918. Maximum Sum Circular Subarray
Description
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.length
1 <= n <= 3 * 104
-3 * 104 <= nums[i] <= 3 * 104
Solutions
Solution 1: Maintain Prefix Maximum
The maximum sum of a circular subarray can be divided into two cases:
- Case 1: The subarray with the maximum sum does not include the circular part, which is the ordinary maximum subarray sum;
- Case 2: The subarray with the maximum sum includes the circular part, which can be transformed into: the total sum of the array minus the minimum subarray sum.
Therefore, we maintain the following variables:
- The minimum prefix sum $pmi$, initially $0$;
- The maximum prefix sum $pmx$, initially $-\infty$;
- The prefix sum $s$, initially $0$;
- The minimum subarray sum $smi$, initially $\infty$;
- The answer $ans$, initially $-\infty$.
Next, we only need to traverse the array $nums$. For the current element $x$ we are traversing, we perform the following update operations:
- Update the prefix sum $s = s + x$;
- Update the answer $ans = \max(ans, s - pmi)$, which is the answer for Case 1 (the prefix sum $s$ minus the minimum prefix sum $pmi$ can give the maximum subarray sum);
- Update $smi = \min(smi, s - pmx)$, which is the minimum subarray sum for Case 2;
- Update $pmi = \min(pmi, s)$, which is the minimum prefix sum;
- Update $pmx = \max(pmx, s)$, which is the maximum prefix sum.
After the traversal, we return the maximum value of $ans$ and $s - smi$ as the answer.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|
1 2 3 4 5 6 7 8 9 10 11 12 |
|