Maximize sum of absolute difference between adjacent elements in Array with sum K

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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++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

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

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