Merge Triplets to Form Target Triplet

Problem

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 <= 10^5
triplets[i].length == target.length == 3
1 <= ai, bi, ci, x, y, z <= 1000

Solution (Java)

class Solution {
    public boolean mergeTriplets(int[][] triplets, int[] target) {
        boolean one = false;
        boolean two = false;
        boolean three = false;
        for (int[] triplet : triplets) {
            if (!one
                    && triplet[0] == target[0]
                    && triplet[1] <= target[1]
                    && triplet[2] <= target[2]) {
                one = true;
            }
            if (!two
                    && triplet[0] <= target[0]
                    && triplet[1] == target[1]
                    && triplet[2] <= target[2]) {
                two = true;
            }
            if (!three
                    && triplet[0] <= target[0]
                    && triplet[1] <= target[1]
                    && triplet[2] == target[2]) {
                three = true;
            }
            if (one && two && three) {
                return true;
            }
        }
        return false;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).