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Posts tagged as “sum”

花花酱 LeetCode 1186. Maximum Subarray Sum with One Deletion

Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elements) with at most one element deletion. In other words, you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.

Note that the subarray needs to be non-empty after deleting one element.

Example 1:

Input: arr = [1,-2,0,3]
Output: 4
Explanation: Because we can choose [1, -2, 0, 3] and drop -2, thus the subarray [1, 0, 3] becomes the maximum value.

Example 2:

Input: arr = [1,-2,-2,3]
Output: 3
Explanation: We just choose [3] and it's the maximum sum.

Example 3:

Input: arr = [-1,-1,-1,-1]
Output: -1
Explanation: The final subarray needs to be non-empty. You can't choose [-1] and delete -1 from it, then get an empty subarray to make the sum equals to 0.

Constraints:

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

Solution: DP

First, handle the special case: all numbers are negative, return the max one.

s0 = max subarray sum ends with a[i]
s1 = max subarray sum ends with a[i] with at most one deletion

whenever s0 or s1 becomes negative, reset them to 0.

Time complexity: O(n)
Space complexity: O(1)

C++

花花酱 LeetCode 1184. Distance Between Bus Stops

A bus has n stops numbered from 0 to n - 1 that form a circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops number i and (i + 1) % n.

The bus goes along both directions i.e. clockwise and counterclockwise.

Return the shortest distance between the given start and destination stops.

Example 1:

Input: distance = [1,2,3,4], start = 0, destination = 1
Output: 1
Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.

Example 2:

Input: distance = [1,2,3,4], start = 0, destination = 2
Output: 3
Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.

Example 3:

Input: distance = [1,2,3,4], start = 0, destination = 3
Output: 4
Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.

Constraints:

  • 1 <= n <= 10^4
  • distance.length == n
  • 0 <= start, destination < n
  • 0 <= distance[i] <= 10^4

Solution: Summation

  1. compute the total sum
  2. compute the sum from s to d, c
  3. ans = min(c, sum – c)

Time complexity: O(d-s)
Space complexity: O(1)

C++

花花酱 LeetCode 985. Sum of Even Numbers After Queries

Problem

We have an array A of integers, and an array queries of queries.

For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries.  Your answer array should have answer[i] as the answer to the i-th query.

Example 1:

Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation: 
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Note:

  1. 1 <= A.length <= 10000
  2. -10000 <= A[i] <= 10000
  3. 1 <= queries.length <= 10000
  4. -10000 <= queries[i][0] <= 10000
  5. 0 <= queries[i][1] < A.length

Solution: Simulation

Time complexity: O(n + |Q|)
Space complexity: O(n)

C++

Python3

花花酱 LeetCode 404. Sum of Left Leaves

Solution: Recursion

Time complexity: O(n)

Space complexity: O(h)

C++

Iterative

 

花花酱 LeetCode 643. Maximum Average Subarray I

Problem

题目大意:找出k长度的子数组的平均值的最大值。

https://leetcode.com/problems/maximum-average-subarray-i/description/

Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value.

Example 1:

Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75

Note:

  1. 1 <= k <= n <= 30,000.
  2. Elements of the given array will be in the range [-10,000, 10,000].

Solution: Sliding Window

Time complexity: O(n)

Space complexity: O(1)

C++

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