LeetCode 0918. Maximum Sum Circular Subarray Solution in Java, C++, Python & More | Explanation + Code

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 -∞;
• The prefix sum s, initially 0;
• The minimum subarray sum smi, initially ∞;
• The answer ans, initially -∞.

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).

PythonJavaC++GoTypeScript

class Solution: def maxSubarraySumCircular(self, nums: List[int]) -> int: pmi, pmx = 0, -inf ans, s, smi = -inf, 0, inf for x in nums: s += x ans = max(ans, s - pmi) smi = min(smi, s - pmx) pmi = min(pmi, s) pmx = max(pmx, s) return max(ans, s - smi)(code-box)

class Solution { public int maxSubarraySumCircular(int[] nums) { final int inf = 1 << 30; int pmi = 0, pmx = -inf; int ans = -inf, s = 0, smi = inf; for (int x : nums) { s += x; ans = Math.max(ans, s - pmi); smi = Math.min(smi, s - pmx); pmi = Math.min(pmi, s); pmx = Math.max(pmx, s); } return Math.max(ans, s - smi); } }(code-box)

class Solution { public: int maxSubarraySumCircular(vector<int>& nums) { const int inf = 1 << 30; int pmi = 0, pmx = -inf; int ans = -inf, s = 0, smi = inf; for (int x : nums) { s += x; ans = max(ans, s - pmi); smi = min(smi, s - pmx); pmi = min(pmi, s); pmx = max(pmx, s); } return max(ans, s - smi); } };(code-box)

func maxSubarraySumCircular(nums []int) int { const inf = 1 << 30 pmi, pmx := 0, -inf ans, s, smi := -inf, 0, inf for _, x := range nums { s += x ans = max(ans, s-pmi) smi = min(smi, s-pmx) pmi = min(pmi, s) pmx = max(pmx, s) } return max(ans, s-smi) }(code-box)

function maxSubarraySumCircular(nums: number[]): number { let [pmi, pmx] = [0, -Infinity]; let [ans, s, smi] = [-Infinity, 0, Infinity]; for (const x of nums) { s += x; ans = Math.max(ans, s - pmi); smi = Math.min(smi, s - pmx); pmi = Math.min(pmi, s); pmx = Math.max(pmx, s); } return Math.max(ans, s - smi); }(code-box)