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# DI String Match

## Description

Given a string `S` that **only** contains "I" (increase) or "D" (decrease), let `N = S.length`.

Return **any** permutation `A` of `[0, 1, ..., N]` such that for all `i = 0, ..., N-1`:

* If `S[i] == "I"`, then `A[i] < A[i+1]`
* If `S[i] == "D"`, then `A[i] > A[i+1]`

**Example 1:**

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

**Example 2:**

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

**Example 3:**

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

**Note:**

1. `1 <= S.length <= 10000`
2. `S` only contains characters `"I"` or `"D"`.

## Solutions

Two passes

```cpp
class Solution {
public:
    vector<int> diStringMatch(string S) {
        int n = 0;
        vector<int> res(S.size() + 1);
        for (int i = S.size() - 1; i >= 0; --i)
            if (S[i] == 'D')
                res[i + 1] = n++;
        res[0] = n++;
        for (int i = 0; i < S.size(); ++i)
            if (S[i] == 'I')
                res[i + 1] = n++;
        return res;
    }
};
```

One pass:

```cpp
class Solution {
public:
    vector<int> diStringMatch(string S) {
        int lo = 0, hi = S.size();
        vector<int> res(S.size() + 1);
        for (int i = S.size() - 1; i >= 0; --i)
            res[i + 1] = S[i] == 'I' ? hi-- : lo++;
        res[0] = lo;
        return res;
    }
};
```

Left to right:

```cpp
class Solution {
public:
    vector<int> diStringMatch(string S) {
        int lo = 0, hi = S.size();
        vector<int> res(S.size() + 1);
        for (int i = 0; i < S.size(); ++i)
            res[i] = S[i] == 'I' ? lo++ : hi--;
        res.back() = lo;
        return res;
    }
};
```