Question

2345. Finding the Number of Visible Mountains

中文文档

Description

You are given a 0-indexed 2D integer array peaks where peaks[i] = [xi, yi] states that mountain i has a peak at coordinates (xi, yi). A mountain can be described as a right-angled isosceles triangle, with its base along the x-axis and a right angle at its peak. More formally, the gradients of ascending and descending the mountain are 1 and -1 respectively.

A mountain is considered visible if its peak does not lie within another mountain (including the border of other mountains).

Return _the number of visible mountains_.

 

Example 1:

Input: peaks = [[2,2],[6,3],[5,4]]
Output: 2
Explanation: The diagram above shows the mountains.
- Mountain 0 is visible since its peak does not lie within another mountain or its sides.
- Mountain 1 is not visible since its peak lies within the side of mountain 2.
- Mountain 2 is visible since its peak does not lie within another mountain or its sides.
There are 2 mountains that are visible.

Example 2:

Input: peaks = [[1,3],[1,3]]
Output: 0
Explanation: The diagram above shows the mountains (they completely overlap).
Both mountains are not visible since their peaks lie within each other.

 

Constraints:

• 1 <= peaks.length <= 105
• peaks[i].length == 2
• 1 <= xi, yi <= 105

Solutions

Solution 1

class Solution:
  def visibleMountains(self, peaks: List[List[int]]) -> int:
    arr = [(x - y, x + y) for x, y in peaks]
    cnt = Counter(arr)
    arr.sort(key=lambda x: (x[0], -x[1]))
    ans, cur = 0, -inf
    for l, r in arr:
      if r <= cur:
        continue
      cur = r
      if cnt[(l, r)] == 1:
        ans += 1
    return ans

class Solution {
  public int visibleMountains(int[][] peaks) {
    int n = peaks.length;
    int[][] arr = new int[n][2];
    Map<String, Integer> cnt = new HashMap<>();
    for (int i = 0; i < n; ++i) {
      int x = peaks[i][0], y = peaks[i][1];
      arr[i] = new int[] {x - y, x + y};
      cnt.merge((x - y) + "" + (x + y), 1, Integer::sum);
    }
    Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
    int ans = 0;
    int cur = Integer.MIN_VALUE;
    for (int[] e : arr) {
      int l = e[0], r = e[1];
      if (r <= cur) {
        continue;
      }
      cur = r;
      if (cnt.get(l + "" + r) == 1) {
        ++ans;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int visibleMountains(vector<vector<int>>& peaks) {
    vector<pair<int, int>> arr;
    for (auto& e : peaks) {
      int x = e[0], y = e[1];
      arr.emplace_back(x - y, -(x + y));
    }
    sort(arr.begin(), arr.end());
    int n = arr.size();
    int ans = 0, cur = INT_MIN;
    for (int i = 0; i < n; ++i) {
      int l = arr[i].first, r = -arr[i].second;
      if (r <= cur) {
        continue;
      }
      cur = r;
      ans += i == n - 1 || (i < n - 1 && arr[i] != arr[i + 1]);
    }
    return ans;
  }
};

func visibleMountains(peaks [][]int) (ans int) {
  n := len(peaks)
  type pair struct{ l, r int }
  arr := make([]pair, n)
  for _, p := range peaks {
    x, y := p[0], p[1]
    arr = append(arr, pair{x - y, x + y})
  }
  sort.Slice(arr, func(i, j int) bool { return arr[i].l < arr[j].l || (arr[i].l == arr[j].l && arr[i].r > arr[j].r) })
  cur := math.MinInt32
  for i, e := range arr {
    l, r := e.l, e.r
    if r <= cur {
      continue
    }
    cur = r
    if !(i < n-1 && l == arr[i+1].l && r == arr[i+1].r) {
      ans++
    }
  }
  return
}

Solution 2

class Solution {
  public int visibleMountains(int[][] peaks) {
    int n = peaks.length;
    int[][] arr = new int[n][2];
    for (int i = 0; i < n; ++i) {
      int x = peaks[i][0], y = peaks[i][1];
      arr[i] = new int[] {x - y, x + y};
    }
    Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
    int ans = 0;
    int cur = Integer.MIN_VALUE;
    for (int i = 0; i < n; ++i) {
      int l = arr[i][0], r = arr[i][1];
      if (r <= cur) {
        continue;
      }
      cur = r;
      if (!(i < n - 1 && arr[i][0] == arr[i + 1][0] && arr[i][1] == arr[i + 1][1])) {
        ++ans;
      }
    }
    return ans;
  }
}