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

Given an array of positive integers of size n. Find the maximum sum of triplet( ai + aj + ak ) such that 0 <= i < j < k < n and ai < aj < ak.

Input: a[] = 2 5 3 1 4 9
Output: 16

Explanation:
All possible triplets are:-
2 3 4 => sum = 9
2 5 9 => sum = 16
2 3 9 => sum = 14
3 4 9 => sum = 16
1 4 9 => sum = 14
Maximum sum = 16


Simple Approach is to traverse for every triplet with three nested ‘for loops’ and find update the sum of all triplets one by one. Time complexity of this approach is O(n3) which is not sufficient for larger value of ‘n’.

Better approach is to make further optimization in above approach. Instead of traversing through every triplets with three nested loops, we can traverse through two nested loops. While traversing through each number(assume as middle element(aj)), find maximum number(ai) smaller than aj preceding it and maximum number(ak) greater than aj beyond it. Now after that, update the maximum answer with calculated sum of ai + aj + ak

CPP

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// C++ program to find maximum triplet sum
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate maximum triplet sum
int maxTripletSum(int arr[], int n)
{
    // Initialize the answer
    int ans = 0;
  
    for (int i = 1; i < n - 1; ++i) {
        int max1 = 0, max2 = 0;
  
        // find maximum value(less than arr[i])
        // from i+1 to n-1
        for (int j = 0; j < i; ++j)
            if (arr[j] < arr[i])
                max1 = max(max1, arr[j]);
  
        // find maximum value(greater than arr[i])
        // from i+1 to n-1
        for (int j = i + 1; j < n; ++j)
            if (arr[j] > arr[i])
                max2 = max(max2, arr[j]);
  
        // store maximum answer
        ans = max(ans, max1 + arr[i] + max2);
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 5, 3, 1, 4, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxTripletSum(arr, n);
    return 0;
}

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Java

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// Java program to find maximum triplet sum
  
import java.io.*;
import java.math.*;
  
class GFG {
  
    // Function to calculate maximum triplet sum
    static int maxTripletSum(int arr[], int n)
    {
        // Initialize the answer
        int ans = 0;
  
        for (int i = 1; i < n - 1; ++i) {
            int max1 = 0, max2 = 0;
  
            // find maximum value(less than arr[i])
            // from i+1 to n-1
            for (int j = 0; j < i; ++j)
                if (arr[j] < arr[i])
                    max1 = Math.max(max1, arr[j]);
  
            // find maximum value(greater than arr[i])
            // from i+1 to n-1
            for (int j = i + 1; j < n; ++j)
                if (arr[j] > arr[i])
                    max2 = Math.max(max2, arr[j]);
  
            // store maximum answer
            ans = Math.max(ans, max1 + arr[i] + max2);
        }
  
        return ans;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 2, 5, 3, 1, 4, 9 };
        int n = arr.length;
        System.out.println(maxTripletSum(arr, n));
    }
}
  
// This code is contributed by Nikita Tiwari.

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Python3

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# Python 3 program to
# find maximum triplet sum
  
# Function to calculate
# maximum triplet sum
def maxTripletSum(arr, n) :
  
    # Initialize the answer
    ans = 0
   
    for i in range(1, (n - 1)) :
        max1 = 0
        max2 = 0
   
        # find maximum value(less than arr[i])
        # from i + 1 to n-1
        for j in range(0, i) :
            if (arr[j] < arr[i]) :
                max1 = max(max1, arr[j])
   
        # find maximum value(greater than arr[i])
        # from i + 1 to n-1
        for j in range((i + 1), n) :
            if (arr[j] > arr[i]) :
                max2 = max(max2, arr[j])
   
        # store maximum answer
        ans = max(ans, max1 + arr[i] + max2)
   
    return ans
  
  
# Driver code
  
arr = [ 2, 5, 3, 1, 4, 9 ]
n = len(arr)
print(maxTripletSum(arr, n))
  
  
# This code is contributed
# by Nikita Tiwari.

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C#

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// C# program to find maximum triplet sum
using System;
  
class GFG {
  
    // Function to calculate maximum triplet sum
    static int maxTripletSum(int[] arr, int n)
    {
          
        // Initialize the answer
        int ans = 0;
  
        for (int i = 1; i < n - 1; ++i)
        {
            int max1 = 0, max2 = 0;
  
