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
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 si 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 */ |
Output:
6 4 5 4 4 5 6
Time Complexity – O(N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
