Merge Triplets to Form Target Triplet

Problem

A triplet is an array of three integers. You are given a 2D integer array triplets, where triplets[i] = [ai, bi, ci] describes the ith triplet. You are also given an integer array target = [x, y, z] that describes the triplet you want to obtain.

To obtain target, you may apply the following operation on triplets any number of times (possibly zero):

Choose two indices (0-indexed) i and j (i != j) and update triplets[j] to become [max(ai, aj), max(bi, bj), max(ci, cj)].
- For example, if triplets[i] = [2, 5, 3] and triplets[j] = [1, 7, 5], triplets[j] will be updated to [max(2, 1), max(5, 7), max(3, 5)] = [2, 7, 5].

Return true if it is possible to obtain the target triplet [x, y, z] as an element of triplets, or false otherwise.

Example 1:

Input: triplets = [[2,5,3],[1,8,4],[1,7,5]], target = [2,7,5]
Output: true
Explanation: Perform the following operations:
- Choose the first and last triplets [[2,5,3],[1,8,4],[1,7,5]]. Update the last triplet to be [max(2,1), max(5,7), max(3,5)] = [2,7,5]. triplets = [[2,5,3],[1,8,4],[2,7,5]]
The target triplet [2,7,5] is now an element of triplets.

Example 2:

Input: triplets = [[3,4,5],[4,5,6]], target = [3,2,5]
Output: false
Explanation: It is impossible to have [3,2,5] as an element because there is no 2 in any of the triplets.

Example 3:

Input: triplets = [[2,5,3],[2,3,4],[1,2,5],[5,2,3]], target = [5,5,5]
Output: true
Explanation: Perform the following operations:
- Choose the first and third triplets [[2,5,3],[2,3,4],[1,2,5],[5,2,3]]. Update the third triplet to be [max(2,1), max(5,2), max(3,5)] = [2,5,5]. triplets = [[2,5,3],[2,3,4],[2,5,5],[5,2,3]].
- Choose the third and fourth triplets [[2,5,3],[2,3,4],[2,5,5],[5,2,3]]. Update the fourth triplet to be [max(2,5), max(5,2), max(5,3)] = [5,5,5]. triplets = [[2,5,3],[2,3,4],[2,5,5],[5,5,5]].
The target triplet [5,5,5] is now an element of triplets.

Constraints:

1 <= triplets.length <= 10^5
triplets[i].length == target.length == 3
1 <= ai, bi, ci, x, y, z <= 1000

Solution

Approach: Greedy Filtering

The key insight is that we can only use triplets where all values are ≤ target values. Then check if we can achieve each target value from valid triplets.

Implementation

class Solution:
    def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:
        good = set()

        for triplet in triplets:
            # Skip triplets with any value > target
            if (triplet[0] > target[0] or
                triplet[1] > target[1] or
                triplet[2] > target[2]):
                continue

            # Track which target indices this triplet can contribute to
            for i in range(3):
                if triplet[i] == target[i]:
                    good.add(i)

        # Check if we found all three target values
        return len(good) == 3

Approach 2: Track Maximum Per Position

class Solution:
    def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:
        current = [0, 0, 0]

        for triplet in triplets:
            # Only merge if all values are within bounds
            if all(triplet[i] <= target[i] for i in range(3)):
                for i in range(3):
                    current[i] = max(current[i], triplet[i])

        return current == target

Complexity Analysis

Time Complexity: O(n), where n is number of triplets. Single pass through array.
Space Complexity: O(1), constant extra space (set has max size 3).

Key Insights

Max Operation: The merge operation takes maximum of each position independently.
Monotonic Growth: Once we merge, values can only increase (or stay same), never decrease.
Filtering Invalid: If any triplet has a value > corresponding target value, we can never use it (would make result too large).
Independent Positions: Each position in the target can be achieved independently from valid triplets.
Existence Check: We just need to check if each target value exists in at least one valid triplet.
Greedy Validity: We can greedily consider all valid triplets - taking max at each position will give best possible result.