# Check if Array can be reordered such that adjacent difference of a pair is K times of the other

• Last Updated : 31 May, 2022

Given an array arr[] of size N and a positive integer K, the task is to check if the array can be reordered such that each pair of consecutive differences differ by a factor of K i.e.,

• arr[i] − arr[i − 1] = K*(arr[i + 1] − arr[i]), or
• arr[i + 1] − arr[i] = K*(arr[i] − arr[i − 1])

Note: Different conditions can hold at different indices, the only restriction is that at each index, at least one of the given conditions must hold.

Examples:

Input: arr[] = {16, 19, 18, 21, 24, 22}, K = 2
Output: Yes
Explanation: After Sorting, arr[] = {16, 18, 19, 21, 22, 24}
For index 1, arr[i] − arr[i − 1] = K * (arr[i + 1] − arr[i]) conditions holds.
Since, arr[i] − arr[i − 1] = 2, K * (arr[i + 1] − arr[i]) = 2*1 = 2.
Similarly, for index 3, above condition hold true.
For, index 2 and 4, arr[i+1]−arr[i] = K*(arr[i]−arr[i−1]) holds.

Input: arr[] = {5, 4, 7, 6}, K = 5
Output: No

Approach: The problem can be solved based on the following idea:

As the value of K is positive, so the difference between any adjacent pair should be of only one type – either positive or negative. This type of arrangement is possible only if the array is sorted in ascending or descending order.
Therefore, check the difference between adjacent elements in sorted order to find if a arrangement exists or not.

To solve the problem based on the above idea, follow the steps mentioned below to implement the approach:

• Sort the array in increasing order.
• Run a loop from i = 1 to N-2
• If any one of the below conditions holds true then continue
•  (next – curr) == k*(curr – prev)
•  (curr – prev) == k*(next – curr)
• Otherwise, print No and return from the function.
• If the loop ends and the condition holds true for all the pairs of adjacent differences then return true.

Below is the implementation of the above approach:

## C++

 `// C++ code to implement the above approach` `#include ``using` `namespace` `std;` `// Function to check the conditions``string checkArray(``int``* arr, ``int` `n, ``int` `k)``{``    ``// Sort the array in increasing order``    ``sort(arr, arr + n);` `    ``// Run a loop from index 1 to N -2``    ``for` `(``int` `i = 1; i <= n - 2; i++) {` `        ``// Store previous element in prev``        ``int` `prev = arr[i - 1];` `        ``// Store current element in curr``        ``int` `curr = arr[i];` `        ``// Store next element in next``        ``int` `next = arr[i + 1];` `        ``// If any conditions holds true``        ``// then continue``        ``if` `(((next - curr) == k * (curr - prev))``            ``|| ((curr - prev) == k * (next - curr))) {``            ``continue``;``        ``}` `        ``// Else print No and return``        ``else` `{``            ``return` `"No"``;``        ``}``    ``}` `    ``// We reach here only if the condition is valid``    ``// for all index except 0 and N-1``    ``return` `"Yes"``;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 16, 19, 18, 21, 24, 22 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `K = 2;` `    ``// Function call``    ``cout << checkArray(arr, N, K);``    ``return` `0;``}`

## Java

 `// Java code to implement the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ``// Function to check the conditions``    ``public` `static` `String checkArray(``int` `arr[], ``int` `n, ``int` `k)``    ``{``        ``// Sort the array in increasing order``        ``Arrays.sort(arr);` `        ``// Run a loop from index 1 to N -2``        ``for` `(``int` `i = ``1``; i <= n - ``2``; i++) {` `            ``// Store previous element in prev``            ``int` `prev = arr[i - ``1``];` `            ``// Store current element in curr``            ``int` `curr = arr[i];` `            ``// Store next element in next``            ``int` `next = arr[i + ``1``];` `            ``// If any conditions holds true``            ``// then continue``            ``if` `(((next - curr) == k * (curr - prev))``                ``|| ((curr - prev) == k * (next - curr))) {``                ``continue``;``            ``}` `            ``// Else print No and return``            ``else` `{``                ``return` `"No"``;``            ``}``        ``}` `        ``// We reach here only if the condition is valid``        ``// for all index except 0 and N-1``        ``return` `"Yes"``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``16``, ``19``, ``18``, ``21``, ``24``, ``22` `};``        ``int` `N = arr.length;``        ``int` `K = ``2``;` `        ``// Function call``        ``System.out.print(checkArray(arr, N, K));``    ``}``}` `// This code is contributed by Rohit Pradhan`

## Python3

 `# Python code to implement the approach` `# Function to check the conditions``def` `checkArray(arr, n, k):` `    ``# Sort the array in increasing order``    ``arr.sort();` `    ``# Run a loop from index 1 to N -2``    ``for` `i ``in` `range``(``1``, n ``-` `1``):` `        ``# Store previous element in prev``        ``prev ``=` `arr[i ``-` `1``];` `        ``# Store current element in curr``        ``curr ``=` `arr[i];` `        ``# Store next element in next``        ``next` `=` `arr[i ``+` `1``];` `        ``# If any conditions holds true``        ``# then continue``        ``if` `(((``next` `-` `curr) ``=``=` `k ``*` `(curr ``-` `prev))``            ``or` `((curr ``-` `prev) ``=``=` `k ``*` `(``next` `-` `curr))):``            ``continue``        ` `        ``# Else print No and return``        ``else``:``            ``return` `"No"``;``        ` `    ``# We reach here only if the condition is valid``    ``# for all index except 0 and N-1``    ``return` `"Yes"``;` `# Driver Code``arr ``=` `[ ``16``, ``19``, ``18``, ``21``, ``24``, ``22` `];``N ``=` `len``(arr)``K ``=` `2``;` `# Function call``print``(checkArray(arr, N, K));` `# This code is contributed by Saurabh Jaiswal`

## C#

 `// C# code to implement the above approach``using` `System;``public` `class` `GFG``{` `  ``// Function to check the conditions``  ``public` `static` `string` `checkArray(``int``[] arr, ``int` `n, ``int` `k)``  ``{` `    ``// Sort the array in increasing order``    ``Array.Sort(arr);` `    ``// Run a loop from index 1 to N -2``    ``for` `(``int` `i = 1; i <= n - 2; i++) {` `      ``// Store previous element in prev``      ``int` `prev = arr[i - 1];` `      ``// Store current element in curr``      ``int` `curr = arr[i];` `      ``// Store next element in next``      ``int` `next = arr[i + 1];` `      ``// If any conditions holds true``      ``// then continue``      ``if` `(((next - curr) == k * (curr - prev))``          ``|| ((curr - prev) == k * (next - curr))) {``        ``continue``;``      ``}` `      ``// Else print No and return``      ``else` `{``        ``return` `"No"``;``      ``}``    ``}` `    ``// We reach here only if the condition is valid``    ``// for all index except 0 and N-1``    ``return` `"Yes"``;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int``[] arr = { 16, 19, 18, 21, 24, 22 };``    ``int` `N = arr.Length;``    ``int` `K = 2;` `    ``// Function call``    ``Console.WriteLine(checkArray(arr, N, K));``  ``}``}` `// this code is a contributed by phasing17`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(N)
Auxiliary Space: O(1)

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