1899. Merge Triplets to Form Target Triplet 

Description

A triplet is an array of three integers. You are given a 2D integer array triplets, where triplets[i] = [a_i, b_i, c_i] describes the i^th 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(a_i, a_j), max(b_i, b_j), max(c_i, c_j)].

<ul>
	<li>For example, if <code>triplets[i] = [2, 5, 3]</code> and <code>triplets[j] = [1, 7, 5]</code>, <code>triplets[j]</code> will be updated to <code>[max(2, 1), max(5, 7), max(3, 5)] = [2, 7, 5]</code>.</li>
</ul>
</li>

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 <= a_i, b_i, c_i, x, y, z <= 1000

Solutions

Solution 1: Greedy

Let $\textit{target} = [x, y, z]$. We need to determine whether there exists a triplet $[a, b, c]$ such that $a \leq x$, $b \leq y$, and $c \leq z$.

We can divide all triplets into two categories:

Triplets that satisfy $a \leq x$, $b \leq y$, and $c \leq z$.
Triplets that do not satisfy $a \leq x$, $b \leq y$, and $c \leq z$.

For the first category, we can take the maximum values of $a$, $b$, and $c$ from these triplets to form a new triplet $[d, e, f]$.

For the second category, we can ignore these triplets because they cannot help us achieve the target triplet.

Finally, we just need to check whether $[d, e, f]$ is equal to $\textit{target}$. If it is, return $\textit{true}$; otherwise, return $\textit{false}$.

Time complexity is $O(n)$, where $n$ is the length of the array $\textit{triplets}$. Space complexity is $O(1)$.

Python3

class Solution:
    def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:
        x, y, z = target
        d = e = f = 0
        for a, b, c in triplets:
            if a <= x and b <= y and c <= z:
                d = max(d, a)
                e = max(e, b)
                f = max(f, c)
        return [d, e, f] == target

Java

class Solution {
    public boolean mergeTriplets(int[][] triplets, int[] target) {
        int x = target[0], y = target[1], z = target[2];
        int d = 0, e = 0, f = 0;
        for (var t : triplets) {
            int a = t[0], b = t[1], c = t[2];
            if (a <= x && b <= y && c <= z) {
                d = Math.max(d, a);
                e = Math.max(e, b);
                f = Math.max(f, c);
            }
        }
        return d == x && e == y && f == z;
    }
}

C++

class Solution {
public:
    bool mergeTriplets(vector<vector<int>>& triplets, vector<int>& target) {
        int x = target[0], y = target[1], z = target[2];
        int d = 0, e = 0, f = 0;
        for (auto& t : triplets) {
            int a = t[0], b = t[1], c = t[2];
            if (a <= x && b <= y && c <= z) {
                d = max(d, a);
                e = max(e, b);
                f = max(f, c);
            }
        }
        return d == x && e == y && f == z;
    }
};

Go

func mergeTriplets(triplets [][]int, target []int) bool {
	x, y, z := target[0], target[1], target[2]
	d, e, f := 0, 0, 0
	for _, t := range triplets {
		a, b, c := t[0], t[1], t[2]
		if a <= x && b <= y && c <= z {
			d = max(d, a)
			e = max(e, b)
			f = max(f, c)
		}
	}
	return d == x && e == y && f == z
}

TypeScript

function mergeTriplets(triplets: number[][], target: number[]): boolean {
    const [x, y, z] = target;
    let [d, e, f] = [0, 0, 0];
    for (const [a, b, c] of triplets) {
        if (a <= x && b <= y && c <= z) {
            d = Math.max(d, a);
            e = Math.max(e, b);
            f = Math.max(f, c);
        }
    }
    return d === x && e === y && f === z;
}

Rust

impl Solution {
    pub fn merge_triplets(triplets: Vec<Vec<i32>>, target: Vec<i32>) -> bool {
        let [x, y, z]: [i32; 3] = target.try_into().unwrap();
        let (mut d, mut e, mut f) = (0, 0, 0);

        for triplet in triplets {
            if let [a, b, c] = triplet[..] {
                if a <= x && b <= y && c <= z {
                    d = d.max(a);
                    e = e.max(b);
                    f = f.max(c);
                }
            }
        }

        [d, e, f] == [x, y, z]
    }
}

Scala

object Solution {
    def mergeTriplets(triplets: Array[Array[Int]], target: Array[Int]): Boolean = {
        val Array(x, y, z) = target
        var (d, e, f) = (0, 0, 0)

        for (Array(a, b, c) <- triplets) {
            if (a <= x && b <= y && c <= z) {
                d = d.max(a)
                e = e.max(b)
                f = f.max(c)
            }
        }

        d == x && e == y && f == z
    }
}