July 19, 2023 - Step 1: Initialize two variables i and counter. i represents the position in the modified character array where we will store the compressed characters.

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 ...

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.

This compresses to "a3b2a2". Note that each group is independent even if two groups have the same character. # # #   · # Constraints: # # # 1 <= chars.length <= 2000 · # chars[i] is a lower-case English letter, upper-case English letter, digit, or symbol. # # · · # · # @lc app=leetcode id=443 lang=python3 · # · # [443] String Compression ·

December 28, 2023 - LeetCode — 1531. String Compression II Hard | Python Solution | By Bhupesh Singh Rathore (aka CRUIO) Run-length encoding is a string compression method that works by replacing consecutive identical …

# 443. String Compression · class Solution: def compress(self, chars): """ :type chars: List[str] :rtype: int · """ here = 0 # start replacing from here · i = 0 # iterator · while i < len(chars): #iterate through array ·

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

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.

February 4, 2026 - You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews. Given an array of characters chars, compress it using the following algorithm: Begin with an empty string s.

🏋️ Python / Modern C++ Solutions of All 3977 LeetCode Problems (Weekly Update) - LeetCode-Solutions/Python/string-compression.py at master · kamyu104/LeetCode-Solutions

September 20, 2024 - class Solution: def compress(self, chars: List[str]) -> int: if len(chars)==1: return 1 read,write=0,0 n=len(chars) while read<n: char=chars[read] count=0 while read<n and char==chars[read]: read+=1 count+=1 chars[write]=char write+=1 if count>1: for digit in str(count): chars[write]=digit write+=1 return write · Python · String ·

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, ...

Explanation: The compressed string is "122" and the first 3 characters of the input array should be ["1","2","2"]. ... Please upgrade to NeetCode Pro to view company tags. ... Python Python 3.14.2 Java Java 21 C++ C++ 23 JavaScript Node 22.08.0 TypeScript TS 5.7 C# C# 12 Go Go 1.23.5 Kotlin Kotlin 2.1.10 Swift Swift 6.1.0 Rust Rust 1.93.1 Auto

May 16, 2025 - class Solution: def compress(self, chars: List[str]) -> int: result = [] for i in range(len(chars)): currentChar = chars[i] result.append(currentChar) countCurrentChar = 1 j = i + 1 while j < len(chars) and currentChar == chars[j]: countCurrentChar += 1 j += 1 if (countCurrentChar > 1): result.append(str(countCurrentChar)) i = j # This line is the trap! chars[:] = result · Why this won’t work: That i = j update is literally doing nothing. In Python, a for loop using range() automatically handles the index - even if you update i inside the loop, the next value will still come from the range.

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.