In-depth solution and explanation for LeetCode 443. String Compression in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis. Better than official and forum solutions.

String Compression - Given an array of characters chars, compress it using the following algorithm: Begin with an empty string s. For each group of consecutive repeating characters in chars: * If the group's length is 1, append the character ...

December 4, 2023 - String Compression (LeetCode Problem 443): Given an array of characters chars, compress it using the following algorithm: Begin with an empty string s.

December 26, 2024 - Your count is (j - i), and you use i = j on line 28 to move along. index++ on lines 19 and 25, the part I found most confusing, are there because you are effectively inserting in place and using index to keep track your “new string.” The std::to_string stuff could have been better, but it was my closest approximation of Kevin’s Java code.

LeetCode Solutions in C++23, Java, Python, MySQL, and TypeScript.

January 30, 2024 - The String Compression problem is one of the famous leetcode problems in Data structure and algorithms that involves finding and returning the length of a new string formed after compressing the original one.

In-depth solution and explanation for LeetCode 1531. String Compression II in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis. Better than official and forum solutions.

public int compress(char[] chars) { int lo = 0; int hi = 0; while (hi < chars.length) { char c = chars[hi]; int count = 0; while (hi < chars.length && chars[hi] == c) { hi += 1; count += 1; } chars[lo] = c; lo += 1; if (count != 1) { for (char digit: String.valueOf(count).toCharArray()) { chars[lo] = digit; lo += 1; } } } return lo; } }

Begin with an empty string s. For each group of consecutive repeating characters in chars: If the group's length is 1, append the character to s. Otherwise, append the character followed by the group's length.

public class Solution { public String compressedString(String word) { StringBuilder builder = new StringBuilder(); char last = word.charAt(0); int count = 1; for (int i = 1, l = word.length(); i < l; i++) { if (word.charAt(i) == last) { count++; if (count == 10) { builder.append(9).append(last); count = 1; } } else { builder.append(count).append(last); last = word.charAt(i); count = 1; } } builder.append(count).append(last); return builder.toString(); } }

Begin with an empty string s. For each group of consecutive repeating characters in chars: If the group's length is 1, append the character to s. Otherwise, append the character followed by the group's length. The compressed string s should not be returned separately, but instead, be stored ...

LeetCode style: Given a character array chars, compress it in-place and return the new length. class Solution { public int compress(char[] chars) { int write = 0; int i = 0; while (i < chars.length) { char current = chars[i]; int count = 0; ...

Given a string s and an integer k. You need to delete at most k characters from s such that the run-length encoded version of s has minimum length. Find the minimum length of the run-length encoded version of s after deleting at most k characters.

LeetCode Solutions in C++23, Java, Python, MySQL, and TypeScript.

December 14, 2023 - Begin with an empty string s. For each group of consecutive repeating characters in chars: If the group’s length is 1, append the character to s. Otherwise, append the character followed by the group’s length. The compressed string s should not be returned separately, but instead, be stored in the input character array chars.

public int compress(char[] chars) { int anchor = 0, write = 0; for (int read = 0; read < chars.length; read++) { if (read + 1 == chars.length || chars[read + 1] != chars[read]) { chars[write++] = chars[anchor]; if (read > anchor) { for (char c: ("" + (read - anchor + 1)).toCharArray()) { chars[write++] = c; } } anchor = read + 1; } } return write; } https://leetcode.com/problems/string-compression/discuss/92559/Simple-Easy-to-Understand-Java-solution ·

We can use two pointers to find the start and end positions of each consecutive character, calculate the length of the consecutive characters, and then append the character and length to the string ... class Solution: def compressString(self, ...

Given a string s and an integer k. You need to delete at most k characters from s such that the run-length encoded version of s has minimum length. Find the minimum length of the run-length encoded version of s after deleting at most k characters.