LeetCode in Kotlin

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

class Solution {
    fun diStringMatch(s: String): IntArray {
        val arr = IntArray(s.length + 1)
        var max = s.length
        for (i in s.indices) {
            if (s[i] == 'D') {
                arr[i] = max
                max--
            }
        }
        run {
            var i = s.length - 1
            while (i >= 0 && max > 0) {
                if (s[i] == 'I' && arr[i + 1] == 0) {
                    arr[i + 1] = max
                    max--
                }
                i--
            }
        }
        var i = 0
        while (i < arr.size && max > 0) {
            if (arr[i] == 0) {
                arr[i] = max
                max--
            }
            i++
        }
        return arr
    }
}

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