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Commit f24708f

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updating all READMEs
1 parent dbaab48 commit f24708f

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11 files changed

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‎.dev/logs.json

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}
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},
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"443": {
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"sols": {}
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"len": 2,
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"sols": {
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"string_compression.rs": "5fb302855760a7a4d0ea4295ed9a857d55bc05720be8684b9b6fc3d354b4dc92"
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}
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"sols": {
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"sort_an_array.go": "6c1d42703394ec5fdc2c84b9104e3d421203cbe7e8e2f1a50735a8de6a62fa0a"
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"sols": {
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"defanging_an_ip_address.rs": "ce3c86c9d7374f52d4e425f6d3b59a5f00044500e044ea321a747df3248716c8"
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}
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"build_array_from_permutation.rs": "b9c95f880619e12a98dcec4b79ce9da4a529e86c877a25220d13f30b45d41944"
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"concatenation_of_array.rs": "6cf40b85b961b1afc615c836527ab8e7930882df779261c3cd4b78e61f125bcc"
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‎README.md

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[**_Rejection Sampling_**][rejection sampling]  • 
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[**_Biconnected Component_**][biconnected component]
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### **Total Problems Solved: _150_**
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### **Total Problems Solved: _156_**
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---
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‎TOPICWISE.md

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‎docs/index.md

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[**_Rejection Sampling_**][rejection sampling]  • 
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[**_Biconnected Component_**][biconnected component]
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### **Total Problems Solved: _150_**
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### **Total Problems Solved: _156_**
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## Contributors
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# 443. String Compression [![share]](https://leetcode.com/problems/string-compression/)
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![][medium]
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## Problem Statement
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<p>Given an array of characters <code>chars</code>, compress it using the following algorithm:</p>
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<p>Begin with an empty string <code>s</code>. For each group of <strong>consecutive repeating characters</strong> in <code>chars</code>:</p>
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<ul>
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<li>If the group's length is <code>1</code>, append the character to <code>s</code>.</li>
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<li>Otherwise, append the character followed by the group's length.</li>
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</ul>
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<p>The compressed string <code>s</code> <strong>should not be returned separately</strong>, but instead, be stored <strong>in the input character array <code>chars</code></strong>. Note that group lengths that are <code>10</code> or longer will be split into multiple characters in <code>chars</code>.</p>
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<p>After you are done <strong>modifying the input array,</strong> return <em>the new length of the array</em>.</p>
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<p>You must write an algorithm that uses only constant extra space.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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```
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Input: chars = ["a","a","b","b","c","c","c"]
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Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
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Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
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```
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<p><strong class="example">Example 2:</strong></p>
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```
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Input: chars = ["a"]
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Output: Return 1, and the first character of the input array should be: ["a"]
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Explanation: The only group is "a", which remains uncompressed since it's a single character.
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```
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<p><strong class="example">Example 3:</strong></p>
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````
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Input: chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
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Output: Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
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Explanation: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".```
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= chars.length &lt;= 2000</code></li>
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<li><code>chars[i]</code> is a lowercase English letter, uppercase English letter, digit, or symbol.</li>
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</ul>
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<details>
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<summary>
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#### _Click to open Hints_
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</summary>
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- How do you know if you are at the end of a consecutive group of characters?
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</details>
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## Solutions
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### [_Rust_](string_compression.rs)
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```rs [Rust]
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impl Solution {
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pub fn compress(chars: &mut Vec<char>) -> i32 {
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let (mut i, n) = (0, chars.len());
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let mut new_len = 0;
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while i < n {
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let mut j = i;
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// increment j until we find a different character or reach the end
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while j < n && chars[j] == chars[i] {
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j += 1;
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}
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// place the character at the new position
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// e.g. if we have aabbccc, we place a at the start of the array
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chars[new_len] = chars[i];
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new_len += 1;
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// if the length of the group of characters is greater than 1
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// i.e. suppose if new_len is 12, we need to place 12 as characters [..., '1','2', ...]
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if j - i > 1 {
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for c in (j - i).to_string().chars() {
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chars[new_len] = c;
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new_len += 1;
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}
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}
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// place i at same position as j,
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// i.e. the start of the next group of characters
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i = j;
