LeetCode in Kotlin
1262. Greatest Sum Divisible by Three
Medium
Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three.
Example 1:
Input: nums = [3,6,5,1,8]
Output: 18
Explanation: Pick numbers 3, 6, 1 and 8 their sum is 18 (maximum sum divisible by 3).
Example 2:
Input: nums = [4]
Output: 0
Explanation: Since 4 is not divisible by 3, do not pick any number.
Example 3:
Input: nums = [1,2,3,4,4]
Output: 12
Explanation: Pick numbers 1, 3, 4 and 4 their sum is 12 (maximum sum divisible by 3).
Constraints:
1 <= nums.length <= 4 * 10^41 <= nums[i] <= 10^4
Solution
class Solution {
fun maxSumDivThree(nums: IntArray): Int {
var sum = 0
var smallestNumWithMod1 = 10001
var secondSmallestNumWithMod1 = 10002
var smallestNumWithMod2 = 10001
var secondSmallestNumWithMod2 = 10002
for (i in nums) {
sum += i
if (i % 3 == 1) {
if (i <= smallestNumWithMod1) {
val temp = smallestNumWithMod1
smallestNumWithMod1 = i
secondSmallestNumWithMod1 = temp
} else if (i < secondSmallestNumWithMod1) {
secondSmallestNumWithMod1 = i
}
}
if (i % 3 == 2) {
if (i <= smallestNumWithMod2) {
val temp = smallestNumWithMod2
smallestNumWithMod2 = i
secondSmallestNumWithMod2 = temp
} else if (i < secondSmallestNumWithMod2) {
secondSmallestNumWithMod2 = i
}
}
}
if (sum % 3 == 0) {
return sum
} else if (sum % 3 == 2) {
val min = Math.min(smallestNumWithMod2, smallestNumWithMod1 + secondSmallestNumWithMod1)
return sum - min
} else if (sum % 3 == 1) {
val min = Math.min(smallestNumWithMod1, smallestNumWithMod2 + secondSmallestNumWithMod2)
return sum - min
}
return sum
}
}