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2345. Finding the Number of Visible Mountains

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

  • 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;
    }
};

  • 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


  • 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
}

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