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Sum of all even frequency elements in Matrix

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Given a NxM matrix of integers containing duplicate elements. The task is to find the sum of all even occurring elements in the given matrix. That is the sum of all such elements whose frequency is even in the matrix.

Examples

Input : mat[] = {{1, 1, 2},
                {2, 3, 3},
                {4, 5, 3}}
Output : 18
The even occurring elements are 1, 2 and their number
of occurrences are 2, 2 respectively. Therefore,
sum = 1+1+2+2 = 6.

Input : mat[] = {{10, 20},
                 {40, 40}}
Output : 80

Approach:

  • Traverse the matrix and use a map 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 and check if it is even, then add this element it’s frequency times to sum.

Below is the implementation of the above approach: 

C++




// C++ program to find sum of all even
// frequency elements in a Matrix
 
#include <bits/stdc++.h>
using namespace std;
 
#define N 3 // Rows
#define M 3 // Columns
 
// Function to find sum of all even
// frequency elements in a Matrix
int sumOddOccurring(int arr[N][M])
{
 
    // Store frequency of elements
    // in matrix
    unordered_map<int, int> mp;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            mp[arr[i][j]]++;
        }
    }
 
    // Sum even frequency elements
    int sum = 0;
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        if (itr->second % 2 == 0) {
            int x = itr->second;
            sum += (itr->first) * (itr->second);
        }
    }
 
    return sum;
}
 
// Driver Code
int main()
{
 
    int mat[N][M] = { { 1, 2, 3 },
                      { 1, 3, 2 },
                      { 1, 5, 6 } };
 
    cout << sumOddOccurring(mat) << endl;
 
    return 0;
}


Java




// Java program to find sum of all even
// frequency elements in a Matrix
import java.util.*;
 
class GFG
{
 
static final int N = 3; // Rows
static final int M = 3; // Columns
 
// Function to find sum of all even
// frequency elements in a Matrix
static int sumOddOccurring(int arr[][])
{
 
    // Store frequency of elements
    // in matrix
    Map<Integer,
        Integer> mp = new HashMap<Integer,
                                  Integer>();
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            if(mp.get(arr[i][j]) == null)
                mp.put(arr[i][j], 1);
            else
                mp.put(arr[i][j],
                      (mp.get(arr[i][j]) + 1));
        }
    }
 
    // Sum even frequency elements
    int sum = 0;
    Set< Map.Entry<Integer,
                   Integer> > st = mp.entrySet();
 
    for (Map.Entry< Integer, Integer> me:st)
    {
        if (me.getValue() % 2 == 0)
        {
            int x = me.getValue();
            sum += (me.getKey()) * (me.getValue());
        }
    }
    return sum;
}
 
// Driver Code
public static void main(String args[])
{
    int mat[][] = {{ 1, 2, 3 },
                   { 1, 3, 2 },
                   { 1, 5, 6 }};
 
    System.out.print(sumOddOccurring(mat));
}
}
 
// This code is contributed by Arnab Kundu


Python3




# Python3 program to find sum of all even
# frequency elements in a Matrix
import sys
 
N = 3 # Rows
M = 3 # Columns
 
# Function to find sum of all even
# frequency elements in a Matrix
def sumOddOccurring(arr):
 
    # Store frequencies of elements
    # in matrix
    mp = dict()
    for i in range(N):
        for j in range(M):
            if arr[i][j] in mp:
                mp[arr[i][j]] += 1
            else:
                mp[arr[i][j]] = 1
 
    # Sum of even frequency elements
    s = 0
    for i in mp:
        if mp[i] % 2 == 0:
            x = mp[i]
            s += i * mp[i]
 
    return s
 
# Driver code
if __name__ == "__main__":
    mat = [[1, 2, 3],
           [1, 3, 2],
           [1, 5, 6]]
 
    print(sumOddOccurring(mat))
 
# This code is contributed by
# sanjeev2552


C#




// C# program to find sum of all even
// frequency elements in a Matrix
using System;
using System.Collections.Generic;
 
class Sol
{
 
static readonly int N = 3; // Rows
static readonly int M = 3; // Columns
 
// Function to find sum of all even
// frequency elements in a Matrix
static int sumOddOccurring(int [,]arr)
{
 
    // Store frequency of elements
    // in matrix
    Dictionary<int, int> mp = new Dictionary<int,int>();
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            if(!mp.ContainsKey(arr[i, j]))
                mp.Add(arr[i, j], 1);
            else{
                var val = mp[arr[i, j]];
                mp.Remove(arr[i, j]);
                mp.Add(arr[i, j], val + 1);
            }
        }
    }
 
    // Sum even frequency elements
    int sum = 0;
    foreach(KeyValuePair<int, int> entry in mp)
    {
        if(entry.Value % 2 == 0){
            sum += entry.Key * entry.Value;
        }
    }
 
    return sum;
}
 
// Driver Code
public static void Main(String []args)
{
 
    int [,]mat = { { 1, 2, 3 },
                    { 1, 3, 2 },
                    { 1, 5, 6 } };
 
    Console.Write( sumOddOccurring(mat) );
 
}
}
 
/* This code contributed by PrinciRaj1992 */


Javascript




<script>
 
// JavaScript program to find sum of all even
// frequency elements in a Matrix
 
var N = 3; // Rows
var M = 3; // Columns
 
// Function to find sum of all even
// frequency elements in a Matrix
function sumOddOccurring(arr)
{
 
    // Store frequency of elements
    // in matrix
    var mp = new Map();
    for (var i = 0; i < N; i++)
    {
        for (var j = 0; j < M; j++)
        {
            if(!mp.has(arr[i][j]))
                mp.set(arr[i][j], 1);
            else{
                var val = mp.get(arr[i][j]);
                mp.delete(arr[i][j]);
                mp.set(arr[i][j], val + 1);
            }
        }
    }
 
    // Sum even frequency elements
    var sum = 0;
    mp.forEach((value, key) => {
        if(value % 2 == 0){
            sum += key * value;
        }
    });
 
    return sum;
}
 
// Driver Code
var mat = [[1, 2, 3 ],
                [1, 3, 2 ],
                [1, 5, 6 ]];
document.write( sumOddOccurring(mat) );
 
 
 
</script>


Output

10

Complexity Analysis:

  • Time Complexity: O(N x M) 
  • Auxiliary Complexity: O(N x M)


Last Updated : 06 Sep, 2022
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