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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

Solution 1

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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