Description
918. Maximum Sum Circular Subarray (Medium)
Given a circular integer array nums of length n,
return the maximum possible sum of a non-empty subarray of
nums.
A circular array means the end of the array connects to the beginning of the
array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the
previous element of nums[i] is nums[(i - 1 + n) % n].
A subarray may only include each element of the fixed buffer nums
at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does
not exist i <= k1, k2 <= j with k1 % n == k2 % n.
Example 1:
Input: nums = [1,-2,3,-2] Output: 3 Explanation: Subarray [3] has maximum sum 3.
Example 2:
Input: nums = [5,-3,5] Output: 10 Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.
Example 3:
Input: nums = [-3,-2,-3] Output: -2 Explanation: Subarray [-2] has maximum sum -2.
Constraints:
n == nums.length1 <= n <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104
Solution
We define $dp[i]$ to represent the maximum sum of subarrays that end at $nums[i]$. Then, we can discuss two cases:
- The subarray is a continuous segment, i.e., $tail \geq head$.
- The subarray is divided into two segments, i.e., $tail < head$.
In each case, we can update the $dp[i]$ value accordingly to find the maximum sum of subarrays that end at each element $nums[i]$.
Code
class Solution {
public:
int maxSubarraySumCircular(vector<int> &nums) {
int n = nums.size();
if (n == 1) {
return nums[0];
}
vector<int> dp(n, INT_MIN);
int sum = accumulate(nums.begin(), nums.end(), 0);
vector<int> normal(n, INT_MIN);
dp[0] = nums[0];
for (int i = 1; i < n; ++i) {
dp[i] = max(nums[i], nums[i] + dp[i - 1]);
}
vector<int> min_sum(n, 0);
min_sum[n - 1] = nums[n - 1];
for (int i = n - 2; i >= 0; --i) {
min_sum[i] = min(min_sum[i + 1] + nums[i], nums[i]);
}
int res = INT_MIN;
for (int i = 0; i < n - 1; ++i) {
dp[i] = max(dp[i], sum - min_sum[i + 1]);
res = max(res, dp[i]);
}
return max(res, dp[n - 1]);
}
};