LeetCode #3572 — MEDIUM

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

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy

Example 1

[1,2,1,3,2]
[5,3,4,6,2]

Example 2

[1,2,1,2]
[4,5,6,7]

Core Insight

What unlocks the optimal approach

For each unique <code>x</code>, keep only the maximum <code>y</code>; all other pairs with the same <code>x</code> are redundant.
Sort the pairs by <code>x</code> so that identical <code>x</code> values form contiguous blocks, then take the maximum <code>y</code> from each block.
Alternatively, use a map (or dictionary) from <code>x</code> to its largest <code>y</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3572: Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values
class Solution {
    public int maxSumDistinctTriplet(int[] x, int[] y) {
        int n = x.length;
        int[][] arr = new int[n][0];
        for (int i = 0; i < n; i++) {
            arr[i] = new int[] {x[i], y[i]};
        }
        Arrays.sort(arr, (a, b) -> b[1] - a[1]);
        int ans = 0;
        Set<Integer> vis = new HashSet<>();
        for (int i = 0; i < n; ++i) {
            int a = arr[i][0], b = arr[i][1];
            if (vis.add(a)) {
                ans += b;
                if (vis.size() == 3) {
                    return ans;
                }
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #3572: Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values
func maxSumDistinctTriplet(x []int, y []int) int {
	n := len(x)
	arr := make([][2]int, n)
	for i := 0; i < n; i++ {
		arr[i] = [2]int{x[i], y[i]}
	}
	sort.Slice(arr, func(i, j int) bool {
		return arr[i][1] > arr[j][1]
	})
	ans := 0
	vis := make(map[int]bool)
	for i := 0; i < n; i++ {
		a, b := arr[i][0], arr[i][1]
		if !vis[a] {
			vis[a] = true
			ans += b
			if len(vis) == 3 {
				return ans
			}
		}
	}
	return -1
}

# Accepted solution for LeetCode #3572: Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values
class Solution:
    def maxSumDistinctTriplet(self, x: List[int], y: List[int]) -> int:
        arr = [(a, b) for a, b in zip(x, y)]
        arr.sort(key=lambda x: -x[1])
        vis = set()
        ans = 0
        for a, b in arr:
            if a in vis:
                continue
            vis.add(a)
            ans += b
            if len(vis) == 3:
                return ans
        return -1

// Accepted solution for LeetCode #3572: Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values
impl Solution {
    pub fn max_sum_distinct_triplet(x: Vec<i32>, y: Vec<i32>) -> i32 {
        let n = x.len();
        let mut arr: Vec<(i32, i32)> = (0..n).map(|i| (x[i], y[i])).collect();
        arr.sort_by(|a, b| b.1.cmp(&a.1));
        let mut vis = std::collections::HashSet::new();
        let mut ans = 0;
        for (a, b) in arr {
            if vis.insert(a) {
                ans += b;
                if vis.len() == 3 {
                    return ans;
                }
            }
        }
        -1
    }
}

// Accepted solution for LeetCode #3572: Maximize Y‑Sum by Picking a Triplet of Distinct X‑Values
function maxSumDistinctTriplet(x: number[], y: number[]): number {
    const n = x.length;
    const arr: [number, number][] = [];
    for (let i = 0; i < n; i++) {
        arr.push([x[i], y[i]]);
    }
    arr.sort((a, b) => b[1] - a[1]);
    const vis = new Set<number>();
    let ans = 0;
    for (let i = 0; i < n; i++) {
        const [a, b] = arr[i];
        if (!vis.has(a)) {
            vis.add(a);
            ans += b;
            if (vis.size === 3) {
                return ans;
            }
        }
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.