Merge Triplets To Form Target Triplet Leetcode Problem 1899 [Python]

Now, we’re going to tackle a problem from LeetCode’s once latest contest: Merge Triplets To Form Target Triplet.

This problem falls into the medium difficulty category and belongs to the Greedy algorithm domain.

It’s a challenging puzzle that has been featured by tech giants like Facebook, Google, and Amazon.

If you’d like to dive into the details or practice it on your own, you can find the problem description and test cases on the official LeetCode website: Merge Triplets to Form Target Triplet.

Problem Overview

Let’s start by understanding the problem.

We are given a list of triplets, where each triplet is an array of three integers.

Each triplet is described as [ai, bi, ci], where ai, bi, and ci represent the values at three different positions in the triplet.

Our goal is to determine if we can obtain a specific target triplet [x, y, z] using the given triplets.

To achieve this, we are allowed to perform the following operation any number of times (including zero):

Choose two indices (0-indexed) i and j (i != j) from the list of triplets.
Update triplets[j] to become [max(ai, aj), max(bi, bj), max(ci, cj)].

In simpler terms, we can take two triplets, compare their values at each position, and choose the maximum value for each position to create a new triplet.

Our task is to return true if it is possible to obtain the target triplet [x, y, z] as one of the elements in the list of triplets.

Otherwise, we return false.

Let’s illustrate this with a few examples.

Example 1:

Input: triplets = [[2, 5, 3], [1, 8, 4], [1, 7, 5]], target = [2, 7, 5]
Output: true

Explanation: We can perform the following operations:

Choose the first and last triplets: [2, 5, 3] and [1, 7, 5].

Update the last triplet to become [max(2, 1), max(5, 7), max(3, 5)] = [2, 7, 5].

Now, the target triplet [2, 7, 5] is present in the list 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 triplet in the list that contains a 3 in the first position.

Example 3:

Input: triplets = [[2, 5, 3], [2, 3, 4], [1, 2, 5], [5, 2, 3]], target = [5, 5, 5]
Output: true

Explanation: We can perform the following operations:

Choose the first and third triplets: [2, 5, 3] and [1, 2, 5].

Update the third triplet to become [max(2, 1), max(5, 2), max(3, 5)] = [2, 5, 5].

Choose the third and fourth triplets: [2, 5, 5] and [5, 2, 3].

Update the fourth triplet to become [max(2, 5), max(5, 2), max(5, 3)] = [5, 5, 5].

Now, the target triplet [5, 5, 5] is present in the list of triplets.

Constraints:

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

Now that we have a clear understanding of the problem, let’s explore the approach to solving it efficiently.

Understanding the Constraints

Before we dive into the code, let’s take a moment to understand the constraints provided in the problem statement:

The number of triplets, triplets.length, can be as large as 105. This means we need an efficient solution that doesn’t have a high time complexity.
Each triplet has three values, and the target triplet also has three values.

These values range from 1 to 1000. Given these constraints, we need to design an algorithm that works efficiently for large input sizes.

Efficient Python Code Solution

Now, let’s implement an efficient Python solution for this problem.

View on GitHub ➹

def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -&gt; bool:
    # Create a set to keep track of valid positions in the target
    good = set()

    for t in triplets:
        # Check if any value in the triplet exceeds the corresponding target value
        if t[0] &gt; target[0] or t[1] &gt; target[1] or t[2] &gt; target[2]:
            continue
        # Check each position in the triplet
        for i, v in enumerate(t):
            # If the value matches the target, add the position to the set
            if v == target[i]:
                good.add(i)
    # If we can obtain all three target values, return true; otherwise, return false
    return len(good) == 3

Let’s break down this code step by step.

1. Set Initialization

We start by creating a set called good.

This set will help us keep track of which positions in the target triplet we can obtain using the given triplets.

The set good will store the indices (0, 1, or 2) that represent the positions in the target.

2. Filtering Invalid Triplets

We iterate through each triplet in the input list, triplets.

For each triplet, we check if any of its values exceed the corresponding values in the target.

If we find a value in the triplet that’s greater than the target value at the same position, we skip this triplet using the continue statement.

This is because such triplets cannot help us achieve the target.

3. Checking for Matching Values

For the remaining triplets (those that haven’t been skipped), we iterate through each position in the triplet using enumerate.

We check if the value in the triplet matches the value in the target at the same position.

If there is a match, it means this triplet contributes to achieving that position in the target, so we add the position (0, 1, or 2) to the good set.

4. Final Check

After processing all the triplets, we check the length of the good set.

If it contains all three indices (0,

1, and 2), it means we can obtain all three target values using the available triplets.

In this case, we return true.

If the length is less than 3, it means we can’t obtain all the target values, so we return false.

This algorithm guarantees an O(n) time complexity, where ‘n’ is the number of triplets in the input list.

It efficiently filters out invalid triplets and checks for matching values in each position, making it a straightforward and fast solution.

Reasoning Behind Our Approach

The key observation that makes this problem surprisingly easy is recognizing that any triplet with a value greater than the target at any position is useless for forming the target triplet.

These triplets will only push the values further from the target, making it impossible to reach the target.

By filtering out such triplets and focusing on the triplets with values equal to or less than the target at each position, we simplify the problem.

This reduction allows us to check if we can obtain each position’s value in the target from the remaining triplets.

If we can, it’s guaranteed that we can form the target triplet.

In the code, we keep track of which positions in the target we can achieve using the good set.

If we can obtain all three positions, we return true; otherwise, we return false.

Conclusion

In this blog post, we’ve explored the LeetCode problem Merge Triplets To Form Target Triplet We’ve examined the problem statement, understood its requirements, and discussed the efficient Python solution that solves it.

By making a key observation about which triplets are useful and filtering out the rest, we simplified the problem.

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Happy coding!