String Compression · Medium · 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 ...

1class Solution { 2 public int compress(char[] chars) { 3 // writeIndex: position to write the compressed result 4 // arrayLength: total length of the input array 5 int writeIndex = 0; 6 int arrayLength = chars.length; 7 8 // Use two pointers to traverse the array 9 // currentIndex: start of current group of same characters 10 // nextIndex: points to the next different character 11 int currentIndex = 0; 12 13 while (currentIndex < arrayLength) { 14 // Find the end of current group of identical characters 15 int nextIndex = currentIndex + 1; 16 while (nextIndex < arrayLength && chars[nextIndex]

December 26, 2024 - Basically, the problem is similar to its counterpart, only you have to do it in place, you do not include the count if the count is 1, a count with multiple digits like “13” has to be separated by digits, you have to return the size, and…well…frankly, I suspect that 3000 people found the description confusing as hell. If I were the ruler of Leetcode, I might have added the following: Clarification 1: The count you return represents the length of your new string.

July 19, 2023 - Step 12: Finally, return the value of i. This value represents the new length of the modified character array, which now contains the compressed characters. The time complexity is O(n), where n is length of chars. Because we are iterating through an array at once. The space complexity is O(1), as we are using constant extra space. ... class Solution: def compress(self, chars: List[str]) -> int: i = 0 counter = 1 for j in range(1, len(chars) + 1): if j < len(chars) and chars[j] == chars[j-1]: counter += 1 else: chars[i] = chars[j-1] i = i + 1 if counter > 1: for k in str(counter): chars[i] = k i += 1 counter = 1 return i

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

Solution: Always validate input constraints and handle edge cases: ... def get_compression_length(frequency): if frequency == 1: return 1 if frequency < 10: return 2 if frequency < 100: return 3 if frequency == 100: # Explicitly handle the upper bound return 4 # Handle unexpected cases return ...

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.

String Compression III · Medium ... is not empty, use the following operation: Remove a maximum length prefix of word made of a single character c repeating at most 9 times....

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 ...

class Solution { public int compress(char[] chars) { int k = 0, n = chars.length; for (int i = 0, j = i + 1; i < n;) { while (j < n && chars[j] == chars[i]) { ++j; } chars[k++] = chars[i]; if (j - i > 1) { String cnt = String.valueOf(j - i); for (char c : cnt.toCharArray()) { chars[k++] = c; } } i = j; } return k; } }

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

class Solution: def compressedString(self, word: str) -> str: g = groupby(word) ans = [] for c, v in g: k = len(list(v)) while k: x = min(9, k) ans.append(str(x) + c) k -= x return "".join(ans) class Solution { public String compressedString(String word) { StringBuilder ans = new StringBuilder(); int n = word.length(); for (int i = 0; i < n;) { int j = i + 1; while (j < n && word.charAt(j) == word.charAt(i)) { ++j; } int k = j - i; while (k > 0) { int x = Math.min(9, k); ans.append(x).append(word.charAt(i)); k -= x; } i = j; } return ans.toString(); } }

April 1, 2022 - Leetcode: https://leetcode.com/problems/string-compression/ Problem: 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 ...

To obtain the compressed representation, we replace each segment of equal characters in the string with the number of characters in the segment followed by the character(e.x we replace segment "CCCC" with 4C).

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 ·

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

September 23, 2026 - Compress it in place using this rule: look at each group of consecutive equal characters. If a group has only one character, write just that character. If a group has more than one character, write the character followed by the group's size ...

January 30, 2024 - The simplest way to solve the String Compression problem is by using two pointers. One pointer for iterating through the original character array and one for keeping track of the current position in the compressed array.

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.

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. ... Input: s = "aaabcccd", k = 2 Output: 4 ...