find-permutation.js

/**
 * Find Permutation
 *
 * By now, you are given a secret signature consisting of character 'D' and 'I'.
 * 'D' represents a decreasing relationship between two numbers,
 * 'I' represents an increasing relationship between two numbers.
 * And our secret signature was constructed by a special integer array,
 * which contains uniquely all the different number from 1 to n (n is the length of the secret signature plus 1).
 * For example, the secret signature "DI" can be constructed by array [2,1,3] or [3,1,2],
 * but won't be constructed by array [3,2,4] or [2,1,3,4], which are both illegal constructing special string that
 * can't represent the "DI" secret signature.
 *
 * On the other hand, now your job is to find the lexicographically smallest permutation of [1, 2, ... n]
 * could refer to the given secret signature in the input.
 *
 * Example 1:
 *
 * Input: "I"
 * Output: [1,2]
 *
 * Explanation: [1,2] is the only legal initial spectial string can construct secret signature "I",
 * where the number 1 and 2 construct an increasing relationship.
 *
 * Example 2:
 *
 * Input: "DI"
 * Output: [2,1,3]
 *
 * Explanation: Both [2,1,3] and [3,1,2] can construct the secret signature "DI",
 * but since we want to find the one with the smallest lexicographical permutation, you need to output [2,1,3]
 *
 * Note:
 *
 * - The input string will only contain the character 'D' and 'I'.
 * - The length of input string is a positive integer and will not exceed 10,000
 */

/**
 * @param {string} s
 * @return {number[]}
 */
const findPermutation = s => {
  const result = Array(s.length + 1);
  const stack = [];

  let j = 0;
  for (let i = 0; i < s.length; i++) {
    stack.push(i + 1);

    if (s[i] === 'I') {
      while (stack.length) {
        result[j++] = stack.pop();
      }
    }
  }

  stack.push(s.length + 1);
  while (stack.length) {
    result[j++] = stack.pop();
  }

  return result;
};

export { findPermutation };