计算矩阵中的多数元素

给定一个包含重复元素的NxM整数矩阵。任务是找到给定矩阵中所有多数出现元素的计数，其中多数元素是那些频率大于或等于(N * M)/ 2的元素。

例子：

Input : mat[] = {{1, 1, 2},
                {2, 3, 3},
                {4, 3, 3}}
Output : 1
The majority elements is 3 and, its frequency is 4.

Input : mat[] = {{20, 20},
                 {40, 40}}
Output : 2

方法：

• 遍历矩阵并使用C++中的映射存储矩阵元素的频率，以使映射的关键是矩阵元素，值是其在矩阵中的频率。
• 然后，遍历该图以找到具有计数变量的元素的频率以对多数元素进行计数，并检查其是否等于或大于(N * M)/ 2。如果为true，则增加计数。

下面是上述方法的实现：

C++

// C++ program to find count of all
// majority elements in a Matrix
  
#include 
using namespace std;
  
#define N 3 // Rows
#define M 3 // Columns
  
// Function to find count of all
// majority elements in a Matrix
int majorityInMatrix(int arr[N][M])
{
  
    unordered_map mp;
  
    // Store frequency of elements
    // in matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            mp[arr[i][j]]++;
        }
    }
  
    // loop to iteratre through map    
    int countMajority = 0;
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
  
        // check if frequency is greater than
        // or equal to (N*M)/2
        if (itr->second >= ((N * M) / 2)) {
            countMajority++;
        }
    }
  
    return countMajority;
}
  
// Driver Code
int main()
{
  
    int mat[N][M] = { { 1, 2, 2 },
                      { 1, 3, 2 },
                      { 1, 2, 6 } };
  
    cout << majorityInMatrix(mat) << endl;
  
    return 0;
}

Java

// Java program to find count of all
// majority elements in a Matrix
import java.util.*;
  
class GFG
{
    static int N = 3; // Rows
    static int M = 3; // Columns
      
    // Function to find count of all
    // majority elements in a Matrix
    static int majorityInMatrix(int arr[][])
    {
      
        HashMap mp = 
                new HashMap();
      
        // Store frequency of elements
        // in matrix
        for (int i = 0; i < N; i++) 
        {
            for (int j = 0; j < M; j++) 
            {
                if(mp.containsKey(arr[i][j]))
                    mp.put(arr[i][j], mp.get(arr[i][j]) + 1 );
                else
                    mp.put(arr[i][j], 1);
            }
        }
      
        // loop to iteratre through map 
        int countMajority = 0;
          
        Iterator> itr = 
                                mp.entrySet().iterator(); 
          
        while(itr.hasNext())
        {
            // check if frequency is greater than
            // or equal to (N*M)/2
            HashMap.Entry entry = itr.next();
              
            if (entry.getValue() >= ((N * M) / 2)) 
            {
                countMajority++;
            }
        }
      
        return countMajority;
    }
      
    // Driver Code
    public static void main (String[] args) 
    {
      
        int mat[][] = { { 1, 2, 2 },
                        { 1, 3, 2 },
                        { 1, 2, 6 } };
      
        System.out.println(majorityInMatrix(mat));
    }
}
  
// This code is contributed by ihritik

Python3

# Python 3 program to find count of all
# majority elements in a Matrix
N = 3 # Rows
M = 3 # Columns
  
# Function to find count of all
# majority elements in a Matrix
def majorityInMatrix(arr):
      
    # we take length equal to max 
    # value in array
    mp = {i:0 for i in range(7)}
  
    # Store frequency of elements
    # in matrix
    for i in range(len(arr)):
        for j in range(len(arr)):
            mp[arr[i][j]] += 1
      
    # loop to iteratre through map 
    countMajority = 0
    for key, value in mp.items():
          
        # check if frequency is greater than
        # or equal to (N*M)/2
        if (value >= (int((N * M) / 2))):
            countMajority += 1
      
    return countMajority
  
# Driver Code
if __name__ == '__main__':
    mat = [[1, 2, 2],
           [1, 3, 2],
           [1, 2, 6]]
    print(majorityInMatrix(mat))
  
# This code is contributed by
# Shashank_Sharma

C#

// C# program to find count of all
// majority elements in a Matrix
using System;
using System.Collections.Generic;
  
class GFG
{
    static int N = 3; // Rows
    static int M = 3; // Columns
      
    // Function to find count of all
    // majority elements in a Matrix
    static int majorityInMatrix(int [ , ]arr)
    {
      
        Dictionary mp = 
                new Dictionary();
      
        // Store frequency of elements
        // in matrix
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < M; j++)
            {
                if(mp.ContainsKey(arr[i, j]))
                    mp[arr[i, j]]++;
                else
                    mp[arr[i, j]] = 1;
            }
        }
      
        // loop to iteratre through map 
        int countMajority = 0;
        Dictionary.KeyCollection keyColl = 
                                            mp.Keys;
          
        foreach( int i in keyColl)
        {     
            // check if frequency is greater than
            // or equal to (N*M)/2
              
            if ( mp[i] >= ((N * M) / 2)) 
            {
                countMajority++;
            }
        }
      
        return countMajority;
    }
      
    // Driver Code
    public static void Main () 
    {
      
        int [, ] mat = { { 1, 2, 2 },
                        { 1, 3, 2 },
                        { 1, 2, 6 } };
      
        Console.WriteLine(majorityInMatrix(mat));
    }
}
  
// This code is contributed by ihritik

输出：

1

时间复杂度： O(N x M)
辅助空间： O(NXM)

进一步优化：
我们可以使用Moore的表决算法来解决O(1)额外空间中的上述问题。我们可以简单地将矩阵元素视为一维数组。