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

## Solutions

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}