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Posts tagged as “prefix sum”

花花酱 LeetCode 437. Path Sum III

Problem

题目大意:给你一棵二叉树,返回单向的路径和等于sum的路径数量。

https://leetcode.com/problems/path-sum-iii/description/

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

Solution 1: Recursion

Time complexity: O(n^2)

Space complexity: O(n)

C++

Solution 2: Running Prefix Sum

Same idea to 花花酱 LeetCode 560. Subarray Sum Equals K

Time complexity: O(n)

Space complexity: O(h)

C++

Related Problem

 

花花酱 LeetCode 560. Subarray Sum Equals K

Problem

题目大意:给你一个数组,问有多少子数组的和为k。

https://leetcode.com/problems/subarray-sum-equals-k/description/

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input:nums = [1,1,1], k = 2
Output: 2

Note:

  1. The length of the array is in range [1, 20,000].
  2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

Solution -1: Brute Force

For every pair of i,j, check sum(nums[i:j]) in O(j-i) = O(n)

Time complexity: O(n^3) TLE

Space complexity: O(1)

Solution 0: Brute Force + Prefix sun

Precompute the prefix sum and check sum of nums[i:j] in O(1)

Time complexity: O(n^2)

Space complexity: O(n)

C++

Solution 1: Running Prefix sum

Keep tracking the prefix sums and their counts.

s -> count: how many arrays nums[0:j] (j < i) that has sum of s

cur_sum = sum(nums[0:i])

check how many arrays nums[0:j] (j < i) that has sum (cur_sum – k)

then there are the same number of arrays nums[j+1: i] that have sum k.

Time complexity: O(n)

Space complexity: O(n)

C++

 

花花酱 307. Range Sum Query – Mutable

题目大意:给你一个数组,让你求一个范围之内所有元素的和,数组元素可以更改。

Problem:

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

The update(i, val) function modifies nums by updating the element at index i to val.

Example:

Note:

  1. The array is only modifiable by the update function.
  2. You may assume the number of calls to update and sumRange function is distributed evenly.

Idea:

Fenwick Tree

Solution:

C++

Time complexity:

init O(nlogn)

query: O(logn)

update: O(logn)

C++

Java

Python3

Solution 2: Segment Tree

C++