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
Explanation:
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 = 4Input: arr[] = { 4, 5, 4, 5, 6, 6, 6 }
Output: 5 5 5 5 4 4 4
Approach:
- Take a hash map, which will store the frequency of all the elements in the array.
- Now, traverse once again.
- Now, replace all the elements by the difference of the total size of the array and its frequency.
- Print the modified array.
Below is the implementation of the above approach:
C++
// C++ program to Replace each element// by the difference of the total size// of the array and its frequency#include <bits/stdc++.h>using namespace std;// Function to replace the elementsvoid 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 codeint main(){ int arr[] = { 1, 2, 5, 2, 2, 5, 4 }; int n = sizeof(arr) / sizeof(arr[0]); ReplaceElements(arr, n); // Print the modified array. for (int i = 0; i < n; ++i) cout << arr[i] << " "; return 0;} |
Java
// Java program to Replace each element// by the difference of the total size// of the array and its frequencyimport 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<Integer, Integer> 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
# Python3 program to Replace each element# by the difference of the total size# of the array and its frequency# Function to replace the elementsdef 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 codearr = [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#
// C# program to Replace each element// by the difference of the total size// of the array and its frequencyusing 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 */ |
Javascript
<script>// JavaScript program to Replace each element// by the difference of the total size// of the array and its frequency// Function to replace the elementsfunction ReplaceElements(arr, n) { // Hash map which will store the // frequency of the elements of the array. let mp = new Map(); for(let i = 0; i < n; i++) { // Increment the frequency // of the element by 1. if (!mp.has(arr[i])) { mp.set(arr[i], 1); } else { mp.set(arr[i], mp.get(arr[i]) + 1); } } // Replace every element by its frequency for(let i = 0; i < n; ++i) { arr[i] = n - mp.get(arr[i]); }}// Driver Codelet arr = [ 1, 2, 5, 2, 2, 5, 4 ];let n = arr.length;ReplaceElements(arr, n);// Print the modified array.for(let i = 0; i < n; ++i){ document.write(arr[i] + " ");}// This code is contributed by code_hunt</script> |
Output
6 4 5 4 4 5 6
Time Complexity: O(N) where N is traversing all elements in the array.
Auxiliary Space: O(N) where N is for map to store the frequency of array elements.
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