Maximum Sum Circular Subarray

Description

Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.

Here, a circular array means the end of the array connects to the beginning of the array.  (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)

Also, a subarray may only include each element of the fixed buffer A at most once.  (Formally, for a subarray C[i], C[i+1], ..., C[j], there does not exist i <= k1, k2 <= j with k1 % A.length = k2 % A.length.)

 

Example 1:

Input: [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3

Example 2:

Input: [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10

Example 3:

Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4

Example 4:

Input: [3,-2,2,-3]
Output: 3
Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3

Example 5:

Input: [-2,-3,-1]
Output: -1
Explanation: Subarray [-1] has maximum sum -1

 

Note:

1. -30000 <= A[i] <= 30000
2. 1 <= A.length <= 30000

Solution(javascript)

/** 对着着53题 单调队列 O(n)
 *
 * @param {number[]} A
 * @return {number}
 */
// const maxSubarraySumCircular = function (A = []) {
//   let max = A[0]
//   const deque = [0]
//   let start = 0
//   const sum = [A[0]]
//   for (let i = 1; i < 2 * A.length; i++) {
//     sum[i] = sum[i - 1] + A[i % A.length]
//   }
//   for (let i = 1; i < 2 * A.length; i++) {
//     while (i - deque[start] > A.length) {
//       start += 1
//     }
//     max = Math.max(sum[i] - sum[deque[start]], max)

//     while (sum[i] < sum[deque[deque.length - 1]]) {
//       deque.pop()
//     }
//     deque.push(i)
//   }

//   return max
// }

// https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass
// 类似 53 题
const maxSubarraySumCircular = function (A = []) {
  let max = A[0]
  let min = A[0]
  let currentMax = max
  let currentMin = min
  let sum = A[0]
  for (let i = 1; i < A.length; i++) {
    currentMax = A[i] + Math.max(currentMax, 0)
    max = Math.max(max, currentMax)
    currentMin = A[i] + Math.min(currentMin, 0)
    min = Math.min(min, currentMin)
    sum += A[i]
  }

  return max < 0 ? max : Math.max(max, sum - min)
}