# Find element with highest frequency in given nested Array

Given an array arr[] of N integers. The task is to create a frequency array freq[] of the given array arr[] and find the maximum element of the frequency array. If two elements have the same frequency in the array freq[], then return the element which has a smaller value.

Examples:

Input: arr[] = {1, 1, 1, 2, 3, 2, 2, 3, 5, 5, 5, 5, 4, 4, 4, 4, 4};
Output:
Explanation:
frequency of elements is given by:
val -> freq[]
1 -> 3
2 -> 3
3 -> 2
4 -> 5
5 -> 4
Element 3 has the maximum frequency in the frequency array.

Input: arr[] = { 3, 5, 15, 51, 15, 14, 14, 14, 14, 4};
Output:
Explanation:
frequency of elements is given by:
val -> freq[]
3 -> 1
4 -> 1
5 -> 1
14 -> 4
15 -> 2
51 -> 1
Element 1 has the maximum frequency in the frequency array.

Approach:

1. Store the frequency of the elements of arr[] in a map say map1, with elements of arr[] as key and their frequency as value.
2. Now, store the frequency of elements of map1 in some other map say map2.
3. Traverse map2 to get the highest element.
4. If there is multiple highest element than the element which has lower value is print.

Below is the implementation of the above approach:

## C++14

 `// C++14 program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to get the highest` `// frequency of frequency array` `int` `findElement(``int` `a[], ``int` `n)` `{` `    ``// To find the maximum frequency` `    ``// initialize it with INT_MIN` `    ``int` `mx = INT_MIN;` `    ``int` `ans = 0;`   `    ``// Initialize maps to store the` `    ``// count of element in array` `    ``map<``int``, ``int``> map1, map2;`   `    ``// Store the frequency of` `    ``// element of array in map1` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``map1[a[i]]++;` `    ``}`   `    ``// Storing the frequency i.e.,` `    ``// (x.second) which is count of` `    ``// element in array` `    ``for` `(``auto` `x : map1) {` `        ``map2[x.second]++;` `    ``}`   `    ``for` `(``auto` `x : map2) {`   `        ``// Check if the fequency of` `        ``// element is greater than mx` `        ``if` `(x.second > mx) {` `            ``mx = x.second;`   `            ``// Store the value to check` `            ``// when frequency is same` `            ``ans = x.first;` `        ``}`   `        ``// If frequency of 2 element is` `        ``// same than storing minimum value` `        ``else` `if` `(x.second == mx) {` `            ``ans = min(ans, x.first);` `        ``}` `    ``}`   `    ``// Return the highest frequency` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given array arr[]` `    ``int` `arr[] = { 1, 1, 1, 2, 3, 2, 2, 3, 5,` `                  ``5, 5, 5, 4, 4, 4, 4, 4 };`   `    ``// Size of the array` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// Function Call` `    ``cout << findElement(arr, n) << endl;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to get the highest` `// frequency of frequency array` `static` `int` `findElement(``int` `a[], ``int` `n)` `{` `    `  `    ``// To find the maximum frequency` `    ``// initialize it with INT_MIN` `    ``int` `mx = Integer.MIN_VALUE;` `    ``int` `ans = ``0``;`   `    ``// Initialize maps to store the` `    ``// count of element in array` `    ``Map map1 = ``new` `HashMap<>(),` `                          ``map2 = ``new` `HashMap<>();`   `    ``// Store the frequency of` `    ``// element of array in map1` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{` `        ``map1.put(a[i], map1.getOrDefault(a[i], ``0``) + ``1``);` `    ``}`   `    ``// Storing the frequency i.e.,` `    ``// (x.second) which is count of` `    ``// element in array` `    ``for``(Integer x : map1.values()) ` `    ``{` `        ``map2.put(x, map2.getOrDefault(x, ``0``) + ``1``);` `    ``}`   `    ``for``(Map.Entry x : map2.entrySet())` `    ``{` `        `  `        ``// Check if the fequency of` `        ``// element is greater than mx` `        ``if` `(x.getValue() > mx)` `        ``{` `            ``mx = x.getValue();`   `            ``// Store the value to check` `            ``// when frequency is same` `            ``ans = x.getKey();` `        ``}`   `        ``// If frequency of 2 element is` `        ``// same than storing minimum value` `        ``else` `if` `(x.getValue() == mx)` `        ``{` `            ``ans = Math.min(ans, x.getKey());` `        ``}` `    ``}`   `    ``// Return the highest frequency` `    ``return` `ans;` `}`   `// Driver code` `public` `static` `void` `main (String[] args) ` `{` `    `  `    ``// Given array arr[]` `    ``int` `arr[] = { ``1``, ``1``, ``1``, ``2``, ``3``, ``2``, ``2``, ``3``, ``5``,` `                  ``5``, ``5``, ``5``, ``4``, ``4``, ``4``, ``4``, ``4` `};` `    `  `    ``// Size of the array` `    ``int` `n = arr.length;` `    `  `    ``// Function call` `    ``System.out.println(findElement(arr, n));` `}` `}`   `// This code is contributed by offbeat`

## Python3

 `# Python3 program for the above approach` `import` `sys`   `# Function to get the highest` `# frequency of frequency array` `def` `findElement(a, n):` `    `  `    ``# To find the maximum frequency` `    ``# initialize it with INT_MIN` `    ``mx ``=` `-``sys.maxsize ``-` `1` `    ``ans ``=` `0`   `    ``# Initialize maps to store the` `    ``# count of element in array` `    ``map1 ``=` `{}` `    ``map2 ``=` `{}`   `    ``# Store the frequency of` `    ``# element of array in map1` `    ``for` `i ``in` `a:` `        ``map1[i] ``=` `map1.get(i, ``0``) ``+` `1`   `    ``# Storing the frequency i.e.,` `    ``# (x.second) which is count of` `    ``# element in array` `    ``for` `x ``in` `map1:` `        ``map2[map1[x]] ``=` `map2.get(map1[x], ``0``) ``+` `1`   `    ``for` `x ``in` `map2:`   `        ``# Check if the fequency of` `        ``# element is greater than mx` `        ``if` `(map2[x] > mx):` `            ``mx ``=` `map2[x]`   `            ``# Store the value to check` `            ``# when frequency is same` `            ``ans ``=` `x`   `        ``# If frequency of 2 element is` `        ``# same than storing minimum value` `        ``elif` `(map2[x] ``=``=` `mx):` `            ``ans ``=` `min``(ans, x)`   `    ``# Return the highest frequency` `    ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Given array arr[]` `    ``arr ``=` `[ ``1``, ``1``, ``1``, ``2``, ``3``, ``2``, ``2``, ``3``, ` `            ``5``, ``5``, ``5``, ``5``, ``4``, ``4``, ``4``, ``4``, ``4``]`   `    ``# Size of the array` `    ``n ``=` `len``(arr)`   `    ``# Function call` `    ``print``(findElement(arr, n))`   `# This code is contributed by mohit kumar 29`

Output:

```3

```

Time Complexity: O(N)
Auxiliary Space: O(N)

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Improved By : mohit kumar 29, offbeat

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