查找一个数组是否是另一个数组的子集|新增方法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。这两个数组均未排序。可以假定两个数组中的元素都是不同的。

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1(简单) ：
使用两个循环：外循环一个接一个地选择arr2 []的所有元素。内循环线性搜索外循环拾取的元素。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[]");
    }
}

Python 3

# 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(使用排序和二进制搜索) ：

• 对arr1 []进行排序，结果为O(mLogm)
• 对于arr2 []的每个元素，请按已排序的arr1 []进行二进制搜索。
• 如果找不到该元素，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 
/* Fucntion 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[]");
    }
}

Python3

# 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

的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
?>

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，则将变得很复杂，而以上代码中的情况并非如此。在上面的代码中使用了快速排序，最坏情况下，快速排序的时间复杂度为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 []，它们的取值为O(mLogm + nLogn)
• 使用Merge类型的进程查看排序后的arr1 []中是否存在排序后的arr2 []的所有元素。

感谢Parthsarthi提出了这种方法。
下图是上述方法的模拟：

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

Python3

# 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.

的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.
?>

输出

arr2[] is subset of arr1[] 

时间复杂度： O(mLogm + nLogn)比方法2更好。在上面的代码中，使用了快速排序，最坏情况下，快速排序的时间复杂度为O(n ^ 2)

方法4(使用散列)

• 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element，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

Python3

# 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

输出

arr2[] is subset of arr1[] 

方法5(使用集)

• 插入第一个数组的集合；这样我们才能知道数组中的元素。
• 插入第一个数组元素后，保存数组的大小。
• 插入到第二个数组的相同集合中。
• 检查集合的大小是否仍然相同，如果是，则为true，否则为false。

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

Python3

# 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

输出

arr2[] is subset of arr1[] 

时间复杂度： O(m + n)，因为我们使用unordered_set并在其中进行插入，如果我们要使用有序集合，则插入会花费

log n将TC增加到O(mlogm + nlogn);但是在这种方法中顺序并不重要。

方法6(使用频率表)

• 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，则降低频率；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++ 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

Python3

# 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

输出

arr2[] is subset of arr1[] 

请注意，当我们在arr2 []中有重复项时，方法1，方法2，方法4和方法5不处理情况。例如，{1，4，4，2}不是{1，4，2}的子集，但是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。