Given an array arr[] of size N, the task is to find 4 indices i, j, k, l such that 0 <= i, j, k, l < N and the value of arr[i]%arr[j] – arr[k]%arr[l] is maximum. Print the maximum difference. If it doesn’t exist, then print -1.
Examples:
Input: N=8, arr[] = {1, 2, 4, 6, 8, 3, 5, 7}
Output: 7
Explanation: Choosing elements 1, 2, 7, 8 and 2%1 – 7%8 gives the maximum result possible.Input: N=3, arr[] = {1, 50, 101}
Output: -1
Explanation: Since, there are 3 elements only so there’s no possible answer.
Naive Approach: The brute force idea would be to check all the possible combinations and then find the maximum difference.
Time Complexity: O(N4)
Auxiliary Space: O(1)
Efficient Approach: The idea is based on the observation that on sorting the array in ascending order, choose the first pair from the left-side, i.e, the minimum 2 values and the second pair from the right side, i.e, the maximum 2 values gives the answer. Further, arr[i+1]%arr[i] is always less than equal to arr[i]%arr[i+1]. So, minimize the first pair value and maximize the second pair value. Follow the steps below to solve the problem:
- If the size of the array is less than 4, then return -1.
- Sort the array arr[] in ascending order.
- Initialize the variable first as arr[1]%arr[0] and second as arr[N-2]%arr[N-1].
- After performing the above steps, print the value of second-first as the answer.
Below is the implementation of the above approach.
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the required // maximum difference void maxProductDifference(vector< int >& arr)
{ // Base Case
if (arr.size() < 4) {
cout << "-1\n" ;
return ;
}
// Sort the array
sort(arr.begin(), arr.end());
// First pair
int first = arr[1] % arr[0];
// Second pair
int second = arr[arr.size() - 2]
% arr[arr.size() - 1];
// Print the result
cout << second - first;
return ;
} // Driver Code int main()
{ vector< int > arr = { 1, 2, 4, 6, 8, 3, 5, 7 };
maxProductDifference(arr);
return 0;
} |
// Java program for the above approach import java.io.*;
import java.util.Arrays;
class GFG
{ // Function to find the required
// maximum difference
static void maxProductDifference( int [] arr)
{
// Base Case
if (arr.length < 4 ) {
System.out.println( "-1" );
return ;
}
// Sort the array
Arrays.sort(arr);
// First pair
int first = arr[ 1 ] % arr[ 0 ];
// Second pair
int second
= arr[arr.length - 2 ] % arr[arr.length - 1 ];
// Print the result
System.out.println(second - first);
return ;
}
// Driver Code
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 4 , 6 , 8 , 3 , 5 , 7 };
maxProductDifference(arr);
}
} // This code is contributed by Potta Lokesh |
# python program for the above approach # Function to find the required # maximum difference def maxProductDifference(arr):
# Base Case
if ( len (arr) < 4 ):
print ( "-1" )
return
# Sort the array
arr.sort()
# First pair
first = arr[ 1 ] % arr[ 0 ]
# Second pair
second = arr[ len (arr) - 2 ] % arr[ len (arr) - 1 ]
# Print the result
print (second - first)
# Driver Code if __name__ = = "__main__" :
arr = [ 1 , 2 , 4 , 6 , 8 , 3 , 5 , 7 ]
maxProductDifference(arr)
# This code is contributed by rakeshsahni
|
// C# program for the above approach using System;
public class GFG
{ // Function to find the required
// maximum difference
static void maxProductDifference( int [] arr)
{
// Base Case
if (arr.Length < 4) {
Console.WriteLine( "-1" );
return ;
}
// Sort the array
Array.Sort(arr);
// First pair
int first = arr[1] % arr[0];
// Second pair
int second
= arr[arr.Length - 2] % arr[arr.Length - 1];
// Print the result
Console.WriteLine(second - first);
return ;
}
// Driver Code
public static void Main(String[] args)
{
int [] arr = { 1, 2, 4, 6, 8, 3, 5, 7 };
maxProductDifference(arr);
}
} // This code is contributed by shikhasingrajput |
<script> // Javascript program for the above approach // Function to find the required // maximum difference function maxProductDifference(arr)
{ // Base Case
if (arr.length < 4) {
document.write( "-1<br>" );
return ;
}
// Sort the array
arr.sort();
// First pair
let first = arr[1] % arr[0];
// Second pair
let second = arr[arr.length - 2] % arr[arr.length - 1];
// Print the result
document.write(second - first);
return ;
} // Driver Code let arr = [1, 2, 4, 6, 8, 3, 5, 7]; maxProductDifference(arr); // This code is contributed by gfgking. </script> |
7
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)