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1899. Merge Triplets to Form Target Triplet

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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)].

  <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 <= ai, bi, ci, x, y, z <= 1000

Solutions

Python3

class Solution:
    def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:
        maxA = maxB = maxC = 0
        tA, tB, tC = target
        for a, b, c in triplets:
            if a <= tA and b <= tB and c <= tC:
                maxA = max(maxA, a)
                maxB = max(maxB, b)
                maxC = max(maxC, c)
        return (maxA, maxB, maxC) == (tA, tB, tC)

Java

class Solution {
    public boolean mergeTriplets(int[][] triplets, int[] target) {
        int maxA = 0, maxB = 0, maxC = 0;
        for (int[] triplet : triplets) {
            int a = triplet[0], b = triplet[1], c = triplet[2];
            if (a <= target[0] && b <= target[1] && c <= target[2]) {
                maxA = Math.max(maxA, a);
                maxB = Math.max(maxB, b);
                maxC = Math.max(maxC, c);
            }
        }
        return maxA == target[0] && maxB == target[1] && maxC == target[2];
    }
}

TypeScript

function mergeTriplets(triplets: number[][], target: number[]): boolean {
    let [x, y, z] = target; // 目标值
    let [i, j, k] = [0, 0, 0]; // 最大值
    for (let triplet of triplets) {
        let [a, b, c] = triplet; // 当前值
        if (a <= x && b <= y && c <= z) {
            i = Math.max(i, a);
            j = Math.max(j, b);
            k = Math.max(k, c);
        }
    }
    return i == x && j == y && k == z;
}

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