Given an array arr consisting of N unique integers, the task is to rearrange the array such that element at index i of array should not be mean of adjacent elements (i.e., of index i-1 and i+1). Any possible rearrangement can be returned.
Example:
Input: arr = [5, 4, 3, 2, 1]
Output: [5, 3, 4, 2, 1]
Explanation: In the input array:
Mean(5, 3) = (5 + 3)/2 = 4,
Mean(4, 2) = (4+ 2 )/2 = 3,
Mean(3, 1) = (3 + 1)/2 = 2.
After rearranging the array as [5, 3, 4, 2, 1], now no element is the mean of adjacent elements: (5 + 4)/2 ? 3, (3 + 2)/2 ? 4, (4 + 1)/2 ? 2Input: arr = [6, 9, 12, 25, 50 75]
Output: [6, 12, 9, 25, 50, 75 ]
Approach: The main observation to solve this problem is that for 3 numbers a, b, and c to satisfy the condition that b shouldn’t be the mean of a and c, [a, b, c] mustn’t be sorted. Therefore, this problem can be solved by following steps:
- Iterate over the array from 1 to (N-1)
- Check whether (arr[i – 1] + arr[i + 1]) / 2 == arr[i])
- If the condition is satisfied swap the elements arr[i] and arr[i+1]
Below is the implementation of the above approach:
C++
// C++ code for above implementation#include <bits/stdc++.h>using namespace std;// Function to rearrange the arrayvoid Rearrange(int arr[], int N){ // Iterating for array for (int i = 1; i < (N - 1); i++) { // Checking whether the element i // is mean of i-1 and i+1 if ((arr[i - 1] + arr[i + 1]) / 2 == arr[i]) { // Rearrange by swapping arr[i] and arr[i+1] swap(arr[i], arr[i + 1]); } } // Printing the output array for (int i = 0; i < N; i++) { cout << arr[i] << " "; }}// Driver codeint main(){ int arr[] = { 6, 9, 12, 25, 50, 75 }; int N = sizeof(arr) / sizeof(int); // calling the function Rearrange(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to rearrange the array static void Rearrange(int arr[], int N){ // Iterating for array for (int i = 1; i < (N - 1); i++) { // Checking whether the element i // is mean of i-1 and i+1 if ((arr[i - 1] + arr[i + 1]) / 2 == arr[i]) { // Rearrange by swapping arr[i] and arr[i+1] int temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; } } // Printing the output array for (int i = 0; i < N; i++) { System.out.print(arr[i] +" "); }} // Driver code public static void main (String[] args) { int arr[] = { 6, 9, 12, 25, 50, 75 }; int N = arr.length; // calling the function Rearrange(arr, N); }}// This code is contributed by Potta Lokesh |
C#
// C# program for the above approach
using System;
class GFG {
// Function to rearrange the array
static void Rearrange(int []arr, int N)
{
// Iterating for array
for (int i = 1; i < (N – 1); i++) {
// Checking whether the element i
// is mean of i-1 and i+1
if ((arr[i – 1] + arr[i + 1]) / 2 == arr[i]) {
// Rearrange by swapping arr[i] and arr[i+1]
int temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
}
}
// Printing the output array
for (int i = 0; i < N; i++) {
Console.Write(arr[i] +” “);
}
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 6, 9, 12, 25, 50, 75 };
int N = arr.Length;
// calling the function
Rearrange(arr, N);
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python3 program for the above approach# Function to rearrange the arraydef Rearrange(arr, N) : # Iterating for array for i in range(1, N - 1) : # Checking whether the element i # is mean of i-1 and i+1 if ((arr[i - 1] + arr[i + 1]) // 2 == arr[i]) : # Rearrange by swapping arr[i] and arr[i+1] arr[i], arr[i + 1] = arr[i + 1], arr[i]; # Printing the output array for i in range(N) : print(arr[i],end= " " ) # Driver codeif __name__ == "__main__" : arr = [ 6, 9, 12, 25, 50, 75 ]; N = len(arr); # calling the function Rearrange(arr, N); # This code is contributed by AnkThon |
Javascript
<script>// JavaScript program for the above approach// Function to rearrange the arrayfunction Rearrange(arr, N){ // Iterating for array for (var i = 1; i < (N - 1); i++) { // Checking whether the element i // is mean of i-1 and i+1 if ((arr[i - 1] + arr[i + 1]) / 2 == arr[i]) { // Rearrange by swapping arr[i] and arr[i+1] var temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; } } // Printing the output array for (var i = 0; i < N; i++) { document.write(arr[i] +" "); }}// Driver codevar arr = [ 6, 9, 12, 25, 50, 75 ];var N = arr.length;// calling the functionRearrange(arr, N); // This code is contributed by shivanisinghss2110</script> |
6 12 9 25 75 50
Time Complexity: O(N)
Auxiliary Space: O(1)
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