LeetCode in Kotlin

3572. Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values

Medium

You are given two integer arrays x and y, each of length n. You must choose three distinct indices i, j, and k such that:

• x[i] != x[j]
• x[j] != x[k]
• x[k] != x[i]

Your goal is to maximize the value of y[i] + y[j] + y[k] under these conditions. Return the maximum possible sum that can be obtained by choosing such a triplet of indices.

If no such triplet exists, return -1.

Example 1:

Input: x = [1,2,1,3,2], y = [5,3,4,6,2]

Output: 14

Explanation:

• Choose i = 0 (x[i] = 1, y[i] = 5), j = 1 (x[j] = 2, y[j] = 3), k = 3 (x[k] = 3, y[k] = 6).
• All three values chosen from x are distinct. 5 + 3 + 6 = 14 is the maximum we can obtain. Hence, the output is 14.

Example 2:

Input: x = [1,2,1,2], y = [4,5,6,7]

Output: -1

Explanation:

• There are only two distinct values in x. Hence, the output is -1.

Constraints:

• n == x.length == y.length
• 3 <= n <= 105
• 1 <= x[i], y[i] <= 106

Solution

class Solution {
    fun maxSumDistinctTriplet(x: IntArray, y: IntArray): Int {
        var index = -1
        var max = -1
        var sum = 0
        for (i in y.indices) {
            if (y[i] > max) {
                max = y[i]
                index = i
            }
        }
        sum += max
        if (max == -1) {
            return -1
        }
        var index2 = -1
        max = -1
        for (i in y.indices) {
            if (y[i] > max && x[i] != x[index]) {
                max = y[i]
                index2 = i
            }
        }
        sum += max
        if (max == -1) {
            return -1
        }
        max = -1
        for (i in y.indices) {
            if (y[i] > max && x[i] != x[index] && x[i] != x[index2]) {
                max = y[i]
            }
        }
        if (max == -1) {
            return -1
        }
        sum += max
        return sum
    }
}

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