LeetCode in Kotlin
2748. Number of Beautiful Pairs
Easy
You are given a 0-indexed integer array nums. A pair of indices i, j where 0 <= i < j < nums.length is called beautiful if the first digit of nums[i] and the last digit of nums[j] are coprime.
Return the total number of beautiful pairs in nums.
Two integers x and y are coprime if there is no integer greater than 1 that divides both of them. In other words, x and y are coprime if gcd(x, y) == 1, where gcd(x, y) is the greatest common divisor of x and y.
Example 1:
Input: nums = [2,5,1,4]
Output: 5
Explanation: There are 5 beautiful pairs in nums:
When i = 0 and j = 1: the first digit of nums[0] is 2, and the last digit of nums[1] is 5. We can confirm that 2 and 5 are coprime, since gcd(2,5) == 1.
When i = 0 and j = 2: the first digit of nums[0] is 2, and the last digit of nums[2] is 1. Indeed, gcd(2,1) == 1.
When i = 1 and j = 2: the first digit of nums[1] is 5, and the last digit of nums[2] is 1. Indeed, gcd(5,1) == 1.
When i = 1 and j = 3: the first digit of nums[1] is 5, and the last digit of nums[3] is 4. Indeed, gcd(5,4) == 1.
When i = 2 and j = 3: the first digit of nums[2] is 1, and the last digit of nums[3] is 4. Indeed, gcd(1,4) == 1.
Thus, we return 5.
Example 2:
Input: nums = [11,21,12]
Output: 2
Explanation: There are 2 beautiful pairs:
When i = 0 and j = 1: the first digit of nums[0] is 1, and the last digit of nums[1] is 1. Indeed, gcd(1,1) == 1.
When i = 0 and j = 2: the first digit of nums[0] is 1, and the last digit of nums[2] is 2. Indeed, gcd(1,2) == 1.
Thus, we return 2.
Constraints:
2 <= nums.length <= 1001 <= nums[i] <= 9999nums[i] % 10 != 0
Solution
class Solution {
fun countBeautifulPairs(nums: IntArray): Int {
var beautifulPairs = 0
var i = 0
var j = 1
while (i < nums.size - 1) {
val firstDigit = getFirstDigit(nums[i])
while (j < nums.size) {
val lastDigit = nums[j] % 10
val botDigitsAreEqualAndNot1 = firstDigit == lastDigit && firstDigit > 1
val botDigitsAreDivisibleBy2 = firstDigit % 2 == 0 && lastDigit % 2 == 0
val botDigitsAreDivisibleBy3 = firstDigit % 3 == 0 && lastDigit % 3 == 0
if (!botDigitsAreEqualAndNot1 && !botDigitsAreDivisibleBy2 && !botDigitsAreDivisibleBy3) {
beautifulPairs++
}
j++
}
i++
j = i + 1
}
return beautifulPairs
}
private fun getFirstDigit(num: Int): Int {
var n = num
var digit = 0
while (n > 0) {
digit = n % 10
n /= 10
}
return digit
}
}