You are given a one dimensional array that may contain both positive and negative integers, find the sum of contiguous subarray of numbers which has the largest sum.
For example, if the given array is {-2, -5, 6, -2, -3, 1, 5, -6}, then the maximum subarray sum is 7 (see highlighted elements).
The naive method is to run two loops. The outer loop picks the beginning element, the inner loop finds the maximum possible sum with first element picked by outer loop and compares this maximum with the overall maximum. Finally, return the overall maximum. The time complexity of the Naive method is O(n^2).
Using Divide and Conquer approach, we can find the maximum subarray sum in O(nLogn) time. Following is the Divide and Conquer algorithm.
- Divide the given array in two halves
- Return the maximum of following three
- Maximum subarray sum in left half (Make a recursive call)
- Maximum subarray sum in right half (Make a recursive call)
- Maximum subarray sum such that the subarray crosses the midpoint
The lines 2.a and 2.b are simple recursive calls. How to find maximum subarray sum such that the subarray crosses the midpoint? We can easily find the crossing sum in linear time. The idea is simple, find the maximum sum starting from mid point and ending at some point on left of mid, then find the maximum sum starting from mid + 1 and ending with some point on right of mid + 1. Finally, combine the two and return the maximum among left, right and combination of both.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h> using namespace std; // A utility function to find maximum of two integers int max(int a, int b) { return (a > b) ? a : b; } // A utility function to find maximum of three integers int max(int a, int b, int c) { return max(max(a, b), c); } // Find the maximum possible sum in arr[] such that arr[m] // is part of it int maxCrossingSum(int arr[], int l, int m, int h) { // Include elements on left of mid. int sum = 0; int left_sum = INT_MIN; for (int i = m; i >= l; i--) { sum = sum + arr[i]; if (sum > left_sum) left_sum = sum; } // Include elements on right of mid sum = 0; int right_sum = INT_MIN; for (int i = m; i <= h; i++) { sum = sum + arr[i]; if (sum > right_sum) right_sum = sum; } // Return sum of elements on left and right of mid // returning only left_sum + right_sum will fail for // [-2, 1] return max(left_sum + right_sum - arr[m], left_sum, right_sum); } // Returns sum of maximum sum subarray in aa[l..h] int maxSubArraySum(int arr[], int l, int h) { // Invalid Range: low is greater than high if (l > h) return INT_MIN; // Base Case: Only one element if (l == h) return arr[l]; // Find middle point int m = (l + h) / 2; /* Return maximum of following three possible cases a) Maximum subarray sum in left half b) Maximum subarray sum in right half c) Maximum subarray sum such that the subarray crosses the midpoint */ return max(maxSubArraySum(arr, l, m - 1), maxSubArraySum(arr, m + 1, h), maxCrossingSum(arr, l, m, h)); } /*Driver program to test maxSubArraySum*/int main() { int arr[] = { 2, 3, 4, 5, 7 }; int n = sizeof(arr) / sizeof(arr[0]); int max_sum = maxSubArraySum(arr, 0, n - 1); cout << "Maximum contiguous sum is " << max_sum; return 0; } |
Java
// A Divide and Conquer based Java // program for maximum subarray sum // problem import java.util.*; class GFG { // Find the maximum possible sum in arr[] // such that arr[m] is part of it static int maxCrossingSum(int arr[], int l, int m, int h) { // Include elements on left of mid. int sum = 0; int left_sum = Integer.MIN_VALUE; for (int i = m; i >= l; i--) { sum = sum + arr[i]; if (sum > left_sum) left_sum = sum; } // Include elements on right of mid sum = 0; int right_sum = Integer.MIN_VALUE; for (int i = m; i <= h; i++) { sum = sum + arr[i]; if (sum > right_sum) right_sum = sum; } // Return sum of elements on left // and right of mid // returning only left_sum + right_sum will fail for // [-2, 1] return Math.max(left_sum + right_sum - arr[m], Math.max(left_sum, right_sum)); } // Returns