Sum of all odd frequency elements in a Matrix

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

Examples: 

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

Input : mat[] = {{10, 20},
                 {40, 40}}
Output : 30

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 and check if it is odd, if it is odd, 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 odd
// 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 odd
// frequency elements in a Matrix
int sumOddOccurring(int arr[N][M])
{

    // Store frequencies of elements
    // in matrix
    map<int, int> mp;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            mp[arr[i][j]]++;
        }
    }

    // Sum of odd frequency elements
    int sum = 0;
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        if (itr->second % 2 != 0) {
            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 odd
// frequency elements in a Matrix

import java.util.*;

class GFG
{

    static int N = 3; // Rows
    static int M = 3; // Columns

    // Function to find sum of all odd
    // frequency elements in a Matrix
    static int sumOddOccurring(int arr[][])
    {
        // Store frequencies of elements
        // in matrix
        Map<Integer, Integer> mp = new HashMap<>();
        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);
                }
            }
        }

        int sum = 0;

        // Sum of odd frequency elements
        for (Map.Entry<Integer, Integer> itr : mp.entrySet())
        {
            if (itr.getValue() % 2 != 0)
            {
                sum += (itr.getKey()) * (itr.getValue());
            }
        }

        return sum;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int mat[][] = {{1, 2, 3},
        {1, 3, 2},
        {1, 5, 6}};

        System.out.println(sumOddOccurring(mat));
    }
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find sum of all odd
# frequency elements in a Matrix

# Function to find sum of all odd
# frequency elements in a Matrix
def sumOddOccurring(mat):

    # Store frequencies of elements
    # in matrix
    mp = {}
    n, m = len(mat), len(mat[0])
    for i in range(n):
        for j in range(m):
            if mat[i][j] in mp:
                mp[(mat[i][j])] = mp.get(mat[i][j]) + 1
            else:
                mp[(mat[i][j])] = 1

    # Sum of odd frequency elements
    _sum = 0
    for i in range(n):
        for j in range(m):
            if mp.get(mat[i][j]) % 2 == 1:
                _sum+=mat[i][j]
    return _sum

# Driver Code
if __name__=='__main__':
    mat=[[1,2,3],[1,3,2],[1,5,6]]
    print(sumOddOccurring(mat))

# This code is Contributed by Vikash Kumar 37

C#

// C# program to find sum of all odd
// frequency 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 sum of all odd
    // frequency elements in a Matrix
    static int sumOddOccurring(int [,]arr)
    {
        // Store frequencies 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]))
                {
                    var v = mp[arr[i,j]];
                    mp.Remove(arr[i,j]);
                    mp.Add(arr[i,j], ++v);
                }
                else
                {
                    mp.Add(arr[i,j], 1);
                }
            }
        }

        int sum = 0;

        // Sum of odd frequency elements
        foreach(KeyValuePair<int, int> itr in mp)
        {
            if (itr.Value % 2 != 0)
            {
                sum += (itr.Key) * (itr.Value);
            }
        }

        return sum;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int [,]mat = {{1, 2, 3},
        {1, 3, 2},
        {1, 5, 6}};

        Console.WriteLine(sumOddOccurring(mat));
    }
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript program to find sum of all odd
// frequency elements in a Matrix
var N = 3; // Rows
var M = 3; // Columns
// Function to find sum of all odd
// frequency elements in a Matrix
function sumOddOccurring(arr)
{
    // Store frequencies 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]))
            {
                var v = mp.get(arr[i][j]);
                mp.delete(arr[i][j]);
                mp.set(arr[i][j], ++v);
            }
            else
            {
                mp.set(arr[i][j], 1);
            }
        }
    }
    var sum = 0;

    // Sum of odd frequency elements
    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

14

Complexity Analysis:

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