# Replace each element by the difference of the total size of the array and frequency of that element

Given an array of integers, the task is to replace every element by the difference of the total size of the array and its frequency.

Examples:

```Input: arr[] = { 1, 2, 5, 2, 2, 5, 4 }
Output: 6 4 5 4 4 5 6
Size of the array is 7.
The frequency of 1 is 1. So replace it by 7-1 = 6
The frequency of 2 is 3. So replace it by 7-3 = 4

Input: arr[] = { 4, 5, 4, 5, 6, 6, 6 }
Output: 5 5 5 5 4 4 4
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Take a hash map, which will store the frequency of all the elements in the array.
2. Now, traverse once again.
3. Now, replace all the elements by the difference of the total size of the array and its frequency.
4. Print the modified array.

Below si the implementation of the above approach:

 `// C++ program to Replace each element ` `// by the difference of the total size ` `// of the array and its frequency ` `#include ` `using` `namespace` `std; ` ` `  `// Function to replace the elements ` `void` `ReplaceElements(``int` `arr[], ``int` `n) ` `{ ` `    ``// Hash map which will store the ` `    ``// frequency of the elements of the array. ` `    ``unordered_map<``int``, ``int``> mp; ` ` `  `    ``for` `(``int` `i = 0; i < n; ++i)  ` ` `  `        ``// Increment the frequency ` `        ``// of the element by 1. ` `        ``mp[arr[i]]++; ` `     `  ` `  `    ``// Replace every element by its frequency ` `    ``for` `(``int` `i = 0; i < n; ++i)  ` `        ``arr[i] = n - mp[arr[i]]; ` ` `  `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 5, 2, 2, 5, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``ReplaceElements(arr, n); ` ` `  `    ``// Print the modified array. ` `    ``for` `(``int` `i = 0; i < n; ++i) ` `        ``cout << arr[i] << ``" "``; ` ` `  `    ``return` `0; ` `} `

 `// Java program to Replace each element ` `// by the difference of the total size ` `// of the array and its frequency ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to replace the elements ` `    ``static` `void` `ReplaceElements(``int` `arr[], ``int` `n)  ` `    ``{ ` `        ``// Hash map which will store the ` `        ``// frequency of the elements of the array. ` `        ``HashMap mp = ``new` `HashMap<>(); ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` ` `  `            ``// Increment the frequency ` `            ``// of the element by 1. ` `            ``if` `(!mp.containsKey(arr[i]))  ` `            ``{ ` `                ``mp.put(arr[i], ``1``); ` `            ``}  ` `            ``else`  `            ``{ ` `                ``mp.put(arr[i], mp.get(arr[i]) + ``1``); ` `            ``} ` `        ``} ` ` `  `        ``// Replace every element by its frequency ` `        ``for` `(``int` `i = ``0``; i < n; ++i)  ` `        ``{ ` `            ``arr[i] = n - mp.get(arr[i]); ` `        ``} ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``1``, ``2``, ``5``, ``2``, ``2``, ``5``, ``4``}; ` `        ``int` `n = arr.length; ` ` `  `        ``ReplaceElements(arr, n); ` ` `  `        ``// Print the modified array. ` `        ``for` `(``int` `i = ``0``; i < n; ++i) ` `        ``{ ` `            ``System.out.print(arr[i] + ``" "``); ` `        ``} ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

 `# Python3 program to Replace each element ` `# by the difference of the total size ` `# of the array and its frequency ` ` `  `# Function to replace the elements ` `def` `ReplaceElements(arr, n): ` `     `  `    ``# Hash map which will store the ` `    ``# frequency of the elements of the array. ` `    ``mp ``=` `dict``() ` ` `  `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Increment the frequency ` `        ``# of the element by 1. ` `        ``mp[arr[i]] ``=` `mp.get(arr[i], ``0``) ``+` `1` ` `  `    ``# Replace every element by its frequency ` `    ``for` `i ``in` `range``(n): ` `        ``arr[i] ``=` `n ``-` `mp[arr[i]] ` ` `  `# Driver code ` `arr ``=` `[``1``, ``2``, ``5``, ``2``, ``2``, ``5``, ``4``] ` `n ``=` `len``(arr) ` ` `  `ReplaceElements(arr, n) ` ` `  `# Print the modified array. ` `for` `i ``in` `range``(n): ` `    ``print``(arr[i], end ``=` `" "``) ` ` `  `# This code is contributed by mohit kumar `

 `// C# program to Replace each element ` `// by the difference of the total size ` `// of the array and its frequency ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to replace the elements ` `    ``static` `void` `ReplaceElements(``int` `[]arr, ``int` `n)  ` `    ``{ ` `        ``// Hash map which will store the ` `        ``// frequency of the elements of the array. ` `        ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>(); ` ` `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` ` `  `            ``// Increment the frequency ` `            ``// of the element by 1. ` `            ``if` `(!mp.ContainsKey(arr[i]))  ` `            ``{ ` `                ``mp.Add(arr[i], 1); ` `            ``}  ` `            ``else` `            ``{ ` `                ``var` `a = mp[arr[i]] + 1; ` `                ``mp.Remove(arr[i]); ` `                ``mp.Add(arr[i], a); ` `            ``} ` `        ``} ` ` `  `        ``// Replace every element by its frequency ` `        ``for` `(``int` `i = 0; i < n; ++i)  ` `        ``{ ` `            ``arr[i] = n - mp[arr[i]]; ` `        ``} ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]arr = {1, 2, 5, 2, 2, 5, 4}; ` `        ``int` `n = arr.Length; ` ` `  `        ``ReplaceElements(arr, n); ` ` `  `        ``// Print the modified array. ` `        ``for` `(``int` `i = 0; i < n; ++i) ` `        ``{ ` `            ``Console.Write(arr[i] + ``" "``); ` `        ``} ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:
```6 4 5 4 4 5 6
```

Time Complexity – O(N)

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