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

Python Code
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class Solution:
    def maxSubarraySumCircular(self, nums: List[int]) -> int:
        s1 = s2 = f1 = f2 = nums[0]
        for num in nums[1:]:
            f1 = num + max(f1, 0)
            f2 = num + min(f2, 0)
            s1 = max(s1, f1)
            s2 = min(s2, f2)
        return s1 if s1 <= 0 else max(s1, sum(nums) - s2)

Java Code
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class Solution {
    public int maxSubarraySumCircular(int[] nums) {
        int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0];
        for (int i = 1; i < nums.length; ++i) {
            total += nums[i];
            f1 = nums[i] + Math.max(f1, 0);
            f2 = nums[i] + Math.min(f2, 0);
            s1 = Math.max(s1, f1);
            s2 = Math.min(s2, f2);
        }
        return s1 > 0 ? Math.max(s1, total - s2) : s1;
    }
}

C++ Code
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class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            total += nums[i];
            f1 = nums[i] + max(f1, 0);
            f2 = nums[i] + min(f2, 0);
            s1 = max(s1, f1);
            s2 = min(s2, f2);
        }
        return s1 > 0 ? max(s1, total - s2) : s1;
    }
};

Go Code
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func maxSubarraySumCircular(nums []int) int {
	s1, s2, f1, f2, total := nums[0], nums[0], nums[0], nums[0], nums[0]
	for i := 1; i < len(nums); i++ {
		total += nums[i]
		f1 = nums[i] + max(f1, 0)
		f2 = nums[i] + min(f2, 0)
		s1 = max(s1, f1)
		s2 = min(s2, f2)
	}
	if s1 <= 0 {
		return s1
	}
	return max(s1, total-s2)
}

TypeScript Code
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function maxSubarraySumCircular(nums: number[]): number {
    let pre1 = nums[0],
        pre2 = nums[0];
    let ans1 = nums[0],
        ans2 = nums[0];
    let sum = nums[0];

    for (let i = 1; i < nums.length; ++i) {
        let cur = nums[i];
        sum += cur;
        pre1 = Math.max(pre1 + cur, cur);
        ans1 = Math.max(pre1, ans1);

        pre2 = Math.min(pre2 + cur, cur);
        ans2 = Math.min(pre2, ans2);
    }
    return ans1 > 0 ? Math.max(ans1, sum - ans2) : ans1;
}

Solution 2

Python Code
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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)

Java Code
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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);
    }
}

C++ Code
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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);
    }
};

Go Code
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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)
}

TypeScript Code
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function maxSubarraySumCircular(nums: number[]): number {
    const inf = 1 << 30;
    let [pmi, pmx] = [0, -inf];
    let [ans, s, smi] = [-inf, 0, inf];
    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);
}