max_sub_arr_dANDq.cpp

// A Divide and Conquer based program for maximum subarray sum problem 
#include <iostream>
#include <limits.h> 
using namespace std ;
// A utility funtion to find maximum of two integers 
int max(int a, int b) { return (a > b)? a : b; } 

// A utility funtion 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[] auch 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+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)); 
} 
int main() 
{ 
int arr[] = {-2, -5, 6, -2, -3, 1, 5, -6}; 
int n = sizeof(arr)/sizeof(arr[0]); 
int max_sum = maxSubArraySum(arr, 0, n-1); 
cout<<"Maximum contiguous sum is "<< max_sum; 

return 0; 
}