LeetCode 443 — String Compression

What does it solve?

Given a character array chars, compress it in-place using run-length encoding. Each group of consecutive identical characters is replaced by the character followed by its count (omit count if 1). Return the new length.

Example: ["a","a","b","b","c","c","c"] → ["a","2","b","2","c","3"], return 6. Counts ≥ 10 are split digit by digit.

Two-pointer approach

// read = scans groups, write = places result
read = 0, write = 0
while read < n:
  ch = chars[read]
  count = 0
  while chars[read] == ch: read++, count++
  chars[write++] = ch
  if count > 1:
    for digit in str(count): chars[write++] = digit

💡 Key insight: The write pointer never overtakes the read pointer — compression always produces ≤ as many characters as it reads. This guarantees in-place modification is safe. Think of it as two pointers on the same array: read scans ahead counting a group, write stamps the compressed result behind it.

Complexity

Time

O(n)

Each character is read exactly once by the read pointer.

Space

O(1)

Only two integer pointers — no extra arrays. Compression is in-place.

Examples & edge cases

Live animation — Read & Write Pointers

Read ptr (active) Read ptr (counting group) Write ptr (writing char) Write ptr (writing digit) Written (compressed)

📖 Input array — read pointer scans →

✏️ Output (in-place) — write pointer places result →

Current char

—

Group count

—

Group range

—

Writes

—

Output so far

nothing written yet

Step — of —

Read ptr

—

position

Write ptr

—

position

Output length

—

chars written

Groups done

compressed

Select an example and press Play to begin.

✓ Compressed!

Speed Medium

Step-by-step table

Step	Action	Read ptr	Write ptr	Char / digit	Group	Output so far

C++ implementation

string_compression.cpp

#include <bits/stdc++.h>
using namespace std;

int compress(vector<char>& chars) {
    int read = 0, write = 0;
    int n = chars.size();

    while (read < n) {
        char ch = chars[read];
        int count = 0;

        // Count the full group of identical chars
        while (read < n && chars[read] == ch) {
            read++;
            count++;
        }

        // Write the character
        chars[write++] = ch;

        // Write count digits only if count > 1
        if (count > 1) {
            string s = to_string(count);
            for (char d : s)
                chars[write++] = d;
        }
    }
    return write;
}

int main() {
    vector<char> chars = {'a','a','b','b','c','c','c'};
    int len = compress(chars);
    cout << "New length: " << len << "\n"; // 6
    // chars = ['a','2','b','2','c','3',...]
    return 0;
}

My Notes

📝 Notes ✅

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