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Find Maximum Sum Strictly Increasing Subarray

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Given an array of positive integers. Find the maximum sum of strictly increasing subarrays. Note that this problem is different from maximum subarray sum and maximum sum increasing subsequence problems.

Examples: 

Input  : arr[] = {1, 2, 3, 2, 5, 1, 7}
Output : 8
Explanation :  Some Strictly increasing subarrays are 
{1, 2, 3} sum = 6, 
{2, 5} sum = 7, 
{1, 7} sum 8 
Maximum Sum = 8 

Input : arr[] = {1, 2, 2, 4}
Output: 6
Explanation : Increasing subarray with maximum sum is 6.

A Simple Solution is to generate all possible subarrays, and for every subarray check if subarray is strictly increasing or not. If subarray is strictly increasing, then we calculate sum & update max_sum. Time complexity O(n2).

An efficient solution of this problem take O(n) time. The idea is keep track of maximum sum and current sum. For every element arr[i], if it is greater than arr[i-1], then we add it to current sum. Else arr[i] is starting point of another potential candidate for maximum sum increasing subarray, so we update current sum as array. But before updating current sum, we update maximum sum if required.

Let input array be 'arr[]' and size of array be 'n'

Initialize : 
max_sum = arr[0]
  // because if array size is 1 than it
  // would return that element.
  // used to store the maximum sum 
current_sum = arr[0] // used to compute current sum 

// Traverse array starting from second element
i goes from 1 to n-1

    // Check if it is strictly increasing then we 
    // update current_sum.
    current_sum = current_sum + arr[i]
    max_sum = max(max_sum, current_sum)
    // Also needed for subarray having last element.
    // else strictly increasing subarray breaks and 
    // arr[i] is starting point of next potential
    // subarray
    max_sum = max(max_sum, current_sum)
    current_sum = arr[i]

return max(max_sum, current_sum)    

Below is the implementation of the above idea. 

C++




// C/C++ program to find the maximum sum of strictly
// increasing subarrays
#include <iostream>
using namespace std;
 
// Returns maximum sum of strictly increasing
// subarrays
int maxsum_SIS(int arr[], int n)
{
    // Initialize max_sum be 0
    int max_sum = arr[0];
 
    // Initialize current sum be arr[0]
    int current_sum = arr[0];
 
    // Traverse array elements after first element.
    for (int i = 1; i < n; i++)
    {
        // update current_sum for
        // strictly increasing subarray
        if (arr[i - 1] < arr[i])
        {
            current_sum = current_sum + arr[i];
            max_sum = max(max_sum, current_sum);
        }
 
        else // strictly increasing subarray break
        {
            // update max_sum and current_sum ;
            max_sum = max(max_sum, current_sum);
            current_sum = arr[i];
        }
    }
 
    return max(max_sum, current_sum);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Maximum sum : " << maxsum_SIS(arr, n);
    return 0;
}


Java




// Java program to find the
// maximum sum of strictly increasing subarrays
 
public class GFG {
 
    // Returns maximum sum
    // of strictly increasing subarrays
    static int maxsum_SIS(int arr[], int n)
    {
        // Initialize max_sum be 0
        int max_sum = arr[0];
 
        // Initialize current sum be arr[0]
        int current_sum = arr[0];
 
        // Traverse array elements after first element.
        for (int i = 1; i < n; i++)
        {
            // update current_sum
            // for strictly increasing subarray
            if (arr[i - 1] < arr[i])
            {
                current_sum = current_sum + arr[i];
                max_sum = Math.max(max_sum, current_sum);
            }
            else // strictly increasing subarray break
            {
                // update max_sum and current_sum ;
                max_sum = Math.max(max_sum, current_sum);
                current_sum = arr[i];
            }
        }
 
        return Math.max(max_sum, current_sum);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 2, 4 };
        int n = arr.length;
        System.out.println("Maximum sum : "
                           + maxsum_SIS(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar


Python3




# Python3 program to find the maximum sum of strictly
# increasing subarrays
 
# Returns maximum sum of strictly increasing
# subarrays
 
 
def maxsum_SIS(arr, n):
    # Initialize max_sum be 0
    max_sum = arr[0]
 
