LeetCode in Kotlin

3168. Minimum Number of Chairs in a Waiting Room

Easy

You are given a string s. Simulate events at each second i:

• If s[i] == 'E', a person enters the waiting room and takes one of the chairs in it.
• If s[i] == 'L', a person leaves the waiting room, freeing up a chair.

Return the minimum number of chairs needed so that a chair is available for every person who enters the waiting room given that it is initially empty.

Example 1:

Input: s = “EEEEEEE”

Output: 7

Explanation:

After each second, a person enters the waiting room and no person leaves it. Therefore, a minimum of 7 chairs is needed.

Example 2:

Input: s = “ELELEEL”

Output: 2

Explanation:

Let’s consider that there are 2 chairs in the waiting room. The table below shows the state of the waiting room at each second.

Second	Event	People in the Waiting Room	Available Chairs
0	Enter	1	1
1	Leave	0	2
2	Enter	1	1
3	Leave	0	2
4	Enter	1	1
5	Enter	2	0
6	Leave	1	1

Example 3:

Input: s = “ELEELEELLL”

Output: 3

Explanation:

Let’s consider that there are 3 chairs in the waiting room. The table below shows the state of the waiting room at each second.

Second	Event	People in the Waiting Room	Available Chairs
0	Enter	1	2
1	Leave	0	3
2	Enter	1	2
3	Enter	2	1
4	Leave	1	2
5	Enter	2	1
6	Enter	3	0
7	Leave	2	1
8	Leave	1	2
9	Leave	0	3

Constraints:

• 1 <= s.length <= 50
• s consists only of the letters 'E' and 'L'.
• s represents a valid sequence of entries and exits.

Solution

import kotlin.math.max

class Solution {
    fun minimumChairs(s: String): Int {
        var count = 0
        var ans = Int.MIN_VALUE
        for (ch in s.toCharArray()) {
            if (ch == 'E') {
                count++
                ans = max(ans, count)
            } else {
                count--
            }
        }
        return ans
    }
}

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