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

Solution 1

Python Code
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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

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

Go Code
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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
}

TypeScript Code
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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;
}

Rust Code
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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
    }
}