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}
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new_len as i32
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}
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}
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````
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### [_..._]()
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```
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```
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<!----------------------------------{ link }--------------------------------->
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[share]: https://graph.org/file/3ea5234dda646b71c574a.png
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[easy]: https://img.shields.io/badge/Difficulty-Easy-bright.svg
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[medium]: https://img.shields.io/badge/Difficulty-Medium-yellow.svg
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[hard]: https://img.shields.io/badge/Difficulty-Hard-red.svg
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# 912. Sort an Array [![share]](https://leetcode.com/problems/sort-an-array/)
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![][medium]
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## Problem Statement
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<p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>
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<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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```
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Input: nums = [5,2,3,1]
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Output: [1,2,3,5]
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Explanation: After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
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```
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<p><strong class="example">Example 2:</strong></p>
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```
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Input: nums = [5,1,1,2,0,0]
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Output: [0,0,1,1,2,5]
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Explanation: Note that the values of nums are not necessairly unique.
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```
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
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<li><code>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
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</ul>
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## Solutions
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### [_Go_](sort_an_array.go)
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```go [Go]
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package main
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func sortArray(nums []int) []int {
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quickSort(nums, 0, len(nums)-1)
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return nums
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}
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func quickSort(nums []int, start, end int) {
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if start >= end {
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return
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}
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pivot := partition(nums, start, end)
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quickSort(nums, start, pivot-1)
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quickSort(nums, pivot+1, end)
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}
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func partition(arr []int, start int, end int) int {
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pivot := arr[end]
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i := start
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for j := start; j < end; j++ {
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if arr[j] < pivot {
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arr[i], arr[j] = arr[j], arr[i]
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i++
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}
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}
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arr[i], arr[end] = arr[end], arr[i]
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return i
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}
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```
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### [_Rust_](sort_an_array.rs)
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```rs [Rust]
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impl Solution {
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pub fn sort_array(nums: Vec<i32>) -> Vec<i32> {
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let len = nums.len() as i32;
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let mut nums = nums;
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Self::quick_sort(&mut nums, 0, len - 1);
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nums
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}
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fn quick_sort(nums: &mut Vec<i32>, start: i32, end: i32) {
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if start >= end {
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return;
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}
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let pivot = Self::partition(nums, start, end);
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Self::quick_sort(nums, start, pivot - 1);
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Self::quick_sort(nums, pivot + 1, end);
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}
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fn partition(nums: &mut Vec<i32>, start: i32, end: i32) -> i32 {
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let pivot_index = Self::choose_pivot(nums, start as usize, end as usize);
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let pivot = nums[pivot_index as usize];
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nums.swap(pivot_index as usize, end as usize);
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let mut i = start;
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for j in start..end {
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if nums[j as usize] < pivot {
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nums.swap(i as usize, j as usize);
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i += 1;
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}
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}
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nums.swap(i as usize, end as usize);
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i
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}
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// this is to optimize the pivot selection
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fn choose_pivot(nums: &mut Vec<i32>, start: usize, end: usize) -> usize {
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let mid = start + (end - start) / 2;
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if nums[start] <= nums[mid] && nums[mid] <= nums[end] {
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mid
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} else if nums[start] <= nums[end] && nums[end] <= nums[mid] {
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end
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} else {
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start
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}
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}
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}
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```
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### [_..._]()
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```
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```
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<!----------------------------------{ link }--------------------------------->
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[share]: https://graph.org/file/3ea5234dda646b71c574a.png
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[easy]: https://img.shields.io/badge/Difficulty-Easy-bright.svg
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[medium]: https://img.shields.io/badge/Difficulty-Medium-yellow.svg
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[hard]: https://img.shields.io/badge/Difficulty-Hard-red.svg

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