    # Initialize current sum be arr[0]
    current_sum = arr[0]
 
    # Traverse array elements after first element.
    for i in range(1, n):
        # update current_sum for strictly increasing subarray
        if (arr[i-1] < arr[i]):
            current_sum = current_sum + arr[i]
            max_sum = max(max_sum, current_sum)
 
        else:
            # strictly increasing subarray break
            # update max_sum and current_sum
            max_sum = max(max_sum, current_sum)
            current_sum = arr[i]
 
    return max(max_sum, current_sum)
 
# Driver code
 
def main():
    arr = [1, 2, 2, 4]
    n = len(arr)
 
    print("Maximum sum : ", maxsum_SIS(arr, n)),
 
 
if __name__ == '__main__':
    main()
 
# This code is contributed by 29AjayKumar


C#




// C# program to find the maximum sum of strictly
// increasing subarrays
using System;
public class GFG {
 
    // Returns maximum sum of strictly increasing
    // subarrays
    static int maxsum_SIS(int[] arr, int n)
    {
        // Initialize max_sum be 0
        int max_sum = arr[0];
 
        // Initialize current sum be arr[0]
        int current_sum = arr[0];
 
        // Traverse array elements after first element.
        for (int i = 1; i < n; i++) {
            // update current_sum for strictly increasing
            // subarray
            if (arr[i - 1] < arr[i]) {
                current_sum = current_sum + arr[i];
                max_sum = Math.Max(max_sum, current_sum);
            }
            else // strictly increasing subarray break
            {
                // update max_sum and current_sum ;
                max_sum = Math.Max(max_sum, current_sum);
                current_sum = arr[i];
            }
        }
 
        return Math.Max(max_sum, current_sum);
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 2, 4 };
        int n = arr.Length;
        Console.WriteLine("Maximum sum : "
                          + maxsum_SIS(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar


PHP




<?php
// PHP program to find the maximum sum of
// strictly increasing subarrays
 
// Returns maximum sum of strictly
// increasing subarrays
function maxsum_SIS($arr , $n)
{
    // Initialize max_sum be 0
    $max_sum = $arr[0];
 
    // Initialize current sum be arr[0]
    $current_sum = $arr[0];
 
    // Traverse array elements after
    // first element.
    for ($i = 1; $i < $n ; $i++)
    {
        // update current_sum for strictly
        // increasing subarray
        if ($arr[$i - 1] < $arr[$i]){
            $current_sum = $current_sum + $arr[$i];
            $max_sum = max($max_sum, $current_sum);
    }
 
        else // strictly increasing
             // subarray break
        {
            // update max_sum and current_sum ;
            $max_sum = max($max_sum, $current_sum);
            $current_sum = $arr[$i];
        }
    }
 
    return max($max_sum, $current_sum);
}
 
// Driver Code
$arr = array(1, 2, 2, 4);
$n = sizeof($arr);
 
echo "Maximum sum : ",
      maxsum_SIS($arr , $n);
 
// This code is contributed by Sachin
?>


Javascript




<script>
 
// Javascript program to find the maximum sum of strictly
// increasing subarrays
 
// Returns maximum sum of strictly increasing
// subarrays
function maxsum_SIS(arr, n)
{
    // Initialize max_sum be 0
    var max_sum = arr[0];
 
    // Initialize current sum be arr[0]
    var current_sum = arr[0];
 
    // Traverse array elements after first element.
    for (var i = 1; i < n; i++)
    {
        // update current_sum for
        // strictly increasing subarray
        if (arr[i - 1] < arr[i])
        {
            current_sum = current_sum + arr[i];
            max_sum = Math.max(max_sum, current_sum);
        }
 
        else // strictly increasing subarray break
        {
            // update max_sum and current_sum ;
            max_sum = Math.max(max_sum, current_sum);
            current_sum = arr[i];
        }
    }
 
    return Math.max(max_sum, current_sum);
}
 
// Driver code
var arr = [ 1, 2, 2, 4 ];
var n = arr.length;
document.write( "Maximum sum : " + maxsum_SIS(arr, n));
 
// This code is contributed by itsok.
</script>


Output

Maximum sum : 6

Time complexity : O(n) 
Auxiliary Space : O(1)

 



Last Updated : 06 Jul, 2022
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