Count majority element in a matrix

Given a NxM matrix of integers containing duplicate elements. The task is to find the count of all majority occurring elements in the given matrix, where majority element are those whose frequency is greater than or equal to (N*M)/2.

Examples:

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

Approach:

  • Traverse the matrix and use a map in C++ to store the frequency of elements of the matrix such that the key of map is the matrix element and value is its frequency in the matrix.
  • Then, traverse the map to find the frequency of elements with a count variable to count majority elements and check if it is equal to or greater than (N*M)/2. If true, then increase the count.

Below is the implementation of the above approach:

C++

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// C++ program to find count of all
// majority elements in a Matrix
  
#include <bits/stdc++.h>
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<int, int> 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;
}

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Java

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// 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<Integer, Integer> mp = 
                new HashMap<Integer, Integer>();
      
        // 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<HashMap.Entry<Integer, Integer>> itr = 
                                mp.entrySet().iterator(); 
          
        while(itr.hasNext())
        {
            // check if frequency is greater than
            // or equal to (N*M)/2
            HashMap.Entry<Integer, Integer> 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

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Python3

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# 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

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C#

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// 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<int, int> mp = 
                new Dictionary<int, int>();
      
        // 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<int, int>.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

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Output:

1

Time Complexity: O(N x M)
Auxiliary Space: O(N X M)

Further Optimization :
We can use Moore’s voting algorithm to solve above problem in O(1) extra space. We can simply consider matrix elements as one dimensional array.



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Improved By : Shashank_Sharma, ihritik