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Maximum Subarray
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Maximum Subarray
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/*
LeetCode 53:
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.
click to show more practice.
More practice:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
Solution:
Language: C++
Tag: DP PreSum
*/
class Solution {
public:
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
int maxSubArray(vector<int>& nums) {
/*
// 1. preSum -- O(1) space + O(n) time
int max_val = INT_MIN;
int sum = 0, min_val = 0;
for (int i = 0; i < nums.size(); ++i) {
sum += nums[i];
max_val = max(max_val, sum - min_val);
min_val = min(min_val, sum);
}
return max_val;
*/
/*
// 2. DP -- O(n) space + O(n) time
int size = nums.size();
//f[i]: the max subarray sum ends in nums[i - 1] in nums[0, 1, ..., i - 1];
int *f = new int[size + 1];
f[0] = 0;
int max_val = INT_MIN;
for (int i = 1; i <= size; ++i) {
f[i] = (f[i - 1] > 0 ? (f[i - 1] + nums[i - 1]) : nums[i - 1]);
max_val = max(max_val, f[i]);
}
return max_val;
*/
/*
// 3. rolling pointer of 2 -- O(n) time + O(1) space
int *f = new int[2];
f[0] = 0;
int max_val = INT_MIN;
for (int i = 1; i <= nums.size(); ++i) {
f[i % 2] = (f[(i - 1) % 2] > 0 ? (f[(i - 1) % 2] + nums[i - 1]) : nums[i - 1]);
max_val = max(max_val, f[i % 2]);
}
return max_val;
*/
// 4. Divide & Conquer ??? -- O(nlogn)time + O(1) space
if (nums.empty()) {
return 0;
}
return helper(nums, 0, nums.size() - 1);
}
int helper(vector<int> &nums, int left, int right) {
if (left == right) {
return nums[left];
}
int mid = left + (right - left) / 2;
int leftSum = helper(nums, left, mid);
int rightSum = helper(nums, mid + 1, right);
int midSum = calMid(nums, left, right);
return max(midSum, max(leftSum, rightSum));
}
int calMid(vector<int> &nums, int left, int right) {
int mid = left + (right - left) / 2;
int leftSum = INT_MIN, rightSum = INT_MIN;
int sum = 0;
for (int i = mid; i >= left; --i) {
sum += nums[i];
leftSum = max(leftSum, sum);
}
sum = 0;
for (int i = mid + 1; i <= right; ++i) {
sum += nums[i];
rightSum = max(rightSum, sum);
}
return (leftSum + rightSum);
}
};