LeetCode in Kotlin

3270. Find the Key of the Numbers

Easy

You are given three positive integers num1, num2, and num3.

The key of num1, num2, and num3 is defined as a four-digit number such that:

• Initially, if any number has less than four digits, it is padded with leading zeros.
• The ith digit (1 <= i <= 4) of the key is generated by taking the smallest digit among the ith digits of num1, num2, and num3.

Return the key of the three numbers without leading zeros (if any).

Example 1:

Input: num1 = 1, num2 = 10, num3 = 1000

Output: 0

Explanation:

On padding, num1 becomes "0001", num2 becomes "0010", and num3 remains "1000".

• The 1st digit of the key is min(0, 0, 1).
• The 2nd digit of the key is min(0, 0, 0).
• The 3rd digit of the key is min(0, 1, 0).
• The 4th digit of the key is min(1, 0, 0).

Hence, the key is "0000", i.e. 0.

Example 2:

Input: num1 = 987, num2 = 879, num3 = 798

Output: 777

Example 3:

Input: num1 = 1, num2 = 2, num3 = 3

Output: 1

Constraints:

• 1 <= num1, num2, num3 <= 9999

Solution

import kotlin.math.min

class Solution {
    fun generateKey(num1: Int, num2: Int, num3: Int): Int {
        val s1 = (
            min(
                num1 / 1000 % 10,
                min(
                    num2 / 1000 % 10,
                    num3 / 1000 % 10,
                ),
            ) * 1000
            )
        val s2 = (
            min(
                num1 / 100 % 10,
                min(
                    num2 / 100 % 10,
                    num3 / 100 % 10,
                ),
            ) * 100
            )
        val s3 =
            (
                min(
                    num1 / 10 % 10,
                    min(
                        num2 / 10 % 10,
                        num3 / 10 % 10,
                    ),
                ) * 10
                )
        val s4 = min(
            num1 % 10,
            min(num2 % 10, num3 % 10),
        )
        return s1 + s2 + s3 + s4
    }
}

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