Move from brute-force thinking to an efficient approach using two pointers strategy.
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:
1, append the character to s.The compressed string s should not be returned separately, but instead, be stored in the input character array chars. Note that group lengths that are 10 or longer will be split into multiple characters in chars.
After you are done modifying the input array, return the new length of the array.
You must write an algorithm that uses only constant extra space.
Note: The characters in the array beyond the returned length do not matter and should be ignored.
Example 1:
Input: chars = ["a","a","b","b","c","c","c"] Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"] Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
Example 2:
Input: chars = ["a"] Output: Return 1, and the first character of the input array should be: ["a"] Explanation: The only group is "a", which remains uncompressed since it's a single character.
Example 3:
Input: chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"] Output: Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"]. Explanation: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".
Constraints:
1 <= chars.length <= 2000chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.Problem summary: 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 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. Note that group lengths that are 10 or longer will be split into multiple characters in chars. After you are done modifying the input array, return the new length of the array. You must write an algorithm that uses only constant extra space. Note: The characters in the array beyond the returned length do not matter and should be ignored.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
["a","a","b","b","c","c","c"]
["a"]
["a","b","b","b","b","b","b","b","b","b","b","b","b"]
count-and-say)encode-and-decode-strings)design-compressed-string-iterator)decompress-run-length-encoded-list)string-compression-iii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #443: String Compression
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;
}
}
// Accepted solution for LeetCode #443: String Compression
func compress(chars []byte) int {
i, k, n := 0, 0, len(chars)
for i < n {
j := i + 1
for j < n && chars[j] == chars[i] {
j++
}
chars[k] = chars[i]
k++
if j-i > 1 {
cnt := strconv.Itoa(j - i)
for _, c := range cnt {
chars[k] = byte(c)
k++
}
}
i = j
}
return k
}
# Accepted solution for LeetCode #443: String Compression
class Solution:
def compress(self, chars: List[str]) -> int:
i, k, n = 0, 0, len(chars)
while i < n:
j = i + 1
while j < n and chars[j] == chars[i]:
j += 1
chars[k] = chars[i]
k += 1
if j - i > 1:
cnt = str(j - i)
for c in cnt:
chars[k] = c
k += 1
i = j
return k
// Accepted solution for LeetCode #443: String Compression
impl Solution {
pub fn compress(chars: &mut Vec<char>) -> i32 {
let (mut i, mut k, n) = (0, 0, chars.len());
while i < n {
let mut j = i + 1;
while j < n && chars[j] == chars[i] {
j += 1;
}
chars[k] = chars[i];
k += 1;
if j - i > 1 {
let cnt = (j - i).to_string();
for c in cnt.chars() {
chars[k] = c;
k += 1;
}
}
i = j;
}
k as i32
}
}
// Accepted solution for LeetCode #443: String Compression
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #443: String Compression
// 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;
// }
// }
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.