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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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

✓ Brute Force Simulation

⏱️ Time: O(2^n) Space: O(2^n)

Repeatedly merge all pairs of triplets and check if any result matches the target. Continue until target is found or no new triplets can be generated.

Greedy Filtering

⏱️ Time: O(n) Space: O(1)

Key insight: if any element in a triplet exceeds the corresponding target element, that triplet is useless. Filter valid triplets and check if we can build each target component.

Brute Force Simulation — Algorithm Steps

Step 1: Generate all possible merges of existing triplets
Step 2: Check if target exists in current set
Step 3: Repeat until no new triplets or target found

Visualization

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Step-by-Step Walkthrough

Initial State

Start with given triplets

Generate Merges

Create all possible merged triplets

Check Target

See if target exists in any generated triplet

Code -

solution.c — C

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

#define MAX_TRIPLETS 10000

typedef struct {
    int data[3];
} Triplet;

typedef struct {
    Triplet* triplets;
    int size;
    int capacity;
} TripletSet;

void parseArray(const char* str, int arr[][3], int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == '[') p++;
        if (*p == ']' || *p == '\0') break;
        
        arr[*size][0] = (int)strtol(p, (char**)&p, 10);
        while (*p == ' ' || *p == ',') p++;
        arr[*size][1] = (int)strtol(p, (char**)&p, 10);
        while (*p == ' ' || *p == ',') p++;
        arr[*size][2] = (int)strtol(p, (char**)&p, 10);
        
        while (*p && *p != '[' && *p != ']') p++;
        if (*p == ']') p++;
        (*size)++;
        
        if (*size >= MAX_TRIPLETS) break;
    }
}

void parseTarget(const char* str, int target[3]) {
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    target[0] = (int)strtol(p, (char**)&p, 10);
    while (*p == ' ' || *p == ',') p++;
    target[1] = (int)strtol(p, (char**)&p, 10);
    while (*p == ' ' || *p == ',') p++;
    target[2] = (int)strtol(p, (char**)&p, 10);
}

bool tripletExists(int triplets[][3], int size, int target[3]) {
    for (int i = 0; i < size; i++) {
        if (triplets[i][0] == target[0] && triplets[i][1] == target[1] && triplets[i][2] == target[2]) {
            return true;
        }
    }
    return false;
}

bool isSubset(int newTriplets[][3], int newSize, int current[][3], int currentSize) {
    for (int i = 0; i < newSize; i++) {
        if (!tripletExists(current, currentSize, newTriplets[i])) {
            return false;
        }
    }
    return true;
}

int max(int a, int b) {
    return a > b ? a : b;
}

bool solution(int triplets[][3], int n, int target[3]) {
    static int current[MAX_TRIPLETS][3];
    int currentSize = n;
    
    for (int i = 0; i < n; i++) {
        current[i][0] = triplets[i][0];
        current[i][1] = triplets[i][1];
        current[i][2] = triplets[i][2];
    }
    
    if (tripletExists(current, currentSize, target)) {
        return true;
    }
    
    while (true) {
        static int newTriplets[MAX_TRIPLETS][3];
        int newSize = 0;
        
        for (int i = 0; i < currentSize; i++) {
            for (int j = i + 1; j < currentSize; j++) {
                int a = current[i][0], b = current[i][1], c = current[i][2];
                int x = current[j][0], y = current[j][1], z = current[j][2];
                int merged[3] = {max(a, x), max(b, y), max(c, z)};
                
                if (merged[0] == target[0] && merged[1] == target[1] && merged[2] == target[2]) {
                    return true;
                }
                
                bool exists = false;
                for (int k = 0; k < newSize; k++) {
                    if (newTriplets[k][0] == merged[0] && newTriplets[k][1] == merged[1] && newTriplets[k][2] == merged[2]) {
                        exists = true;
                        break;
                    }
                }
                
                if (!exists && newSize < MAX_TRIPLETS) {
                    newTriplets[newSize][0] = merged[0];
                    newTriplets[newSize][1] = merged[1];
                    newTriplets[newSize][2] = merged[2];
                    newSize++;
                }
            }
        }
        
        if (isSubset(newTriplets, newSize, current, currentSize)) {
            break;
        }
        
        for (int i = 0; i < newSize && currentSize < MAX_TRIPLETS; i++) {
            if (!tripletExists(current, currentSize, newTriplets[i])) {
                current[currentSize][0] = newTriplets[i][0];
                current[currentSize][1] = newTriplets[i][1];
                current[currentSize][2] = newTriplets[i][2];
                currentSize++;
            }
        }
    }
    
    return false;
}

int main() {
    char line1[10000], line2[1000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    int triplets[MAX_TRIPLETS][3];
    int target[3];
    int n;
    
    parseArray(line1, triplets, &n);
    parseTarget(line2, target);
    
    printf(solution(triplets, n, target) ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Each merge creates new triplets, leading to exponential growth

⚠ Quadratic Growth

Space Complexity

O(2^n)

Need to store all generated triplets which grows exponentially

⚠ Quadratic Space

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Merge Triplets to Form Target Triplet - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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