Given an array arr[] of size N and a number K. The task is to print the array arr[] after K operations such that at every operation, replace every element arr[i] in the array with max – arr[i] where max is the maximum element in the array.
Examples :
Input: arr[] = {4, 8, 12, 16}, K = 4
Output: 0 4 8 12
Explanation:
For the given array arr[] = { 4, 8, 12, 16 }. The array is changed as follows for every operation K:
{ 12, 8, 4, 0 } – K = 1
{ 0, 4, 8, 12 } – K = 2
{ 12, 8, 4, 0 } – K = 3
{ 0, 4, 8, 12 } – K = 4
Input: arr[] = {8, 0, 3, 5}, K = 3
Output: 0 8 5 3
Explanation:
For the given array arr[] = { 8, 0, 3, 5 }. The array is changed as follows for every operation K:
{ 0, 8, 5, 3 } – K = 1
{ 8, 0, 3, 5 } – K = 2
{ 0, 8, 5, 3 } – K = 3
Approach: The idea is to clearly observe the array after every step.
- Initially, since we are subtracting the maximum element from the array, we can be sure that there will be at least one element in the array with zero value. This happens after the first step that is K = 1. Let the array after this step be A[] with maximum element M.
- After this first step, the array becomes stagnant and the values alternatively change from its value A[i] to M – A[i] alternatively.
- For example, let us take the array arr[] = {4, 8, 12, 16}. In this array, the maximum value is 16 and let us assume K = 4.
- At the first step, (i.e.) for K = 1, the array is reduced to {12, 8, 4, 0}. That is, every element arr[i] is replaced with 16 – arr[i]. Let this array be A[] and the maximum element in this array M is 12.
- After the first step, every element in this array A[i] alternatively change between A[i] and 12 – A[i]. That is, for K = 2, the array now becomes {0, 4, 8, 12}.
- Similarly, for the third step, (i.e.) K = 3, the array again becomes {12, 8, 4, 0} which is the same as for K = 1.
- And for the fourth step, (i.e.) K = 4, the array becomes {0, 4, 8, 12} which is the same as for K = 2 and so on.
Therefore, it can be concluded that we only need to check if K is odd or even:
- If K is even: replace every element arr[i] with arr[i] – min where min is the minimum element in the array.
- If K is odd: replace every element arr[i] with max – arr[i] where max is the maximum element in the array.
Below is the implementation of the above approach:
CPP
// C++ program to print the array// after K operations#include <bits/stdc++.h>using namespace std;// Function to print the array// after K operationsvoid printArray(int A[], int n, int K){ // Variables to store the minimum and // the maximum elements of the array int minEle = INT_MAX, maxEle = INT_MIN; // Loop to find the minimum and the // maximum elements of the array for (int i = 0; i < n; i++) { minEle = min(minEle, A[i]); maxEle = max(maxEle, A[i]); } // If K is not equal to 0 if (K != 0) { // If K is odd if (K % 2 == 1) { // Replace every element with // max - arr[i] for (int i = 0; i < n; i++) A[i] = maxEle - A[i]; } // If K is even else { // Replace every element with // A[i] - min for (int i = 0; i < n; i++) A[i] = A[i] - minEle; } } // Printing the array after K operations for (int i = 0; i < n; i++) cout << A[i] << " ";}// Driver codeint main(){ int arr[] = { 4, 8, 12, 16 }; int K = 4; int N = sizeof(arr) / sizeof(arr[0]); printArray(arr, N, K); return 0;} |
Java
