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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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