Rearrange given array such that no array element is same as its index
Last Updated :
24 Feb, 2022
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 index
void 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 Code
int main()
{
int arr[] = { 1, 5, 3, 2, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
rearrangeArray(arr, N);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to rearrange the array a[]
// such that none of the array elements
// is same as its index
static 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 Code
public 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 program for the above approach
# Function to rearrange the array a[]
# such that none of the array elements
# is same as its index
def 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 Code
if __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# program for the above approach
using System;
public class GFG
{
// Function to rearrange the array []a
// such that none of the array elements
// is same as its index
static 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 Code
public 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
<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 index
function 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>
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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