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] ...
Given a circular array C of integers represented by `A`, find the maximum possible sum of a non-empty subarray of C. Here, acircular arraymeans the end of the array connects to the beginning of the array. (Formally,C[i] = A[i]when0 <= i < A.length, andC[i+A.length] = C[...
Given a circular array C of integers represented byA, find the maximum possible sum of a non-empty subarray of C. Here, acircular arraymeans the end of the array connects to the beginning of the array. (Formally,C[i] = A[i]when0 <= i < A.length, andC[i+A.length] = C[i]wh...
918. Maximum Sum Circular Subarray # 题目 # 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. (Fo
The maximum circular sum in the above array is: 29 To solve the problem of finding the maximum circular subarray sum in a given array, the program needs to handle two scenarios: the maximum sum subarray that does not wrap around and the maximum sum subarray that does wrap around the array...
partition the array into exactly K subarrays and calculate their sum. find the minimum possible difference of the maximum sum and the minimum sum. (1<k<=n<=40) for example for N=6 and K=3 and array={5 1 1 1 3 2} the optimal way is to split it into [5][1 1 1][3 2] so...
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1 class Solution { 2 func maxSubarraySumCircular(_ A: [Int]) -> Int { 3 guard A.count > 0 else { 4 return 0 5 } 6 let s = A.reduce(0, +) 7 var M = A[0] 8 var m = A[0] 9 var lastM = A[0] 10 var lastm = A[0] 11 for i in 1 ..< A.count { 12 let...
参考Maximum Sum Circular Subarray,可分为以下三种情况: 最大子数组和在输入数组的中间; 最大子数组和由输入数组的头尾部分组成; 最大子数组和跨越中间跨越多个输入数组。 其中,第三种情况可以看作是第二种情况的结果假设(k-2)*sum(arr),即子数组和大于0时,第三种情况时肯定大于第二种情况的。最终结果取上面...