[leetCode/JS] 14. Longest Common Prefix

문제 설명

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.

 

 

입출력 예

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 <= a~i~, b~i, ci, x, y, z <= 1000

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내 솔루션

• 정말로 스왑하고 다시 검색하고 하다가 느려져서 고민 고민하다가 만들어진 로직.
• traget보다 triplets가 작으면 어짜피 스왑도 불가능 하기 때문에 건너뛰는 방식.

var mergeTriplets = function(triplets, target) {
  let max = [0, 0, 0];
  for(let i = 0; i < triplets.length; i++) {
    if(triplets[i].every((n, idx) => n <= target[idx])){
      max = triplets[i].map((n, i) => Math.max(n, max[i]))
    }
  }
  return JSON.stringify(max) === JSON.stringify(target)
};

 

감상평

• 문제가 겁나 복잡하다. 세상에.