# Count majority element in a matrix

• Last Updated : 06 Sep, 2022

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++

 `// 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<``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 iterator 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 iterator 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 iterator 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<``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 iterator 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`

## Javascript

 ``

Output

```1
```

Complexity Analysis:

• 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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