// Java program to print the array// after K operationsimport java.io.*; class GFG { // Function to print the array // after K operations static void printArray(int[] A, int n, int K) { // Variables to store the minimum and // the maximum elements of the array int minEle = Integer.MAX_VALUE, maxEle = Integer.MAX_VALUE; // Loop to find the minimum and the // maximum elements of the array for (int i = 0; i < n; i++) { minEle = Math.min(minEle, A[i]); maxEle = Math.max(maxEle, A[i]); } // If K is not equal to 0 if (K != 0) { // If K is odd if (K % 2 == 1) { // Replace every element with // max - arr[i] for (int i = 0; i < n; i++) A[i] = maxEle - A[i]; } // If K is even else { // Replace every element with // A[i] - min for (int i = 0; i < n; i++) A[i] = A[i] - minEle; } } // Printing the array after K operations for (int i = 0; i < n; i++) System.out.print(A[i] + " "); } // Driver Code public static void main (String[] args) { int[] arr = { 4, 8, 12, 16 }; int K = 4; int N = arr.length; printArray(arr, N, K); }}// This code is contributed by shivanisinghss2110 |
Python3
# Python3 program to print the array# after K operations# Function to print the array# after K operationsdef printArray(A, n, K): # Variables to store the minimum and # the maximum elements of the array minEle = 10**9 maxEle = -10**9 # Loop to find the minimum and the # maximum elements of the array for i in range(n): minEle = min(minEle, A[i]) maxEle = max(maxEle, A[i]) # If K is not equal to 0 if (K != 0): # If K is odd if (K % 2 == 1): # Replace every element with # max - arr[i] for i in range(n): A[i] = maxEle - A[i] # If K is even else: # Replace every element with # A[i] - min for i in range(n): A[i] = A[i] - minEle # Printing the array after K operations for i in A: print(i, end=" ")# Driver codeif __name__ == '__main__': arr=[4, 8, 12, 16] K = 4 N = len(arr) printArray(arr, N, K)# This code is contributed by mohit kumar 29 |
C#
// C# program to print the array// after K operationsusing System;class GFG{ // Function to print the array // after K operations static void printArray(int[] A, int n, int K) { // Variables to store the minimum and // the maximum elements of the array int minEle = Int32.MaxValue, maxEle = Int32.MinValue; // Loop to find the minimum and the // maximum elements of the array for (int i = 0; i < n; i++) { minEle = Math.Min(minEle, A[i]); maxEle = Math.Max(maxEle, A[i]); } // If K is not equal to 0 if (K != 0) { // If K is odd if (K % 2 == 1) { // Replace every element with // max - arr[i] for (int i = 0; i < n; i++) A[i] = maxEle - A[i]; } // If K is even else { // Replace every element with // A[i] - min for (int i = 0; i < n; i++) A[i] = A[i] - minEle; } } // Printing the array after K operations for (int i = 0; i < n; i++) Console.Write(A[i] + " "); } // Driver code static public void Main () { int[] arr = { 4, 8, 12, 16 }; int K = 4; int N = arr.Length; printArray(arr, N, K); }}// This code is contributed by shubhamsingh10 |
Javascript
<script>// Javascript program to print the array// after K operations// Function to print the array// after K operationsfunction printArray(A, n, K){ // Variables to store the minimum and // the maximum elements of the array var minEle = 100000000, maxEle = -100000000; // Loop to find the minimum and the // maximum elements of the array for (var i = 0; i < n; i++) { minEle = Math.min(minEle, A[i]); maxEle = Math.max(maxEle, A[i]); } // If K is not equal to 0 if (K != 0) { // If K is odd if (K % 2 == 1) { // Replace every element with // max - arr[i] for (var i = 0; i < n; i++) A[i] = maxEle - A[i]; } // If K is even else { // Replace every element with // A[i] - min for (var i = 0; i < n; i++) A[i] = A[i] - minEle; } } // Printing the array after K operations for (var i = 0; i < n; i++) document.write(A[i] + " ");}// Driver codearr = [ 4, 8, 12, 16 ];var K = 4;var N = arr.length;printArray(arr, N, K);</script> |
Output
0 4 8 12
Time Complexity: O(N) where N is the size of the array.
Auxiliary Space: O(1), As constant extra space is used.
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