LeetCode in Kotlin

1871. Jump Game VII

Medium

You are given a 0-indexed binary string s and two integers minJump and maxJump. In the beginning, you are standing at index 0, which is equal to '0'. You can move from index i to index j if the following conditions are fulfilled:

• i + minJump <= j <= min(i + maxJump, s.length - 1), and
• s[j] == '0'.

Return true if you can reach index s.length - 1 in s, or false otherwise.

Example 1:

Input: s = “011010”, minJump = 2, maxJump = 3

Output: true

Explanation:

In the first step, move from index 0 to index 3.

In the second step, move from index 3 to index 5.

Example 2:

Input: s = “01101110”, minJump = 2, maxJump = 3

Output: false

Constraints:

• 2 <= s.length <= 105
• s[i] is either '0' or '1'.
• s[0] == '0'
• 1 <= minJump <= maxJump < s.length

Solution

class Solution {
    fun canReach(s: String, minJump: Int, maxJump: Int): Boolean {
        var j = 0
        val n = s.length
        val li = s.toCharArray()
        var i = 0
        while (i < n) {
            // o == ok
            if (i == 0 || li[i] == 'o') {
                j = Math.max(j, i + minJump)
                while (j < Math.min(n, i + maxJump + 1)) {
                    if (li[j] == '0') {
                        li[j] = 'o'
                    }
                    j++
                }
            }
            if (j > n) {
                break
            }
            i++
        }
        return li[n - 1] == 'o'
    }
}

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