sum of maximum sum subarray // in aa[l..h] static int maxSubArraySum(int arr[], int l, int h) { //Invalid Range: low is greater than high if (l > h) return Integer.MIN_VALUE; // Base Case: Only one element if (l == h) return arr[l]; // Find middle point int m = (l + h) / 2; /* Return maximum of following three possible cases: a) Maximum subarray sum in left half b) Maximum subarray sum in right half c) Maximum subarray sum such that the subarray crosses the midpoint */ return Math.max( Math.max(maxSubArraySum(arr, l, m-1), maxSubArraySum(arr, m + 1, h)), maxCrossingSum(arr, l, m, h)); } /* Driver program to test maxSubArraySum */ public static void main(String[] args) { int arr[] = { 2, 3, 4, 5, 7 }; int n = arr.length; int max_sum = maxSubArraySum(arr, 0, n - 1); System.out.println("Maximum contiguous sum is " + max_sum); } } // This code is contributed by Prerna Saini |
Python
# A Divide and Conquer based program # for maximum subarray sum problem # Find the maximum possible sum in # arr[] auch that arr[m] is part of it def maxCrossingSum(arr, l, m, h): # Include elements on left of mid. sm = 0 left_sum = -10000 for i in range(m, l-1, -1): sm = sm + arr[i] if (sm > left_sum): left_sum = sm # Include elements on right of mid sm = 0 right_sum = -1000 for i in range(m, h + 1): sm = sm + arr[i] if (sm > right_sum): right_sum = sm # Return sum of elements on left and right of mid # returning only left_sum + right_sum will fail for [-2, 1] return max(left_sum + right_sum - arr[m], left_sum, right_sum) # Returns sum of maximum sum subarray in aa[l..h] def maxSubArraySum(arr, l, h): #Invalid Range: low is greater than high if (l > h): return -10000 # Base Case: Only one element if (l == h): return arr[l] # Find middle point m = (l + h) // 2 # Return maximum of following three possible cases # a) Maximum subarray sum in left half # b) Maximum subarray sum in right half # c) Maximum subarray sum such that the # subarray crosses the midpoint return max(maxSubArraySum(arr, l, m-1), maxSubArraySum(arr, m+1, h), maxCrossingSum(arr, l, m, h)) # Driver Code arr = [2, 3, 4, 5, 7] n = len(arr) max_sum = maxSubArraySum(arr, 0, n-1) print("Maximum contiguous sum is ", max_sum) # This code is contributed by Nikita Tiwari. |
C#
// A Divide and Conquer based C# // program for maximum subarray sum // problem using System; class GFG { // Find the maximum possible sum in arr[] // such that arr[m] is part of it static int maxCrossingSum(int[] arr, int l, int m, int h) { // Include elements on left of mid. int sum = 0; int left_sum = int.MinValue; for (int i = m; i >= l; i--) { sum = sum + arr[i]; if (sum > left_sum) left_sum = sum; } // Include elements on right of mid sum = 0; int right_sum = int.MinValue; ; for (int i = m; i <= h; i++) { sum = sum + arr[i]; if (sum > right_sum) right_sum = sum; } // Return sum of elements on left // and right of mid // returning only left_sum + right_sum will fail for // [-2, 1] return Math.Max(left_sum + right_sum -arr[m], Math.Max(left_sum, right_sum)); } // Returns sum of maximum sum subarray // in aa[l..h] static int maxSubArraySum(int[] arr, int l, int h) { //Invalid Range: low is greater than high if (l > h) return int.MinValue; // Base Case: Only one element if (l == h) return arr[l]; // Find middle point int m = (l + h) / 2; /* Return maximum of following three possible cases: a) Maximum subarray sum in left half b) Maximum subarray sum in right half c) Maximum subarray sum such that the subarray crosses the midpoint */ return Math.Max( Math.Max(maxSubArraySum(arr, l, m-1), maxSubArraySum(arr, m + 1, h)), maxCrossingSum(arr, l, m, h)); } /* Driver program to test maxSubArraySum */ public static void Main() { int[] arr = { 2, 3, 4, 5, 7 }; int n = arr.Length; int max_sum = maxSubArraySum(arr, 0, n - 1); Console.Write("Maximum contiguous sum is " + max_sum); } } // This code is contributed by vt_m. |
PHP
