# 2022-05-31 ## [844. Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/) ### Description Given two strings `s` and `t`, return `true` *if they are equal when both are typed into empty text editors*. `'#'` means a backspace character. Note that after backspacing an empty text, the text will continue empty. **Example 1:** ``` Input: s = "ab#c", t = "ad#c" Output: true Explanation: Both s and t become "ac". ``` **Example 2:** ``` Input: s = "ab##", t = "c#d#" Output: true Explanation: Both s and t become "". ``` **Example 3:** ``` Input: s = "a#c", t = "b" Output: false Explanation: s becomes "c" while t becomes "b". ``` **Constraints:** * `1 <= s.length, t.length <= 200` * `s` and `t` only contain lowercase letters and `'#'` characters. **Follow up:** Can you solve it in `O(n)` time and `O(1)` space? ### Solution #### Approach #0 ```go func backspaceCompare(s string, t string) bool { m, n := len(s), len(t) i, j := m-1, n-1 for i >= 0 || j >= 0 { i, j = goBack(s, i), goBack(t, j) if i >= 0 && j >= 0 && s[i] != t[j] { return false } i-- j-- } return i == j } func goBack(s string, i int) int { count := 0 j := i for j >= 0 && (s[j] == '#' || count > 0) { if s[j] != '#' { count-- } else { count++ } j-- } return j } ``` #### Approach #1 ```go func backspaceCompare(s string, t string) bool { return build(s) == build(t) } func build(s string) string { var st []rune for _, ch := range s { if ch == '#' { n := len(st) if n > 0 { st = st[:n-1] } } else { st = append(st, ch) } } return string(st) } ``` ## [986. Interval List Intersections](https://leetcode.com/problems/interval-list-intersections/) ### Description You are given two lists of closed intervals, `firstList` and `secondList`, where `firstList[i] = [start[i], end[i]]` and `secondList[j] = [start[j], end[j]]`. Each list of intervals is pairwise **disjoint** and in **sorted order**. Return *the intersection of these two interval lists*. A **closed interval** `[a, b]` (with `a <= b`) denotes the set of real numbers `x` with `a <= x <= b`. The **intersection** of two closed intervals is a set of real numbers that are either empty or represented as a closed interval. For example, the intersection of `[1, 3]` and `[2, 4]` is `[2, 3]`. **Example 1:** ![](https://img.content.cc/a/2022/05/31/08-42-02-011-cc562333c497dcf782c38a52753cc7cd-286873.png) ``` Input: firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]] Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]] ``` **Example 2:** ``` Input: firstList = [[1,3],[5,9]], secondList = [] Output: [] ``` **Constraints:** * `0 <= firstList.length, secondList.length <= 1000` * `firstList.length + secondList.length >= 1` * `0 <= start[i] < end[i] <= 10^9` * `end[i] < start[i+1]` * `0 <= start[j] < end[j] <= 10^9` * `end[j] < start[j+1]` ### Solution #### Approach #0 ```go func intervalIntersection(firstList [][]int, secondList [][]int) [][]int { i, j := 0, 0 m, n := len(firstList), len(secondList) var ans [][]int for i < m && j < n { a1, b1 := firstList[i][0], firstList[i][1] a2, b2 := secondList[j][0], secondList[j][1] if a1 <= b2 && b1 >= a2 { x, y := a1, b2 if a1 <= a2 { x = a2 } if b1 <= b2 { y = b1 } ans = append(ans, []int{x, y}) } if b1 < b2 { i++ } else { j++ } } return ans } ``` #### Approach #1 ```go func intervalIntersection(firstList [][]int, secondList [][]int) [][]int { i, j := 0, 0 var ans [][]int for i < len(firstList) && j < len(secondList) { low := max(firstList[i][0], secondList[j][0]) high := min(firstList[i][1], secondList[j][1]) if low <= high { ans = append(ans, []int{low, high}) } if firstList[i][1] < secondList[j][1] { i++ } else { j++ } } return ans } func max(a, b int) int { if a > b { return a } return b } func min(a, b int) int { if a < b { return a } return b } ``` ## [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/) ### Description You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`. Find two lines that together with the x-axis form a container, such that the container contains the most water. Return *the maximum amount of water a container can store*. **Notice** that you may not slant the container. **Example 1:** ![](https://img.content.cc/a/2022/05/31/08-43-53-733-9daebb6ebbdb925763fbd31e9a7aa329-db83c2.jpeg) ``` Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49. ``` **Example 2:** ``` Input: height = [1,1] Output: 1 ``` **Constraints:** * `n == height.length` * `2 <= n <= 10^5` * `0 <= height[i] <= 10^4` ### Solution #### Approach #0 ```go func maxArea(height []int) int { n := len(height) i, j := 0, n-1 var ans int for i < j { ans = max(ans, (j-i)*min(height[i], height[j])) if height[i] < height[j] { i++ } else { j-- } } return ans } func max(a, b int) int { if a > b { return a } return b } func min(a, b int) int { if a < b { return a } return b } ``` --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. 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