# 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]
```

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
}
```

Share It, If You Like It.