < m; i++)
    {
      s.add(arr1[i]);
    }
    int p = s.size();
    for (int i = 0; i < n; i++)
    {
      s.add(arr2[i]);
    }
 
    if (s.size() == p)
    {
      System.out.println("arr2[] is subset of arr1[] " + "\n");
    }
    else
    {
      System.out.println("arr2[] is not subset of arr1[] " + "\n" );
    }
  }
}
 
// This code is contributed by avanitrachhadiya2155

蟒蛇3

# Python3 code
arr1 = [ 11, 1, 13, 21, 3, 7 ]
arr2 = [ 11, 3, 7, 1 ]
m = len(arr1)
n = len(arr2)
s = set()
for i in range(m) :
    s.add(arr1[i])
 
p = len(s)
for i in range(n) :
    s.add(arr2[i])
 
if (len(s) == p) :
    print("arr2[] is subset of arr1[] ")
 
else :
    print("arr2[] is not subset of arr1[] ")
     
    # This code is contributed by divyeshrabadiya07.

C＃

using System;
using System.Collections.Generic;
 
public class GFG
{
  static public void Main ()
  {
    int[] arr1 = { 11, 1, 13, 21, 3, 7 };
    int[] arr2 = { 11, 3, 7, 1 };
    int m = arr1.Length;
    int n = arr2.Length;
 
    HashSet s = new HashSet();
    for (int i = 0; i < m; i++)
    {
      s.Add(arr1[i]);
    }
    int p = s.Count;
    for (int i = 0; i < n; i++)
    {
      s.Add(arr2[i]);
    }
 
    if (s.Count == p)
    {
      Console.WriteLine("arr2[] is subset of arr1[] " + "\n");
    }
    else
    {
      Console.WriteLine("arr2[] is not subset of arr1[] " + "\n" );
    }
  }
}
 
// This code is contributed by rag2127

Javascript

 
let arr1=[11, 1, 13, 21, 3, 7];
let arr2=[11, 3, 7, 1 ];
let m=arr1.length;
let n=arr2.length;
let s = new Set();
for (let i = 0; i < m; i++)
{
    s.add(arr1[i]);
}
let p = s.size;
for (let i = 0; i < n; i++)
{
    s.add(arr2[i]);
}
 
if (s.size == p)
{
    document.write("arr2[] is subset of arr1[] " + "<br>");
}
else
{
    document.write("arr2[] is not subset of arr1[] " + "<br>" );
}
 
 
// This code is contributed by ab2127
 

输出

arr2[] is subset of arr1[] 

时间复杂度： O(m+n) 因为我们使用 unordered_set 并在其中插入，如果我们使用有序集插入将花费 log n 将 TC 增加到 O(mlogm+nlogn)；但在这种方法中顺序无关紧要。

方法六（使用频数表）

• 为 arr1[] 的所有元素创建一个频率表。
• 遍历 arr2[] 并在频率表中搜索 arr2[] 的每个元素。如果找到元素降低频率，如果没有找到元素频率则返回0。
• 如果找到所有元素，则返回 1。

下面是上述方法的实现：

C++14

// C++ program to find whether an array
// is subset of another array
#include 
using namespace std;
/* Return true if arr2[] is a subset of arr1[] */
 
bool isSubset(int arr1[], int m,
              int arr2[], int n)
{
    // Create a Frequency Table using STL
    map frequency;
     
    // Increase the frequency of each element
    // in the frequency table.
    for (int i = 0; i < m; i++)
    {
        frequency[arr1[i]]++;
    }
    // Decrease the frequency if the
    // element was found in the frequency
    // table with the frequency more than 0.
    // else return 0 and if loop is
    // completed return 1.
    for (int i = 0; i < n; i++)
    {
        if (frequency[arr2[i]] > 0)
            frequency[arr2[i]]--;
        else
        {
            return false;
        }
    }
    return true;
}
 
// Driver Code
int main()
{
    int arr1[] = { 11, 1, 13, 21, 3, 7 };
    int arr2[] = { 11, 3, 7, 1 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
 
    if (isSubset(arr1, m, arr2, n))
        cout << "arr2[] is subset of arr1[] "
             << "\n";
    else
        cout << "arr2[] is not a subset of arr1[] "
             << "\n";
    return 0;
}
// This code is contributed by Dhawal Sarin.

