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There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [xstart, xend] denotes a balloon whose horizontal diameter stretches between xstart and xend. You do not know the exact y-coordinates of the balloons.

Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with xstart and xend is burst by an arrow shot at x if xstart <= x <= xend. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.

Given the array points, return the minimum number of arrows that must be shot to burst all balloons.

Example 1:


Input: points = [[10,16],[2,8],[1,6],[7,12]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
- Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].

Example 2:


Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.

Example 3:


Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
- Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].

Constraints:

1 <= points.length <= 105

points[i].length == 2

231 <= xstart < xend <= 231 - 1

풀이

재영

결국 우리가 자주 풀었던 문제의 유형이네요!

O(NlogN)이면 거뜬하게 풀 수 있는 문제입니다. 😉


/**
 * @param {number[][]} points
 * @return {number}
 */
const findMinArrowShots = (points) => {
  points.sort(([a1], [b1]) => a1 - b1);
  let last = Infinity;
  let result = 0;

  points.forEach(([start, end]) => {
    if (last >= start) {
      if (!isFinite(last)) result += 1;
      last = Math.min(last, end);
    } else {
      result += 1;
      last = end;
    }
  });

  return result;
};

최적화 아닌 최적화…?

결과적으로 무한대인지에 대한 확인을 last >= start일 때마다 하고 있습니다.

이때, points는 무조건 1보다 크거나 같다는 조건이 있기 때문에, 항상 result는 저 분기처리가 없을 때 1만큼 크다는 결과를 가질 수 있습니다. (첫 번째 터뜨려야 할 지점이 현재는 값에서 생략됐기 때문)

따라서 결과 값에 1을 더하면 (무의미할 정도지만) 더 빠른 계산이 가능하겠군요!


/**
 * @param {number[][]} points
 * @return {number}
 */
const findMinArrowShots = (points) => {
  points.sort(([a1], [b1]) => a1 - b1);
  let last = Infinity;
  let result = 0;

  points.forEach(([start, end]) => {
    if (last >= start) {
      last = Math.min(last, end);
    } else {
      result += 1;
      last = end;
    }
  });

  return result + 1;
};

효성


/**
 * @param {number[][]} points
 * @return {number}
 */
var findMinArrowShots = function (points) {
    points.sort((a, b) => a[0] - b[0]);
    let prev = null;
    let count = 0;

    for (const [start, end] of points) {
        if (prev === null || prev < start) {
            count++;
            prev = end;
        }
        else {
            prev = Math.min(prev, end);
        };
    }
    return count;
};

Minimum Number of Arrows to Burst Balloons

풀이

최적화 아닌 최적화…?