🤖 leetcode 394. Decode String | medium | parsing | javascript

🗒️ Problems

Decode String - LeetCode

Given an encoded string, return its decoded string.

The encoding rule is: k[encoded_string], where the encoded_string
inside the square brackets is being repeated exactly k times.
Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid;
there are no extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and
that digits are only for those repeat numbers, k.
For example, there will not be input like 3a or 2[4].

The test cases are generated so that the length of the output will never exceed 10^5.

🤔 First attempt

• Non nested case like 3[a]2[bc] was not difficult.
• But the nested one like 3[a2[c]] was a little bit tricky.
  • How to parse ? using recursive ? ...

✨ Idea

• Parsing sequence
  • 💡 First find ].
  • Find the matching previous [
  • And find the repeating number.
  • Decode string using the above found and save back to the stack.

🍀 Steps

💡 First find [object Object].

    if (ch === "]") {

Find the matching previous [

• Find the previous matching [ in the stack.

let sub = "";
while (stack.at(-1) !== "[") {
  sub = stack.pop() + sub;
}
// pop "["
stack.pop();

Find the repeating number.

• Repeating number can be more than one character.

    if (ch === "]") {
      let num = ""
      while (!isNaN(stack.at(-1))) {
        num = stack.pop() + num
      }

Decode string using the above found and save back to the stack.

stack.push(sub.repeat(num));

🔥 My Solution

/**
 * @param {string} s
 * @return {string}
 */
var decodeString = function (s) {
  const N = s.length;

  const stack = [];

  for (let i = 0; i < N; i += 1) {
    const ch = s[i];

    if (ch === "]") {
      let sub = "";
      while (stack.at(-1) !== "[") {
        sub = stack.pop() + sub;
      }
      stack.pop();
      let num = "";
      while (!isNaN(stack.at(-1))) {
        num = stack.pop() + num;
      }
      stack.push(sub.repeat(num));
    } else {
      stack.push(ch);
    }
  }

  return stack.join("");
};