Java

// Java program to find whether an array
// is subset of another array
import java.io.*;
import java.util.*;
 
class GFG{
 
// Return true if arr2[] is a subset of arr1[]
static boolean isSubset(int[] arr1, int m,
                        int[] arr2, int n)
{
     
    // Create a Frequency Table using STL
    HashMap frequency = new HashMap();
 
    // Increase the frequency of each element
    // in the frequency table.
    for(int i = 0; i < m; i++)
    {
        frequency.put(arr1[i],
                      frequency.getOrDefault(
                          arr1[i], 0) + 1);
    }
     
    // Decrease the frequency if the
    // element was found in the frequency
    // table with the frequency more than 0.
    // else return 0 and if loop is
    // completed return 1.
    for(int i = 0; i < n; i++)
    {
        if (frequency.getOrDefault(arr2[i], 0) > 0)
            frequency.put(arr2[i],
                          frequency.get(arr1[i]) - 1);
        else
        {
            return false;
        }
    }
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr1 = { 11, 1, 13, 21, 3, 7 };
    int[] arr2 = { 11, 3, 7, 1 };
     
    int m = arr1.length;
    int n = arr2.length;
 
    if (isSubset(arr1, m, arr2, n))
        System.out.println(
            "arr2[] is subset of arr1[] ");
    else
        System.out.println(
            "arr2[] is not a subset of arr1[] ");
}
}
 
// This code is contributed by akhilsaini

蟒蛇3

# Python3 program to find whether an array
# is subset of another array
 
# Return true if arr2[] is a subset of arr1[]
def isSubset(arr1, m, arr2, n):
     
    # Create a Frequency Table using STL
    frequency = {}
 
    # Increase the frequency of each element
    # in the frequency table.
    for i in range(0, m):
        if arr1[i] in frequency:
            frequency[arr1[i]] = frequency[arr1[i]] + 1
        else:
            frequency[arr1[i]] = 1
 
    # Decrease the frequency if the
    # element was found in the frequency
    # table with the frequency more than 0.
    # else return 0 and if loop is
    # completed return 1.
    for i in range(0, n):
        if (frequency[arr2[i]] > 0):
            frequency[arr2[i]] -= 1
        else:
            return False
 
    return True
 
# Driver Code
if __name__ == '__main__':
     
    arr1 = [ 11, 1, 13, 21, 3, 7 ]
    arr2 = [ 11, 3, 7, 1 ]
     
    m = len(arr1)
    n = len(arr2)
 
    if (isSubset(arr1, m, arr2, n)):
        print("arr2[] is subset of arr1[] ")
    else:
        print("arr2[] is not a subset of arr1[] ")
 
# This code is contributed by akhilsaini

C＃

// C# program to find whether an array
// is subset of another array
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
// Return true if arr2[] is a subset of arr1[]
static bool isSubset(int[] arr1, int m,
                     int[] arr2, int n)
{
     
    // Create a Frequency Table using STL
    Dictionary frequency = new Dictionary();
                                                
    // Increase the frequency of each element
    // in the frequency table.
    for(int i = 0; i < m; i++)
    {
        if (frequency.ContainsKey(arr1[i]))
            frequency[arr1[i]] += 1;
        else
            frequency[arr1[i]] = 1;
    }
     
    // Decrease the frequency if the
    // element was found in the frequency
    // table with the frequency more than 0.
    // else return 0 and if loop is
    // completed return 1.
    for(int i = 0; i < n; i++)
    {
        if (frequency[arr2[i]] > 0)
            frequency[arr2[i]] -= 1;
        else
        {
            return false;
        }
    }
    return true;
}
 
// Driver Code
public static void Main()
{
    int[] arr1 = { 11, 1, 13, 21, 3, 7 };
    int[] arr2 = { 11, 3, 7, 1 };
     
    int m = arr1.Length;
    int n = arr2.Length;
 
    if (isSubset(arr1, m, arr2, n))
        Console.WriteLine(
            "arr2[] is subset of arr1[] ");
    else
        Console.WriteLine(
            "arr2[] is not a subset of arr1[] ");
}
}
 
// This code is contributed by akhilsaini

Javascript

// javascript program to find whether an array
// is subset of another array
 
/* Return true if arr2[] is a subset of arr1[] */
 
function isSubset(arr1,m,arr2,n)
{
    // Create a Frequency Table using STL
    let frequency = new Array(arr1);
     
    // Increase the frequency of each element
    // in the frequency table.
    for (let i = 0; i < m; i++)
    {
        frequency[arr1[i]]++;
    }
    // Decrease the frequency if the
    // element was found in the frequency
    // table with the frequency more than 0.
    // else return 0 and if loop is
    // completed return 1.
    for (let i = 0; i < n; i++)
    {
        if (frequency[arr2[i]] > 0)
            frequency[arr2[i]]--;
        else
        {
            return false;
        }
    }
    return true;
}
 
// Driver Code
    let arr1 = [11, 1, 13, 21, 3, 7 ];
    let arr2 = [ 11, 3, 7, 1 ];
    let m = arr1.length;
    let n = arr2.length;
 
    if (isSubset(arr1, m, arr2, n))
        document.write("arr2[] is subset of arr1[] "
            + "\n");
    else
        document.write("arr2[] is not a subset of arr1[] "
            + "\n");
             
            // This code is contributed by vaibhavrabadiya117.

输出

arr2[] is subset of arr1[] 

请注意，方法 1、方法 2、方法 4 和方法 5 不处理 arr2[] 中有重复项的情况。例如，{1, 4, 4, 2} 不是 {1, 4, 2} 的子集，但这些方法会将其打印为子集。

时间复杂度： O(m+n) 优于方法 1,2,3。

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