# Largest subarray with frequency of all elements same

Given an array arr[] of N integers, the task is to find the size of largest subarray with frequency of all elements same.

Examples:

Input: arr[] = {1, 2, 2, 5, 6, 5, 6}
Output: 6
Explanation:
The subarrary = {2, 2, 5, 6, 5, 6} has frequency of every element is 2.

Input: arr[] = {1, 1, 1, 1, 1}
Output: 5
Explanation:
The subarrary = {1, 1, 1, 1, 1} has frequency of every element is 5.

Approach: The idea is to generate all possible subarrays and check for each subarray whether any subarray has frequency of all elements or not. Below are the steps:

1. Generate all possible subarrays.
2. For each subarray, take two maps, one map to stores the frequency of every element and second map stores number of elements with given frequency.
3. If for any subarray, size of second map becomes equal to 1, that means every element have same frequency in the subarray.
4. Return the maximum size of all such subarrays.

Below is the implementation of above approach:

## C++

 `// C++ program for the above approach ` ` `  `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find maximum subarray size ` `int` `max_subarray_size(``int` `N, ``int` `arr[]) ` `{ ` `    ``int` `ans = 0; ` ` `  `    ``// Generating all subarray ` `    ``// i -> starting index ` `    ``// j -> end index ` `    ``for` `(``int` `i = 0; i < N; i++) { ` ` `  `        ``// Map 1 to hash frequency ` `        ``// of all elements in subarray ` `        ``unordered_map<``int``, ``int``> map1; ` ` `  `        ``// Map 2 to hash frequency ` `        ``// of all frequencies of ` `        ``// elements ` `        ``unordered_map<``int``, ``int``> map2; ` ` `  `        ``for` `(``int` `j = i; j < N; j++) { ` ` `  `            ``// ele_count is the previous ` `            ``// frequency of arr[j] ` `            ``int` `ele_count; ` ` `  `            ``// Finding previous frequency of ` `            ``// arr[j] in map 1 ` `            ``if` `(map1.find(arr[j]) ` `                ``== map1.end()) { ` `                ``ele_count = 0; ` `            ``} ` `            ``else` `{ ` `                ``ele_count = map1[arr[j]]; ` `            ``} ` ` `  `            ``// Increasing frequency of arr[j] ` `            ``// by 1 ` `            ``map1[arr[j]]++; ` ` `  `            ``// Check if previous frequency ` `            ``// is present in map 2 ` `            ``if` `(map2.find(ele_count) ` `                ``!= map2.end()) { ` ` `  `                ``// Delete previous frequency ` `                ``// if hash is equal to 1 ` `                ``if` `(map2[ele_count] == 1) { ` `                    ``map2.erase(ele_count); ` `                ``} ` `                ``else` `{ ` ` `  `                    ``// Decrement the hash of ` `                    ``// previous frequency ` `                    ``map2[ele_count]--; ` `                ``} ` `            ``} ` ` `  `            ``// Incrementing hash of new ` `            ``// frequency in map 2 ` `            ``map2[ele_count + 1]++; ` ` `  `            ``// Check if map2 size is 1 ` `            ``// and updating answer ` `            ``if` `(map2.size() == 1) ` `                ``ans = max(ans, j - i + 1); ` `        ``} ` `    ``} ` ` `  `    ``// Return the maximum size of subarray ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array arr[] ` `    ``int` `arr[] = { 1, 2, 2, 5, 6, 5, 6 }; ` ` `  `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``cout << max_subarray_size(N, arr); ` `    ``return` `0; ` `} `

Output:

```6
```

Time Complexity: O(N2)
Auxiliary Space: O(1)

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