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 any one of the below conditions holds true then continue
- 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++ code to implement the above approach #include <bits/stdc++.h> 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[0]);
int K = 2;
// Function call
cout << checkArray(arr, N, K);
return 0;
} |
// 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 |
# 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# 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 |
<script> // JavaScript code to implement the approach // Function to check the conditions function checkArray(arr, n, k)
{ // Sort the array in increasing order
arr.sort();
// Run a loop from index 1 to N -2
for (let i = 1; i <= n - 2; i++) {
// Store previous element in prev
let prev = arr[i - 1];
// Store current element in curr
let curr = arr[i];
// Store next element in next
let 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 let arr = [ 16, 19, 18, 21, 24, 22 ]; let N = arr.length; let K = 2; // Function call document.write(checkArray(arr, N, K), "</br>" );
// This code is contributed by shinjanpatra </script> |
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)