Merge Triplets to Form Target Triplet - Practice Coding Problems

Merge Triplets to Form Target Triplet - Problem

Array Greedy Medium

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.

Input & Output

Example 1 — Basic Merge

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

› Output: true

💡 Note: Merge triplets[0] and triplets[2]: [max(2,1), max(5,7), max(3,5)] = [2,7,5] which equals target

Example 2 — Impossible Target

$ Input: triplets = [[3,4,5],[4,5,6]], target = [3,2,5]

› Output: false

💡 Note: Target[1] = 2, but all triplets have second element ≥ 4, so we can never achieve 2 in position 1

Example 3 — Direct Match

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

› Output: true

💡 Note: Can achieve [5,5,5] by merging multiple triplets: max positions give us 5,5,5

Constraints

1 ≤ triplets.length ≤ 10⁵
triplets[i].length == target.length == 3
1 ≤ a_i, b_i, c_i, x, y, z ≤ 1000

Visualization

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Understanding the Visualization

Input

Given triplets and target triplet

Merge

Combine triplets using max operation

Check

Verify if target is achievable

Key Takeaway

🎯 Key Insight: Only triplets with all elements ≤ target can contribute to the solution

Asked in

G Google 12 M Microsoft 8

Brief HTML summary: The key insight is that any triplet with an element exceeding the target is useless. Best approach is greedy filtering - only consider valid triplets and check if all target components are achievable. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Greedy Filtering	O(n)	O(1)	Filter triplets that can't contribute to target, then check if target is reachable
Brute Force Simulation	O(2^n)	O(2^n)	Try all possible merge operations until target is found or no more progress

Greedy Filtering — Algorithm Steps

Step 1: Filter triplets - keep only those where all elements ≤ target
Step 2: Check if we can achieve each target component from filtered triplets

Visualization

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Step-by-Step Walkthrough

Filter Invalid

Remove triplets with any element > target

Check Components

Verify each target component is achievable

Final Check

All components must be achievable

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

bool solution(int triplets[][3], int n, int target[3]) {
    // Check if we can achieve each target component
    bool canAchieve[3] = {false, false, false};
    
    for (int i = 0; i < n; i++) {
        // Skip triplets that have any element > target
        if (triplets[i][0] > target[0] || triplets[i][1] > target[1] || triplets[i][2] > target[2]) {
            continue;
        }
        
        // Check if this triplet helps achieve any target component
        if (triplets[i][0] == target[0]) canAchieve[0] = true;
        if (triplets[i][1] == target[1]) canAchieve[1] = true;
        if (triplets[i][2] == target[2]) canAchieve[2] = true;
    }
    
    // We need all three components to be achievable
    return canAchieve[0] && canAchieve[1] && canAchieve[2];
}

int main() {
    // Example input
    int triplets[3][3] = {{2,5,3},{1,8,4},{1,7,5}};
    int target[3] = {2,7,5};
    int n = 3;
    
    printf(solution(triplets, n, target) ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through triplets to filter and check each component

✓ Linear Growth

Space Complexity

O(1)

Only store three boolean flags for target components

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Filtering — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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