LeetCode 1899. Merge Triplets to Form Target Triplet Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Greedy

Let target = [x, y, z]. We need to determine whether there exists a triplet [a, b, c] such that a ≤ x, b ≤ y, and c ≤ z.

We can divide all triplets into two categories:

1. Triplets that satisfy a ≤ x, b ≤ y, and c ≤ z.
2. Triplets that do not satisfy a ≤ x, b ≤ y, and c ≤ z.

For the first category, we can take the maximum values of a, b, and c from these triplets to form a new triplet [d, e, f].

For the second category, we can ignore these triplets because they cannot help us achieve the target triplet.

Finally, we just need to check whether [d, e, f] is equal to target. If it is, return true; otherwise, return false.

Time complexity is O(n), where n is the length of the array triplets. Space complexity is O(1).

PythonJavaC++GoTypeScriptRustScala

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

impl Solution { pub fn merge_triplets(triplets: Vec<Vec<i32>>, target: Vec<i32>) -> bool { let [x, y, z]: [i32; 3] = target.try_into().unwrap(); let (mut d, mut e, mut f) = (0, 0, 0); for triplet in triplets { if let [a, b, c] = triplet[..] { if a <= x && b <= y && c <= z { d = d.max(a); e = e.max(b); f = f.max(c); } } } [d, e, f] == [x, y, z] } }(code-box)

object Solution { def mergeTriplets(triplets: Array[Array[Int]], target: Array[Int]): Boolean = { val Array(x, y, z) = target var (d, e, f) = (0, 0, 0) for (Array(a, b, c) <- triplets) { if (a <= x && b <= y && c <= z) { d = d.max(a) e = e.max(b) f = f.max(c) } } d == x && e == y && f == z } }(code-box)