<?php // A Divide and Conquer based program // for maximum subarray sum problem // Find the maximum possible sum in arr[] // such that arr[m] is part of it function maxCrossingSum(&$arr, $l, $m, $h) { // Include elements on left of mid. $sum = 0; $left_sum = PHP_INT_MIN; for ($i = $m; $i >= $l; $i--) { $sum = $sum + $arr[$i]; if ($sum > $left_sum) $left_sum = $sum; } // Include elements on right of mid $sum = 0; $right_sum = PHP_INT_MIN; for ($i = $m; $i <= $h; $i++) { $sum = $sum + $arr[$i]; if ($sum > $right_sum) $right_sum = $sum; } // Return sum of elements on left // and right of mid // returning only left_sum + right_sum will fail for [-2, 1] return max($left_sum + $right_sum - $arr[$m], $left_sum, $right_sum); } // Returns sum of maximum sum // subarray in aa[l..h] function maxSubArraySum(&$arr, $l, $h) { //Invalid Range: low is greater than high if ($l > $h) return PHP_INT_MIN; // Base Case: Only one element if ($l == $h) return $arr[$l]; // Find middle point $m = intval(($l + $h) / 2); /* Return maximum of following three possible cases a) Maximum subarray sum in left half b) Maximum subarray sum in right half c) Maximum subarray sum such that the subarray crosses the midpoint */ return max(maxSubArraySum($arr, $l, $m-1), maxSubArraySum($arr, $m + 1, $h), maxCrossingSum($arr, $l, $m, $h)); } // Driver Code $arr = array(2, 3, 4, 5, 7); $n = count($arr); $max_sum = maxSubArraySum($arr, 0, $n - 1); echo "Maximum contiguous sum is " . $max_sum; // This code is contributed by rathbhupendra, Aadil ?> |
Javascript
<script> // A Divide and Conquer based program for maximum subarray // sum problem // A utility function to find maximum of two integers function max(a,b) { return (a > b) ? a : b; } // A utility function to find maximum of three integers function max(a,b,c) { return Math.max(Math.max(a, b), c); } // Find the maximum possible sum in arr[] auch that arr[m] // is part of it function maxCrossingSum(arr, l, m,h) { // Include elements on left of mid. let sum = 0; let left_sum = Number.MIN_VALUE; for (let i = m; i >= l; i--) { sum = sum + arr[i]; if (sum > left_sum) left_sum = sum; } // Include elements on right of mid sum = 0; let right_sum = Number.MIN_VALUE; for (let i = m; i <= h; i++) { sum = sum + arr[i]; if (sum > right_sum) right_sum = sum; } // Return sum of elements on left and right of mid // returning only left_sum + right_sum will fail for // [-2, 1] return max(left_sum + right_sum - arr[m], left_sum, right_sum); } // Returns sum of maximum sum subarray in aa[l..h] function maxSubArraySum(arr, l,h) { //Invalid Range: low is greater than high if (l > h) return Number.MIN_VALUE; // Base Case: Only one element if (l == h) return arr[l]; // Find middle point let m = parseInt((l + h) / 2, 10); /* Return maximum of following three possible cases a) Maximum subarray sum in left half b) Maximum subarray sum in right half c) Maximum subarray sum such that the subarray crosses the midpoint */ return max(maxSubArraySum(arr, l, m-1), maxSubArraySum(arr, m + 1, h), maxCrossingSum(arr, l, m, h)); } let arr = [ 2, 3, 4, 5, 7 ]; let n = arr.length; let max_sum = maxSubArraySum(arr, 0, n - 1); document.write("Maximum contiguous sum is " + max_sum); // This code is contributed by vaibhavrabadiya117. </script> |
Output
Maximum contiguous sum is 21
Time Complexity: maxSubArraySum() is a recursive method and time complexity can be expressed as following recurrence relation.
T(n) = 2T(n/2) + Θ(n)
Time Complexity : O(nlogn)
Auxiliary Space: O(1).
The above recurrence is similar to Merge Sort and can be solved either using Recurrence Tree method or Master method. It falls in case II of Master Method and solution of the recurrence is Θ(nLogn).
The Kadane’s Algorithm for this problem takes O(n) time. Therefore the Kadane’s algorithm is better than the Divide and Conquer approach, but this problem can be considered as a good example to show power of Divide and Conquer. The above simple approach where we divide the array in two halves, reduces the time complexity from O(n^2) to O(nLogn).
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