# 484. Find Permutation

Find Permutation - LeetCode

A permutation perm of n integers of all the integers in the range [1, n] can be represented as a string s of length n - 1 where:

``````s[i] == 'I' if perm[i] < perm[i + 1], and
s[i] == 'D' if perm[i] > perm[i + 1].Given a string s, reconstruct the lexicographically smallest permutation perm and return it.
``````

Example 1: Input: s = “I” Output: [1,2] Explanation: [1,2] is the only legal permutation that can represented by s, where the number 1 and 2 construct an increasing relationship. Example 2: Input: s = “DI” Output: [2,1,3] Explanation: Both [2,1,3] and [3,1,2] can be represented as “DI”, but since we want to find the smallest lexicographical permutation, you should return [2,1,3]

Constraints:

``````1 <= s.length <= 105
s[i] is either 'I' or 'D'.
``````

• code
``````public class Solution {
public int[] findPermutation(String s) {
int[] res = new int[s.length() + 1];
ArrayDeque <Integer> stack = new ArrayDeque<>();
int j = 0;
for (int i = 1; i <= s.length(); i++) {
if (s.charAt(i - 1) == 'I') {
stack.push(i);
while (!stack.isEmpty())
res[j++] = stack.pop();
} else
stack.push(i);
}
stack.push(s.length() + 1);
while (!stack.isEmpty())
res[j++] = stack.pop();
return res;
}
}
``````
• code
``````class Solution:
def findPermutation(self, s: str) -> List[int]:
s = s + "I"
res = [0] * len(s)
l = 0

for r in range(1, len(s)+1):
if s[r-1] == 'I': # r = 3,
for _ in range(r - l): # 0 <= 3, 1<=2 , 2<=1
res[l] = r # res[0]=3, res[1]=2, res[2]=1
l += 1
r -= 1
return res

``````