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1899. Merge Triplets to Form Target Triplet

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

class Solution:
    def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:
        # Initialize three variables to track the maximum values of each index
        max_x, max_y, max_z = 0, 0, 0

        # Loop through each triplet
        for a, b, c in triplets:
            # Check if the current triplet can potentially contribute without exceeding the target
            if a <= target[0] and b <= target[1] and c <= target[2]:
                # Update the maximum values seen so far that are valid
                max_x = max(max_x, a)
                max_y = max(max_y, b)
                max_z = max(max_z, c)

        # Check if the collected maximum values meet or exceed the target values
        return max_x == target[0] and max_y == target[1] and max_z == target[2]

Thoughts

Easier than the problem looks.

Time Complexity

O(n), where n is the number of triplets. We only need one pass through the list of triplets.

Space Complexity

O(1). We are using a constant amount of extra space regardless of the input size.