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Maximize the value of (A[i]-A[j])*A[k] for any ordered triplet of indices i, j and k

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Given an array A consisting of N positive integers and for any ordered triplet( i, j, k) such that i, j, and k are all distinct and 0 ≤ i, j, k < N, the value of this triplet is (Ai − Aj)⋅Ak, the task is to find maximum value among all distinct ordered triplets.

Note: Two triplets (a, b, c) and (d, e, f) are considered to be different if at least one of the conditions is satisfied such that a ≠ d or b ≠ e or c ≠ f. As an example, (1, 2, 3) and (2, 3, 1) are two different ordered triplets.


Input: A[] =  {1, 1, 3}
Output: 2

Explanation: The desired triplet is (2, 1, 0), which yields the value of (Ai−Aj)⋅Ak = (3 − 1)⋅1 = 2.

Input: A[] =  {3, 4, 4, 1, 2}
Output: 12

Approach: The problem can be solved based on the following observation:

Sort the array A in increasing order. Since we want to maximize the value of (Ai − Aj).Ak, and all elements in A are positive, it is best to maximise both (Ai − Aj) and Ak. There are two options:

  • Select largest and smallest element in A as Ai and Aj, and choose second maximum element in A as Ak. The value is (AN-1 − A0).AN−2
  • Choose the maximum element as Ak and choose the second maximum element, and the minimum element as Ai and Aj, getting a triplet of value (AN-2 − A0).AN-1

Since AN – 2 ≤ AN-1, we can prove that  (AN-1 − A0).AN−2 ≥ (AN-2 − A0).AN-1. Hence, the maximum value we can obtain is  
(AN-1 − A0).AN−2 

Follow the below steps to solve the problem:

  • Sort the array in ascending order.
  • Find the difference between the maximum and minimum element of the sorted array
  • Then multiply the difference with the second maximum element of the sorted array to get the answer.

Below is the implementation of the above approach.




import java.util.*;


public class GFG {





    public static long maxValue(int a[], int n)



        int min = a[0];

        int max = a[n - 1];

        int secmax = a[n - 2];

        long ans = (long)(a[n - 1] - a[0]) * a[n - 2];

        return ans;




    public static void main(String[] args)


        int A[] = { 1, 1, 3 };

        int N = A.length;


        System.out.println(maxValue(A, N));



Time Complexity: O(N * log(N)) 
Auxiliary Space: O(1)


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