Find Permutation

Similar Problems:

- Leetcode: Next Permutation
- Leetcode: Positions of Large Groups
- Tag: #inspiring, #combination, #padplaceholder

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

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 // DDIID // 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 }