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 = 0 // 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] // 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 implementation of above idea.
// 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 = 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];
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 program 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 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 = 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];
} 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 program 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 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 = 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]
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 program 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# 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 = 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];
} 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 program 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 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 = 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 ];
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 ?> |
Output:
Maximum sum : 6
Time complexity : O(n)
Auxiliary Space : O(1)
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