# Find maximum sum of triplets in an array such than i < j < k and a[i] < a[j] < a[k] in C++

C++Server Side ProgrammingProgramming

## Concept

With respect of a given array of positive integers of size n, our task to determine the maximum sum of triplet ( ai + aj + ak ) such that 0 <= i < j < k < n and ai< aj< ak.

## Input

a[] = 3 6 4 2 5 10

## Output

19

## Explanation

All possible triplets are:-
3 4 5 => sum = 12
3 6 10 => sum = 19
3 4 10 => sum = 17
4 5 10 => sum = 19
2 5 10 => sum = 17
Maximum sum = 19

## Method

Now, a simple approach is to visit for every triplet with three nested ‘for loops’ and determine update the sum of all triplets one by one. Here, time complexity of this method is O(n^3) which is not enough for higher value of ‘n’.

Again, we can apply a better approach for making further optimization in above approach. In this method, instead of visiting through every triplet with three nested loops, we can visit through two nested loops.

At the time of visiting through each number(let as middle element(aj )), determine maximum number(ai) less than aj preceding it and maximum number(ak ) larger than aj beyond it. Finally, now, update the maximum answer with calculated sum of ai + aj + ak

## Example

Live Demo

// C++ program to find maximum triplet sum
#include <bits/stdc++.h>
using namespace std;
// Shows function to calculate maximum triplet sum
int maxTripletSum(int arr1[], int n1){
// Used to initialize the answer
int ans1 = 0;
for (int i = 1; i < n1 - 1; ++i) {
int max1 = 0, max2 = 0;
// Determine maximum value(less than arr1[i])
// from i+1 to n1-1
for (int j = 0; j < i; ++j)
if (arr1[j] < arr1[i])
max1 = max(max1, arr1[j]);
// Determine maximum value(greater than arr1[i])
// from i+1 to n1-1
for (int j = i + 1; j < n1; ++j)
if (arr1[j] > arr1[i])
max2 = max(max2, arr1[j]);
if(max1 && max2)
ans1=max(ans1,max1+arr1[i]+max2);
}
return ans1;
}
// Driver code
int main(){
int Arr[] = { 3, 6, 4, 2, 5, 10 };
int N = sizeof(Arr) / sizeof(Arr[0]);
cout << maxTripletSum(Arr, N);
return 0;
}

## Output

19
Published on 25-Jul-2020 08:57:53