An array is given, find length of the subarray having maximum sum.
Examples :
Input : a[] = {1, -2, 1, 1, -2, 1}
Output : Length of the subarray is 2
Explanation: Subarray with consecutive elements
and maximum sum will be {1, 1}. So length is 2
Input : ar[] = { -2, -3, 4, -1, -2, 1, 5, -3 }
Output : Length of the subarray is 5
Explanation: Subarray with consecutive elements
and maximum sum will be {4, -1, -2, 1, 5}.
This problem is mainly a variation of Largest Sum Contiguous Subarray Problem.
The idea is to update starting index whenever sum ending here becomes less than 0.
C++
// C++ program to print length of the largest // contiguous array sum #include<bits/stdc++.h> using namespace std; int maxSubArraySum(int a[], int size) { int max_so_far = INT_MIN, max_ending_here = 0, start =0, end = 0, s=0; for (int i=0; i< size; i++ ) { max_ending_here += a[i]; if (max_so_far < max_ending_here) { max_so_far = max_ending_here; start = s; end = i; } if (max_ending_here < 0) { max_ending_here = 0; s = i + 1; } } return (end - start + 1); } /*Driver program to test maxSubArraySum*/int main() { int a[] = {-2, -3, 4, -1, -2, 1, 5, -3}; int n = sizeof(a)/sizeof(a[0]); cout << maxSubArraySum(a, n); return 0; } |
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Java
// Java program to print length of the largest // contiguous array sum class GFG { static int maxSubArraySum(int a[], int size) { int max_so_far = Integer.MIN_VALUE, max_ending_here = 0,start = 0, end = 0, s = 0; for (int i = 0; i < size; i++) { max_ending_here += a[i]; if (max_so_far < max_ending_here) { max_so_far = max_ending_here; start = s; end = i; } if (max_ending_here < 0) { max_ending_here = 0; s = i + 1; } } return (end - start + 1); } // Driver code public static void main(String[] args) { int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 }; int n = a.length; System.out.println(maxSubArraySum(a, n)); } } |
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Python3
# Python program to print largest contiguous array sum from sys import maxsize # Function to find the maximum contiguous subarray # and print its starting and end index def maxSubArraySum(a,size): max_so_far = -maxsize - 1 max_ending_here = 0 start = 0 end = 0 s = 0 for i in range(0,size): max_ending_here += a[i] if max_so_far < max_ending_here: max_so_far = max_ending_here start = s end = i if max_ending_here < 0: max_ending_here = 0 s = i+1 return (end - start + 1) # Driver program to test maxSubArraySum a = [-2, -3, 4, -1, -2, 1, 5, -3] print(maxSubArraySum(a,len(a))) |
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C#
// C# program to print length of the // largest contiguous array sum using System; class GFG { // Function to find maximum subarray sum static int maxSubArraySum(int []a, int size) { int max_so_far = int.MinValue, max_ending_here = 0,start = 0, end = 0, s = 0; for (int i = 0; i < size; i++) { max_ending_here += a[i]; if (max_so_far < max_ending_here) { max_so_far = max_ending_here; start = s; end = i; } if (max_ending_here < 0) { max_ending_here = 0; s = i + 1; } } return (end - start + 1); } // Driver code public static void Main(String[] args) { int []a = {-2, -3, 4, -1, -2, 1, 5, -3}; int n = a.Length; Console.Write(maxSubArraySum(a, n)); } } // This code is contributed by parashar... |
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PHP
<?php // PHP program for Bresenhamâs // Line Generation Assumptions : // 1) Line is drawn from // left to right. // 2) x1 < x2 and y1 < y2 // 3) Slope of the line is // between 0 and 1. // We draw a line from lower // left to upper right. // function for line generation function bresenham($x1, $y1, $x2, $y2) { $m_new = 2 * ($y2 - $y1); $slope_error_new = $m_new - ($x2 - $x1); for ($x = $x1, $y = $y1; $x <= $x2; $x++) { echo "(" ,$x , "," , $y, ")\n"; // Add slope to increment // angle formed $slope_error_new += $m_new; // Slope error reached limit, // time to increment y and // update slope error. if ($slope_error_new >= 0) { $y++; $slope_error_new -= 2 * ($x2 - $x1); } } } // Driver Code $x1 = 3; $y1 = 2; $x2 = 15; $y2 = 5; bresenham($x1, $y1, $x2, $y2); // This code is contributed by nitin mittal. ?> |
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Output :
5
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