LC 1899 - Merge Triplets to Form Target Triplet

Question

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 <= 105
• triplets[i].length == target.length == 3
• 1 <= ai, bi, ci, x, y, z <= 1000

Links

Question here and solution here

Solution

concept

we first filter out all the triplets that have at least one element greater than the target -> we can’t form the target triplet with these then we check if target exist in the remaining triplets -> step 1 guarantees that we can form the target triplet with the remaining triplets since all elements are less than or equal to the target

code

class Solution:
    def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:
        check = {0: False,
                 1: False,
                 2: False}
        
        for triplet in triplets:
            if triplet[0] > target[0] or triplet[1] > target[1] or triplet[2] > target[2]:
                continue
            if triplet[0] == target[0]:
                check[0] = True
            if triplet[1] == target[1]:
                check[1] = True
            if triplet[2] == target[2]:
                check[2] = True
        
        return all(value == True for value in check.values())
    

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

        for t in triplets:
            if t[0] > target[0] or t[1] > target[1] or t[2] > target[2]:
                continue
            for i, v in enumerate(t):
                if v == target[i]:
                    good.add(i)
        return len(good) == 3

Complexity

time: $O(n)$
space: $O(1)$