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Solution 16: Sequential Byte Compression (String Compression)

Approach Explanation

We use two pointers: read to iterate through the original array and write to update the array in-place.

Step-by-Step Logic

  1. write = 0, i = 0.
  2. While i < len(chars):
    • char = chars[i], count = 0.
    • While i < len(chars) and chars[i] == char:
      • i += 1, count += 1.
    • chars[write] = char, write += 1.
    • If count > 1:
      • For each digit in str(count):
        • chars[write] = digit, write += 1.
  3. Return write.

Complexity

  • Time Complexity: O(N).
  • Space Complexity: O(1).

Code

def compress(chars):
    write = 0
    i = 0
    while i < len(chars):
        char = chars[i]
        count = 0
        while i < len(chars) and chars[i] == char:
            i += 1
            count += 1
        
        chars[write] = char
        write += 1
        
        if count > 1:
            for digit in str(count):
                chars[write] = digit
                write += 1
                
    return write