LeetCode in Kotlin

3986. Number of Elapsed Seconds Between Two Times

Easy

You are given two valid times startTime and endTime, each represented as a string in the format "HH:MM:SS".

Return the number of seconds that have elapsed from startTime to endTime.

Example 1:

Input: startTime = “01:00:00”, endTime = “01:00:25”

Output: 25

Explanation:

endTime is 25 seconds ahead of startTime.

Example 2:

Input: startTime = “12:34:56”, endTime = “13:00:00”

Output: 1504

Explanation:

endTime is 25 minutes and 4 seconds ahead of startTime, which equals 1504 seconds.

Constraints:

• startTime.length == 8
• endTime.length == 8
• startTime and endTime are valid times in the format "HH:MM:SS"
• 00 <= HH <= 23
• 00 <= MM <= 59
• 00 <= SS <= 59
• endTime is not earlier than startTime

Solution

import kotlin.math.pow

class Solution {
    fun secondsBetweenTimes(startTime: String, endTime: String): Int {
        var startSec = 0
        var endSec = 0
        var sSec = 0
        var eSec = 0
        var colon = 2
        for (i in 0..7) {
            if (startTime[i] == ':') {
                startSec = (startSec + sSec * 60.0.pow(colon.toDouble())).toInt()
                endSec = (endSec + eSec * 60.0.pow(colon.toDouble())).toInt()
                colon--
                sSec = 0
                eSec = 0
                continue
            }
            sSec = sSec * 10 + (startTime[i].code - '0'.code)
            eSec = eSec * 10 + (endTime[i].code - '0'.code)
        }
        startSec = (startSec + sSec * 60.0.pow(colon.toDouble())).toInt()
        endSec = (endSec + eSec * 60.0.pow(colon.toDouble())).toInt()
        return endSec - startSec
    }
}

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