# 942. DI String Match

## Description

A permutation perm of n + 1 integers of all the integers in the range [0, n] can be represented as a string s of length n 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 permutation perm and return it. If there are multiple valid permutations perm, return any of them.

Example 1:

Input: s = "IDID"
Output: [0,4,1,3,2]


Example 2:

Input: s = "III"
Output: [0,1,2,3]


Example 3:

Input: s = "DDI"
Output: [3,2,0,1]


Constraints:

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

## Solutions

• class Solution {
public int[] diStringMatch(String s) {
int n = s.length();
int low = 0, high = n;
int[] ans = new int[n + 1];
for (int i = 0; i < n; i++) {
if (s.charAt(i) == 'I') {
ans[i] = low++;
} else {
ans[i] = high--;
}
}
ans[n] = low;
return ans;
}
}


• class Solution {
public:
vector<int> diStringMatch(string s) {
int n = s.size();
int low = 0, high = n;
vector<int> ans(n + 1);
for (int i = 0; i < n; ++i) {
if (s[i] == 'I') {
ans[i] = low++;
} else {
ans[i] = high--;
}
}
ans[n] = low;
return ans;
}
};


• class Solution:
def diStringMatch(self, s: str) -> List[int]:
n = len(s)
low, high = 0, n
ans = []
for i in range(n):
if s[i] == 'I':
ans.append(low)
low += 1
else:
ans.append(high)
high -= 1
ans.append(low)
return ans


• func diStringMatch(s string) []int {
n := len(s)
low, high := 0, n
var ans []int
for i := 0; i < n; i++ {
if s[i] == 'I' {
ans = append(ans, low)
low++
} else {
ans = append(ans, high)
high--
}
}
ans = append(ans, low)
return ans
}

• function diStringMatch(s: string): number[] {
const n = s.length;
const res = new Array(n + 1);
let low = 0;
let high = n;
for (let i = 0; i < n; i++) {
if (s[i] === 'I') {
res[i] = low++;
} else {
res[i] = high--;
}
}
res[n] = low;
return res;
}


• impl Solution {
pub fn di_string_match(s: String) -> Vec<i32> {
let s = s.as_bytes();
let n = s.len();
let mut res = Vec::with_capacity(n + 1);
let (mut low, mut high) = (-1, (n + 1) as i32);
for i in 0..n {
res.push(
if s[i] == b'I' {
low += 1;
low
} else {
high -= 1;
high
}
);
}
res.push(low + 1);
res
}
}