NxM 矩阵的每一行中存在的数组元素的计数

给定 N 行，每行有 M 个元素，还有一个由 L 个数字组成的数组 arr[]，任务是打印矩阵每一行中存在的该数组元素的计数。
例子：

Input: {8 27 39 589 23
        23 34 589 12 45
        939 32 27 12 78
        23 349 48 21 32},  
       
       arr[] = {589, 39, 27}

Output: 1st row - 3
        2nd row - 1
        3rd row - 1 
        4th row - 0
In 1st row, all three elements in array z[] are present
In 2nd row, only 589 in array z[] are present
In 3rd row, only 27 in array z[] are present 
In 4th row, none of the elements are present. 

Input: {1, 2, 3
        4, 5, 6}, 
       
       arr[] = {2, 3, 4}

Output: 1st row - 2
        2nd row - 1

一种天真的方法是对数组 arr[] 中的每个元素进行迭代，并对^第i 行对数组 arr[] 中的每个元素进行线性搜索。计算元素的数量并打印每一行的结果。
时间复杂度： O(N*M*L)
一种有效的方法是迭代矩阵^第i 行中的所有元素。使用哈希表标记所有元素。迭代 Z 数组中的数字数组，检查该数字是否存在于哈希表中。增加存在的每个元素的计数。检查完所有元素后，打印计数。
下面是上述方法的实现：

C++

// C++ program to print the count of
// elements present in the NxM matrix
#include 
using namespace std;
 
// Function to print the count of
// elements present in the NxM matrix
void printCount(int a[][5], int n, int m, int z[], int l)
{
    // iterate in the n rows
    for (int i = 0; i < n; i++) {
        // map to mark elements in N-th row
        unordered_map mp;
 
        // mark all elements in the n-th row
        for (int j = 0; j < m; j++)
            mp[a[i][j]] = 1;
 
        int count = 0;
 
        // check for occurrence of all elements
        for (int j = 0; j < l; j++) {
            if (mp[z[j]])
                count += 1;
        }
 
        // print the occurrence of all elements
        cout << "row" << i + 1 << " = " << count << endl;
    }
}
 
// Driver Code
int main()
{
 
    // NxM matrix
    int a[][5] = { { 8, 27, 39, 589, 23 },
                { 23, 34, 589, 12, 45 },
                { 939, 32, 27, 12, 78 },
                { 23, 349, 48, 21, 32 } };
 
    // elements array
    int arr[] = { 589, 39, 27 };
 
    int n = sizeof(a) / sizeof(a[0]);
 
    int m = 5;
 
    int l = sizeof(arr) / sizeof(arr[0]);
 
    printCount(a, n, m, arr, l);
 
    return 0;
}

Java

// Java program to print the count of
// elements present in the NxM matrix
import java.util.*;
 
class GFG
{
 
// Function to print the count of
// elements present in the NxM matrix
static void printCount(int a[][], int n, int m,
                                int z[], int l)
{
    // iterate in the n rows
    for (int i = 0; i < n; i++)
    {
        // map to mark elements in N-th row
        Map mp = new HashMap<>();
 
        // mark all elements in the n-th row
        for (int j = 0; j < m; j++)
            mp.put(a[i][j], 1);
 
        int count = 0;
 
        // check for occurrence of all elements
        for (int j = 0; j < l; j++)
        {
            if (mp.containsKey(z[j]))
                count += 1;
        }
 
        // print the occurrence of all elements
                System.out.println("row" +(i + 1) + " = " + count);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    // NxM matrix
    int a[][] = { { 8, 27, 39, 589, 23 },
                { 23, 34, 589, 12, 45 },
                { 939, 32, 27, 12, 78 },
                { 23, 349, 48, 21, 32 } };
 
    // elements array
    int arr[] = { 589, 39, 27 };
 
    int n = a.length;
 
    int m = 5;
 
    int l = arr.length;
 
    printCount(a, n, m, arr, l);
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to print the count of
# elements present in the NxM matrix
 
# Function to print the count of
# elements present in the NxM matrix
def printCount(a, n, m, z, l):
 
    # iterate in the n rows
    for i in range(n):
         
        # map to mark elements in N-th row
        mp = dict()
 
        # mark all elements in the n-th row
        for j in range(m):
            mp[a[i][j]] = 1
 
        count = 0
 
        # check for occurrence of all elements
        for j in range(l):
             
            if z[j] in mp.keys():
                 
                count += 1
         
        # print the occurrence of all elements
        print("row", i + 1, " = ", count )
 
# Driver Code
 
# NxM matrix
a = [[ 8, 27, 39, 589, 23 ],
     [ 23, 34, 589, 12, 45 ],
     [ 939, 32, 27, 12, 78 ],
     [ 23, 349, 48, 21, 32 ]]
 
# elements array
arr = [ 589, 39, 27 ]
 
n = len(a)
 
m = 5
 
l = len(arr)
 
printCount(a, n, m, arr, l)
 
# This code is contributed by mohit kumar 29

C#

// C# program to print the count of
// elements present in the NxM matrix
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to print the count of
// elements present in the NxM matrix
static void printCount(int [,]a, int n, int m,
                                int []z, int l)
{
    // iterate in the n rows
    for (int i = 0; i < n; i++)
    {
        // map to mark elements in N-th row
        Dictionary mp = new Dictionary();
 
        // mark all elements in the n-th row
        for (int j = 0; j < m; j++)
            mp.Add(a[i,j], 1);
 
        int count = 0;
 
        // check for occurrence of all elements
        for (int j = 0; j < l; j++)
        {
            if (mp.ContainsKey(z[j]))
                count += 1;
        }
 
        // print the occurrence of all elements
        Console.WriteLine("row" +(i + 1) + " = " + count);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    // NxM matrix
    int [,]a = { { 8, 27, 39, 589, 23 },
                { 23, 34, 589, 12, 45 },
                { 939, 32, 27, 12, 78 },
                { 23, 349, 48, 21, 32 } };
 
    // elements array
    int []arr = { 589, 39, 27 };
 
    int n = a.GetLength(0);
 
    int m = 5;
 
    int l = arr.Length;
 
    printCount(a, n, m, arr, l);
}
}
 
/* This code is contributed by PrinciRaj1992 */

Javascript

输出：

row1 = 3
row2 = 1
row3 = 1
row4 = 0

时间复杂度： O(N*M)

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