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

Given two integers N and K, the task is to maximise 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++

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// C++ program to maximize the
// sum of absolute differences
// between adjacent elements
#include <bits/stdc++.h>
using namespace std;
  
// Function for maximising 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

22

Time Complexity: O(1).

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Improved By : 29AjayKumar, Code_Mech