2182. Construct String With Repeat Limit

Description

You are given a string s and an integer repeatLimit. Construct a new string repeatLimitedString using the characters of s such that no letter appears more than repeatLimit times in a row. You do not have to use all characters from s.

Return the lexicographically largestrepeatLimitedString possible.

A string a is lexicographically larger than a string b if in the first position where a and b differ, string a has a letter that appears later in the alphabet than the corresponding letter in b. If the first min(a.length, b.length) characters do not differ, then the longer string is the lexicographically larger one.

 

Example 1:

Input: s = "cczazcc", repeatLimit = 3
Output: "zzcccac"
Explanation: We use all of the characters from s to construct the repeatLimitedString "zzcccac".
The letter 'a' appears at most 1 time in a row.
The letter 'c' appears at most 3 times in a row.
The letter 'z' appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.

Example 2:

Input: s = "aababab", repeatLimit = 2
Output: "bbabaa"
Explanation: We use only some of the characters from s to construct the repeatLimitedString "bbabaa". 
The letter 'a' appears at most 2 times in a row.
The letter 'b' appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return "bbabaa".
Note that the string "bbabaaa" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.

 

Constraints:

• 1 <= repeatLimit <= s.length <= 105
• s consists of lowercase English letters.

 

Solutions

Solution: Greedy

• Time complexity: O(26n -> n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string} s
 * @param {number} repeatLimit
 * @return {string}
 */
const repeatLimitedString = function (s, repeatLimit) {
  const BASE_CODE = 'a'.charCodeAt(0);
  const n = s.length;
  const letters = Array.from({ length: 26 }, () => 0);
  const result = [];
  let repeat = 1;

  const getCode = letter => letter.charCodeAt(0) - BASE_CODE;
  const getLargestCode = previousCode => {
    for (let code = 25; code >= 0; code--) {
      const count = letters[code];

      if (!count) continue;
      repeat = code === previousCode ? repeat + 1 : 1;

      if (repeat > repeatLimit) continue;
      return code;
    }
    return -1;
  };

  for (const letter of s) {
    letters[getCode(letter)] += 1;
  }

  for (let index = 0; index < n; index++) {
    const code = getLargestCode(result[index - 1]);

    if (code < 0) break;
    letters[code] -= 1;
    result.push(code);
  }

  return result.reduce((word, code) => {
    return `${word}${String.fromCharCode(code + BASE_CODE)}`;
  }, '');
};