LeetCode in Kotlin
3780. Maximum Sum of Three Numbers Divisible by Three
Medium
You are given an integer array nums.
Your task is to choose exactly three integers from nums such that their sum is divisible by three.
Return the maximum possible sum of such a triplet. If no such triplet exists, return 0.
Example 1:
Input: nums = [4,2,3,1]
Output: 9
Explanation:
The valid triplets whose sum is divisible by 3 are:
(4, 2, 3)with a sum of4 + 2 + 3 = 9.(2, 3, 1)with a sum of2 + 3 + 1 = 6.
Thus, the answer is 9.
Example 2:
Input: nums = [2,1,5]
Output: 0
Explanation:
No triplet forms a sum divisible by 3, so the answer is 0.
Constraints:
3 <= nums.length <= 1051 <= nums[i] <= 105
Solution
import kotlin.math.max
class Solution {
fun maximumSum(nums: IntArray): Int {
val group: Array<IntArray> = Array<IntArray>(3) { IntArray(3) }
for (num in nums) {
val m = num % 3
var max1 = group[m][0]
var max2 = group[m][1]
var max3 = group[m][2]
if (num >= max1) {
max3 = max2
max2 = max1
max1 = num
} else if (num >= max2) {
max3 = max2
max2 = num
} else if (num > max3) {
max3 = num
}
group[m][0] = max1
group[m][1] = max2
group[m][2] = max3
}
var res = 0
for (g in group) {
var sum = 0
var i = 0
while (i < 3) {
if (g[i] == 0) {
sum = 0
break
}
sum += g[i]
i += 1
}
res = max(res, sum)
}
var max = 0
for (g in group) {
if (g[0] == 0) {
max = 0
break
}
max += g[0]
}
res = max(res, max)
return res
}
}