Divide and Conquer | Set 3 (Maximum Subarray Sum) | GeeksforGeeks
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.
Using Divide and Conquer approach, we can find the maximum subarray sum in O(nLogn) time. Following is the Divide and Conquer algorithm.
1) Divide the given array in two halves
2) Return the maximum of following three
….a) Maximum subarray sum in left half (Make a recursive call)
….b) Maximum subarray sum in right half (Make a recursive call)
….c) Maximum subarray sum such that the subarray crosses the midpoint 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 sum point on right of mid + 1.
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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Using Divide and Conquer approach, we can find the maximum subarray sum in O(nLogn) time. Following is the Divide and Conquer algorithm.
1) Divide the given array in two halves
2) Return the maximum of following three
….a) Maximum subarray sum in left half (Make a recursive call)
….b) Maximum subarray sum in right half (Make a recursive call)
….c) Maximum subarray sum such that the subarray crosses the midpoint 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 sum point on right of mid + 1.
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).
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+1; i <= h; i++) { sum = sum + arr[i]; if (sum > right_sum) right_sum = sum; } // Return sum of elements on left and right of mid return left_sum + right_sum;}// Returns sum of maxium sum subarray in aa[l..h]int maxSubArraySum(int arr[], int l, int h){ // 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), maxSubArraySum(arr, m+1, h), maxCrossingSum(arr, l, m, h));}