LeetCode: K-Concatenation Maximum Sum

Posted on February 10, 2020July 26, 2020 by braindenny

K-Concatenation Maximum Sum

Similar Problems:

Given an integer array arr and an integer k, modify the array by repeating it k times.

For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].

Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0.

As the answer can be very large, return the answer modulo 10^9 + 7.

Example 1:

Input: arr = [1,2], k = 3
Output: 9

Example 2:

Input: arr = [1,-2,1], k = 5
Output: 2

Example 3:

Input: arr = [-1,-2], k = 7
Output: 0

Constraints:

1 <= arr.length <= 10^5
1 <= k <= 10^5
-10^4 <= arr[i] <= 10^4

Github: code.dennyzhang.com

Credits To: leetcode.com

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

Solution:

## https://code.dennyzhang.com/k-concatenation-maximum-sum
## Basic Ideas: Kadane
##
##  If we include one repeition of the list, we can include all
##
## Complexity: Time O(n), Space O(1)
class Solution:
    def kConcatenationMaxSum(self, arr: List[int], k: int) -> int:
        n = len(arr)
        totalSum = sum(arr)
        res = 0
        if totalSum >= 0:
            res = totalSum*(max(0, k-2))

        nums = [0]*n
        if k > 1:
            nums = [0]*2*n
            for i in range(n):
                nums[i] = arr[i]
                nums[i+n] = arr[i]
        else:
            for i in range(n):
                nums[i] = arr[i]

        # Kadane
        maxSum = -sys.maxsize
        maxSoFar = 0
        for i in range(len(nums)):
            v = nums[i]
            maxSoFar = max(v, maxSoFar+v)
            maxSum = max(maxSum, maxSoFar)

        res += maxSum
        return max(res, 0) % (10**9+7)

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