Leetcode: Sum Circular Subarray

Maximum Sum Circular Subarray



Similar Problems:


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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


  • Solution:
// Blog link: https://code.dennyzhang.com/maximum-sum-circular-subarray
// Basic Ideas: dynamic programming + leftrightpass
//
// f(i): From A[0] to A[i], find the sum of max subarray which ends with A[i]
// l(i): sum(A[0]+...A[i-1])
// g(i): From A[len(A)-1] to A[i], find the sum of max subarray which starts with A[len(A)-1]
//
// dp(i) = max(f(i), l(i)+g(i))
// res = max(dp[])
// Complexity: Time O(n), Space O(n)
func maxSubarraySumCircular(A []int) int {
    f, l := make([]int, len(A)), make([]int, len(A))
    sum := 0
    res := A[0]
    for i, num := range A {
        l[i] = sum
        if i == 0 || f[i-1] < 0 {
            f[i] = num
        } else {
            f[i] = f[i-1] + num
        }
        if f[i] > res { res = f[i] }
        sum += num
    }
    max, sum := A[len(A)-1], 0
    for i:=len(A)-1; i>=0; i-- {
        if A[i] + sum > max {
            max = A[i] + sum
        }
        if l[i]+max > res {
            res = l[i]+max
        }
        sum += A[i]        
    }
    return res
}
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