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.

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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.


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Constraints:


	1 &lt;= triplets.length &lt;= 105
	triplets[i].length == target.length == 3
	1 &lt;= ai, bi, ci, x, y, z &lt;= 1000


class Solution:
    def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:

        i=0
        while True:
            if i==len(triplets):
                break
            for j in range(3):
                if triplets[i][j]>target[j]:
                    triplets.pop(i)
                    i-=1
                    break
            i+=1

        a = [x[0] for x in triplets]
        b = [x[1] for x in triplets]
        c = [x[2] for x in triplets]

        if target[0] not in a:
            return False
        if target[1] not in b:
            return False
        if target[2] not in c:
            return False
        return True
class Solution {
    public boolean mergeTriplets(int[][] triplets, int[] target) {

        boolean xFound = false, yFound = false, zFound = false;
                
        for(int[] triplet : triplets){
		
            //Current Triplet is target
            if(triplet[0] == target[0] && triplet[1] == target[1] && triplet[2] == target[2])return true;
            
            if(triplet[0] == target[0]){
                if(triplet[1] <= target[1] && triplet[2] <= target[2])
                    if(yFound && zFound)return true;
                    else xFound = true;
            }
            if(triplet[1] == target[1]){
                if(triplet[0] <= target[0] && triplet[2] <= target[2])
                    if(xFound && zFound)return true;
                    else yFound = true;
            }
            if(triplet[2] == target[2]){
                if(triplet[1] <= target[1] && triplet[0] <= target[0])
                    if(yFound && xFound)return true;
                    else zFound = true;
            }        
        }
        return xFound && yFound && zFound;
    }
}
class Solution {
public:
    bool mergeTriplets(vector<vector<int>>& triplets, vector<int>& target) {
        int first = 0, second = 0, third = 0;
        for (auto tr : triplets) {
            if (tr[0] == target[0] && tr[1] <= target[1] && tr[2] <= target[2]) first = 1;
            if (tr[0] <= target[0] && tr[1] == target[1] && tr[2] <= target[2]) second = 1;
            if (tr[0] <= target[0] && tr[1] <= target[1] && tr[2] == target[2]) third = 1;
        }
        return first && second && third;
    }
};
var mergeTriplets = function(triplets, target) {

    let fst = false, snd = false, thrd = false;

    const [t1, t2, t3] = target;

    for(let i = 0; i < triplets.length; i++) {

        const [a, b, c] = triplets[i];

        if(a === t1 && b <= t2 && c <= t3) fst = true;

        if(b === t2 && a <= t1 && c <= t3) snd = true;

        if(c === t3 && a <= t1 && b <= t2) thrd = true;

        if(fst && snd && thrd) return true;
    }
    return false;
};
Merge Triplets to Form Target Triplet