LeetCode-in-Java

942. DI String Match

Easy

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'.

Solution

public class Solution {
    public int[] diStringMatch(String s) {
        int[] arr = new int[s.length() + 1];
        int max = s.length();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == 'D') {
                arr[i] = max;
                max--;
            }
        }
        for (int i = s.length() - 1; i >= 0 && max > 0; i--) {
            if (s.charAt(i) == 'I' && arr[i + 1] == 0) {
                arr[i + 1] = max;
                max--;
            }
        }
        for (int i = 0; i < arr.length && max > 0; i++) {
            if (arr[i] == 0) {
                arr[i] = max;
                max--;
            }
        }

        return arr;
    }
}

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