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function compress(s) {
  // Handle edge case
  if (s.length <= 1) return s;
  
  // Two-pointer approach
  let result = "";
  let count = 1;
  
  for (let i = 1; i <= s.length; i++) {
    // If current character is the same as previous, increment count
    if (i < s.length && s[i] === s[i-1]) {
      count++;
    } else {
      // Add the previous character to result
      result += s[i-1];
      
      // Add the count if it's greater than 1
      if (count > 1) {
        result += count;
      }
      
      // Reset count for the next character
      count = 1;
    }
  }
  
  // Return the shorter string
  return result.length < s.length ? result : s;
}
 
// Alternative approach: Two-pass solution
function compressOptimized(s) {
  // Handle edge case
  if (s.length <= 1) return s;
  
  // First pass: Calculate compressed length
  let compressedLength = 0;
  let count = 1;
  
  for (let i = 1; i <= s.length; i++) {
    if (i < s.length && s[i] === s[i-1]) {
      count++;
    } else {
      compressedLength += 1;  // for the character
      if (count > 1) {
        compressedLength += count.toString().length;  // for the count
      }
      count = 1;
    }
  }
  
  // If compression doesn't save space, return original
  if (compressedLength >= s.length) {
    return s;
  }
  
  // Second pass: Create compressed string
  let result = "";
  count = 1;
  
  for (let i = 1; i <= s.length; i++) {
    if (i < s.length && s[i] === s[i-1]) {
      count++;
    } else {
      result += s[i-1];
      if (count > 1) {
        result += count;
      }
      count = 1;
    }
  }
  
  return result;
}
 
// Test cases
console.log(compress("aabcccccaaa"));  // "a2bc5a3"
console.log(compress("abcdef"));  // "abcdef"
console.log(compress("aaaaabbbbaaaabbddc"));  // "a5b4a4b2d2c"