Given two integers N and K, the task is to maximize the sum of absolute differences between adjacent elements of an array of length N and sum K.
Examples:
Input: N = 5, K = 10
Output: 20
Explanation:
The array arr[] with sum 10 can be {0, 5, 0, 5, 0}, maximizing the sum of absolute difference of adjacent elements ( 5 + 5 + 5 + 5 = 20)Input: N = 2, K = 10
Output: 10
Approach:
To maximize the sum of adjacent elements, follow the steps below:
- If N is 2, the maximum sum possible is K by placing K in 1 index and 0 on the other.
- If N is 1, the maximum sum possible will always be 0.
- For all other values of N, the answer will be 2 * K.
Illustration:
For N = 3, the arrangement {0, K, 0} maximizes the sum of absolute difference between adjacent elements to 2 * K.
For N = 4, the arrangement {0, K/2, 0, K/2} or {0, K, 0, 0} maximizes the required sum of absolute difference between adjacent elements to 2 * K.
Below is the implementation of the above approach:
// C++ program to maximize the // sum of absolute differences // between adjacent elements #include <bits/stdc++.h> using namespace std;
// Function for maximizing the sum int maxAdjacentDifference( int N, int K)
{ // Difference is 0 when only
// one element is present
// in array
if (N == 1) {
return 0;
}
// Difference is K when
// two elements are
// present in array
if (N == 2) {
return K;
}
// Otherwise
return 2 * K;
} // Driver code int main()
{ int N = 6;
int K = 11;
cout << maxAdjacentDifference(N, K);
return 0;
} |
// Java program to maximize the // sum of absolute differences // between adjacent elements import java.util.*;
class GFG{
// Function for maximising the sum static int maxAdjacentDifference( int N, int K)
{ // Difference is 0 when only
// one element is present
// in array
if (N == 1 )
{
return 0 ;
}
// Difference is K when
// two elements are
// present in array
if (N == 2 )
{
return K;
}
// Otherwise
return 2 * K;
} // Driver code public static void main(String[] args)
{ int N = 6 ;
int K = 11 ;
System.out.print(maxAdjacentDifference(N, K));
} } // This code is contributed by 29AjayKumar |
# Python3 program to maximize the # sum of absolute differences # between adjacent elements # Function for maximising the sum def maxAdjacentDifference(N, K):
# Difference is 0 when only
# one element is present
# in array
if (N = = 1 ):
return 0 ;
# Difference is K when
# two elements are
# present in array
if (N = = 2 ):
return K;
# Otherwise
return 2 * K;
# Driver code N = 6 ;
K = 11 ;
print (maxAdjacentDifference(N, K));
# This code is contributed by Code_Mech |
// C# program to maximize the // sum of absolute differences // between adjacent elements using System;
class GFG{
// Function for maximising the sum static int maxAdjacentDifference( int N, int K)
{ // Difference is 0 when only
// one element is present
// in array
if (N == 1)
{
return 0;
}
// Difference is K when
// two elements are
// present in array
if (N == 2)
{
return K;
}
// Otherwise
return 2 * K;
} // Driver code public static void Main(String[] args)
{ int N = 6;
int K = 11;
Console.Write(maxAdjacentDifference(N, K));
} } // This code is contributed by 29AjayKumar |
<script> // JavaScript program to maximize the // sum of absolute differences // between adjacent elements // Function for maximising the sum function maxAdjacentDifference(N, K)
{ // Difference is 0 when only
// one element is present
// in array
if (N == 1)
{
return 0;
}
// Difference is K when
// two elements are
// present in array
if (N == 2)
{
return K;
}
// Otherwise
return 2 * K;
} // Driver Code let N = 6;
let K = 11;
document.write(maxAdjacentDifference(N, K));
// This code is contributed by susmitakundugoaldanga. </script> |
22
Time Complexity: O(1).
Auxiliary Space: O(1)