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| 1 | +// A Divide and Conquer based program for maximum subarray sum problem |
| 2 | +#include <bits/stdc++.h> |
| 3 | + |
| 4 | + |
| 5 | +// A utility funtion to find maximum of two integers |
| 6 | +int max(int a, int b) { return (a > b)? a : b; } |
| 7 | + |
| 8 | +// A utility funtion to find maximum of three integers |
| 9 | +int max(int a, int b, int c) { return max(max(a, b), c); } |
| 10 | + |
| 11 | +// Find the maximum possible sum in arr[] auch that arr[m] is part of it |
| 12 | +int maxCrossingSum(int arr[], int l, int m, int h) |
| 13 | +{ |
| 14 | + // Include elements on left of mid. |
| 15 | + int sum = 0; |
| 16 | + int left_sum = INT_MIN; |
| 17 | + for (int i = m; i >= l; i--) |
| 18 | + { |
| 19 | + sum = sum + arr[i]; |
| 20 | + if (sum > left_sum) |
| 21 | + left_sum = sum; |
| 22 | + } |
| 23 | + |
| 24 | + // Include elements on right of mid |
| 25 | + sum = 0; |
| 26 | + int right_sum = INT_MIN; |
| 27 | + for (int i = m+1; i <= h; i++) |
| 28 | + { |
| 29 | + sum = sum + arr[i]; |
| 30 | + if (sum > right_sum) |
| 31 | + right_sum = sum; |
| 32 | + } |
| 33 | + |
| 34 | + // Return sum of elements on left and right of mid |
| 35 | + return left_sum + right_sum; |
| 36 | +} |
| 37 | + |
| 38 | +// Returns sum of maxium sum subarray in aa[l..h] |
| 39 | +int maxSubArraySum(int arr[], int l, int h) |
| 40 | +{ |
| 41 | +// Base Case: Only one element |
| 42 | +if (l == h) |
| 43 | + return arr[l]; |
| 44 | + |
| 45 | +// Find middle point |
| 46 | +int m = (l + h)/2; |
| 47 | + |
| 48 | +/* Return maximum of following three possible cases |
| 49 | + a) Maximum subarray sum in left half |
| 50 | + b) Maximum subarray sum in right half |
| 51 | + c) Maximum subarray sum such that the subarray crosses the midpoint */ |
| 52 | +return max(maxSubArraySum(arr, l, m), |
| 53 | + maxSubArraySum(arr, m+1, h), |
| 54 | + maxCrossingSum(arr, l, m, h)); |
| 55 | +} |
| 56 | + |
| 57 | +/*Driver program to test maxSubArraySum*/ |
| 58 | +int main() |
| 59 | +{ |
| 60 | +int arr[] = {2, 3, 4, 5, 7}; |
| 61 | +int n = sizeof(arr)/sizeof(arr[0]); |
| 62 | +int max_sum = maxSubArraySum(arr, 0, n-1); |
| 63 | +printf("Maximum contiguous sum is %dn", max_sum); |
| 64 | +getchar(); |
| 65 | +return 0; |
| 66 | +} |
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