Leetcode: Find Permutation

Find Permutation

Similar Problems:

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]


  • 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

Github: code.dennyzhang

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

  • Solution:
// Blog link: https://code.dennyzhang.com/find-permutation
// Basic Ideas:
// Seperate by continuous D
// 123456 -> (123)4(56) -> (321)4(65)
// Complexity: Time O(n), Space O(n)
func findPermutation(s string) []int {
    res := make([]int, len(s)+1)
    for i, _ := range res { res[i] = i+1 }

    start := -1
    // pad one "I" to the end, thus we can deal with the last group of "D"
    for i, ch := range fmt.Sprintf(s, "I") {
        if ch == 'D' {
            if start == -1 { start = i }
        } else {
            // find a continous group of "D": s[start: i]
            if start != -1 {
                l, r := start, i
                for l<r {
                    res[l], res[r] = res[r], res[l]
                    l, r = l+1, r-1
                start = -1
    return res

Share It, If You Like It.

Leave a Reply

Your email address will not be published.