            // find maximum value(less than 
            // arr[i]) from i+1 to n-1
            for (int j = 0; j < i; ++j)
                if (arr[j] < arr[i])
                    max1 = Math.Max(max1, arr[j]);
  
            // find maximum value(greater than
            // arr[i]) from i+1 to n-1
            for (int j = i + 1; j < n; ++j)
                if (arr[j] > arr[i])
                    max2 = Math.Max(max2, arr[j]);
  
            // store maximum answer
            ans = Math.Max(ans, max1 + arr[i] + max2);
        }
  
        return ans;
    }
  
    // Driver code
    public static void Main()
    {
          
        int[] arr = { 2, 5, 3, 1, 4, 9 };
        int n = arr.Length;
          
        Console.WriteLine(maxTripletSum(arr, n));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find maximum triplet sum
  
// Function to calculate maximum triplet sum
function maxTripletSum($arr, $n)
{
      
    // Initialize the answer
    $ans = 0;
  
    for ($i = 1; $i < $n - 1; ++$i
    {
        $max1 = 0; $max2 = 0;
  
        // find maximum value(less than
        // arr[i]) from i+1 to n-1
        for ($j = 0; $j < $i; ++$j)
            if ($arr[$j] < $arr[$i])
                $max1 = max($max1, $arr[$j]);
  
        // find maximum value(greater than
        // arr[i]) from i+1 to n-1
        for ($j = $i + 1; $j < $n; ++$j)
            if ($arr[$j] > $arr[$i])
                $max2 = max($max2, $arr[$j]);
  
        // store maximum answer
        $ans = max($ans, $max1 + $arr[$i] + $max2);
    }
  
    return $ans;
}
  
    // Driver code
    $arr = array(2, 5, 3, 1, 4, 9);
    $n = sizeof($arr);
    echo maxTripletSum($arr, $n);
  
// This code is contributed by nitin mittal.
?>

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Output :

16

Time complexity: O(n2)
Auxiliary space: O(1)

Best and efficient approach is use the concept of maximum suffix-array and binary search.

  • For finding maximum number greater number greater than given number beyond it, we can maintain a maximum suffix-array array such that for any number(suffixi) it would contain maximum number from index i, i+1, … n-1. Suffix array can be calculated in O(n) time.
  • For finding maximum number smaller than the given number preceding it, we can maintain a sorted list of numbers before a given number such we can simply perform a binary search to find a number which is just smaller than the given number. In C++ language, we can perform this by using set associative container of STL library.
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// C++ program to find maximum triplet sum
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to get the lower last min
// value of 'n'
int getLowValue(set<int>& lowValue, int& n)
{
    auto it = lowValue.lower_bound(n);
  
    // Since 'lower_bound' returns the first
    // iterator greater than 'n', thus we
    // have to decrement the pointer for
    // getting the minimum value
    --it;
  
    return (*it);
}
  
// Function to calculate maximum triplet sum
int maxTripletSum(int arr[], int n)
{
    // Initialize suffix-array
    int maxSuffArr[n + 1];
  
    // Set the last element
    maxSuffArr[n] = 0;
  
    // Calculate suffix-array containing maximum
    // value for index i, i+1, i+2, ... n-1 in
    // arr[i]
    for (int i = n - 1; i >= 0; --i)
        maxSuffArr[i] = max(maxSuffArr[i + 1], arr[i]);
  
    int ans = 0;
  
    // Initialize set container
    set<int> lowValue;
  
    // Insert minimum value for first comparison
    // in the set
    lowValue.insert(INT_MIN);
  
    for (int i = 0; i < n - 1; ++i) {
        if (maxSuffArr[i + 1] > arr[i]) {
            ans = max(ans, getLowValue(lowValue,
                                       arr[i])
                               + arr[i] + maxSuffArr[i + 1]);
  
            // Insert arr[i] in set<> for further
            // processing
            lowValue.insert(arr[i]);
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 5, 3, 1, 4, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << maxTripletSum(arr, n);
    return 0;
}

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Output:
16

Time complexity: O(n*log(n))
Auxiliary space: O(n)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeksorg. See your article appearing on the GeeksforGeeks main page and help other Geeks



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Improved By : vt_m, nitin mittal



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