Question

918. Maximum Sum Circular Subarray

中文文档

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

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)

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;
  }
}

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;
  }
};

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)
}

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

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)

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);
  }
}

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);
  }
};

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)
}

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);
}