Given an array arr[] consisting of N distinct integers, the task is to rearrange the array such that no element is same as its index ( 1-based indexing ). If multiple solutions exist, print any one of them.
Examples:
Input: arr[] = {4, 2, 3, 1}
Output: 3 1 4 2
Explanation: The elements at indices {1, 2, 3, 4} are {3, 1, 4, 2} respectively.Input: arr[] = {10, 20, 30, 40, 6}
Output: 6 10 20 30 40
Explanation: The elements at indices {1, 2, 3, 4, 5} are {6, 10, 20, 30, 40} respectively.
Approach: The idea is to use sorting and swap each adjacent pair of indices at any index i if arr[i] is equal to i. This is because, if arr[i] = i holds true, then definitely arr[i + 1] ? i and arr[i] ? i + 1 because arr[i + 1] > arr[i]. If the last element, arr[N] is equal to N, then swap arr[N] and arr[N – 1]. Follow the steps below to solve the problem:
- Sort the array arr[] in the increasing order.
- Traverse the array over the range [0, N – 2] using the variable i and check if arr[i] is the same as (i + 1) or not. If found to be true, then swap arr[i] and arr[i + 1].
- Now, for the last array element, if arr[N] is same as N, then swap arr[N] and arr[N – 1].
- After completing the above steps, print the modified array.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to rearrange the array a[]// such that none of the array elements// is same as its indexvoid rearrangeArray(int a[], int n){ // Sort the array sort(a, a + n); // Traverse the indices [0, N - 2] // of the given array for (int i = 0; i < n - 1; i++) { // Check if the current element // is equal to its index if (a[i] == i + 1) { // If found to be true, swap // current element with the // next element swap(a[i], a[i + 1]); } } // Check if the last element is // same as its index if (a[n - 1] == n) { // If found to be true, swap // current element with the // previous element swap(a[n - 1], a[n - 2]); } // Print the modified array for (int i = 0; i < n; i++) { cout << a[i] << " "; }}// Driver Codeint main(){ int arr[] = { 1, 5, 3, 2, 4 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call rearrangeArray(arr, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to rearrange the array a[]// such that none of the array elements// is same as its indexstatic void rearrangeArray(int a[], int n){ // Sort the array Arrays.sort(a); // Traverse the indices [0, N - 2] // of the given array for(int i = 0; i < n - 1; i++) { // Check if the current element // is equal to its index if (a[i] == i + 1) { // If found to be true, swap // current element with the // next element int temp = a[i]; a[i] = a[i + 1]; a[i + 1] = temp; } } // Check if the last element is // same as its index if (a[n - 1] == n) { // If found to be true, swap // current element with the // previous element int temp = a[n - 1]; a[n - 1] = a[n - 2]; a[n - 2] = temp; } // Print the modified array for(int i = 0; i < n; i++) { System.out.print(a[i] + " "); }}// Driver Codepublic static void main(String args[]){ int arr[] = { 1, 5, 3, 2, 4 }; int N = arr.length; // Function Call rearrangeArray(arr, N);}}// This code is contributed by ipg2016107 |
Python3
# Python3 program for the above approach# Function to rearrange the array a[]# such that none of the array elements# is same as its indexdef rearrangeArray(a, n): # Sort the array a = sorted(a) # Traverse the indices [0, N - 2] # of the given array for i in range(n - 1): # Check if the current element # is equal to its index if (a[i] == i + 1): # If found to be true, swap # current element with the # next element a[i], a[i + 1] = a[i + 1], a[i] # Check if the last element is # same as its index if (a[n - 1] == n): # If found to be true, swap # current element with the # previous element a[n - 1], a[n - 2] = a[n - 2], a[n - 1] # Print the modified array for i in range(n): print(a[i], end = " ")# Driver Codeif __name__ == '__main__': arr = [1, 5, 3, 2, 4] N = len(arr) # Function Call rearrangeArray(arr, N) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;public class GFG{ // Function to rearrange the array []a// such that none of the array elements// is same as its indexstatic void rearrangeArray(int []a, int n){ // Sort the array Array.Sort(a); // Traverse the indices [0, N - 2] // of the given array for(int i = 0; i < n - 1; i++) { // Check if the current element // is equal to its index if (a[i] == i + 1) { // If found to be true, swap // current element with the // next element int temp = a[i]; a[i] = a[i + 1]; a[i + 1] = temp; } } // Check if the last element is // same as its index if (a[n - 1] == n) { // If found to be true, swap // current element with the // previous element int temp = a[n - 1]; a[n - 1] = a[n - 2]; a[n - 2] = temp; } // Print the modified array for(int i = 0; i < n; i++) { Console.Write(a[i] + " "); }}// Driver Codepublic static void Main(String []args){ int []arr = { 1, 5, 3, 2, 4 }; int N = arr.Length; // Function Call rearrangeArray(arr, N);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript program to implement// the above approach// Function to rearrange the array a[]// such that none of the array elements// is same as its indexfunction rearrangeArray(a, n){ // Sort the array a.sort(); // Traverse the indices [0, N - 2] // of the given array for(let i = 0; i < n - 1; i++) { // Check if the current element // is equal to its index if (a[i] == i + 1) { // If found to be true, swap // current element with the // next element let temp = a[i]; a[i] = a[i + 1]; a[i + 1] = temp; } } // Check if the last element is // same as its index if (a[n - 1] == n) { // If found to be true, swap // current element with the // previous element let temp = a[n - 1]; a[n - 1] = a[n - 2]; a[n - 2] = temp; } // Print the modified array for(let i = 0; i < n; i++) { document.write(a[i] + " "); }}// Driver code let arr = [ 1, 5, 3, 2, 4 ]; let N = arr.length; // Function Call rearrangeArray(arr, N); // This code is contributed by splevel62.</script> |
Output:
2 1 4 5 3
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
