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Formatted question description: https://leetcode.ca/all/2237.html

2237. Count Positions on Street With Required Brightness

Description

You are given an integer n. A perfectly straight street is represented by a number line ranging from 0 to n - 1. You are given a 2D integer array lights representing the street lamp(s) on the street. Each lights[i] = [positioni, rangei] indicates that there is a street lamp at position positioni that lights up the area from [max(0, positioni - rangei), min(n - 1, positioni + rangei)] (inclusive).

The brightness of a position p is defined as the number of street lamps that light up the position p. You are given a 0-indexed integer array requirement of size n where requirement[i] is the minimum brightness of the ith position on the street.

Return the number of positions i on the street between 0 and n - 1 that have a brightness of at least requirement[i].

 

Example 1:

Input: n = 5, lights = [[0,1],[2,1],[3,2]], requirement = [0,2,1,4,1]
Output: 4
Explanation:
- The first street lamp lights up the area from [max(0, 0 - 1), min(n - 1, 0 + 1)] = [0, 1] (inclusive).
- The second street lamp lights up the area from [max(0, 2 - 1), min(n - 1, 2 + 1)] = [1, 3] (inclusive).
- The third street lamp lights up the area from [max(0, 3 - 2), min(n - 1, 3 + 2)] = [1, 4] (inclusive).

-   Position 0 is covered by the first street lamp. It is covered by 1 street lamp which is greater than requirement[0].
-   Position 1 is covered by the first, second, and third street lamps. It is covered by 3 street lamps which is greater than requirement[1].
-   Position 2 is covered by the second and third street lamps. It is covered by 2 street lamps which is greater than requirement[2].
-   Position 3 is covered by the second and third street lamps. It is covered by 2 street lamps which is less than requirement[3].
-   Position 4 is covered by the third street lamp. It is covered by 1 street lamp which is equal to requirement[4].

Positions 0, 1, 2, and 4 meet the requirement so we return 4.

Example 2:

Input: n = 1, lights = [[0,1]], requirement = [2]
Output: 0
Explanation:
- The first street lamp lights up the area from [max(0, 0 - 1), min(n - 1, 0 + 1)] = [0, 0] (inclusive).
- Position 0 is covered by the first street lamp. It is covered by 1 street lamp which is less than requirement[0].
- We return 0 because no position meets their brightness requirement.

 

Constraints:

  • 1 <= n <= 105
  • 1 <= lights.length <= 105
  • 0 <= positioni < n
  • 0 <= rangei <= 105
  • requirement.length == n
  • 0 <= requirement[i] <= 105

Solutions

  • class Solution {
    public int meetRequirement(int n, int[][] lights, int[] requirement) {
        int[] d = new int[100010];
        for (int[] e : lights) {
            int i = Math.max(0, e[0] - e[1]);
            int j = Math.min(n - 1, e[0] + e[1]);
            ++d[i];
            --d[j + 1];
        }
        int s = 0;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            s += d[i];
            if (s >= requirement[i]) {
                ++ans;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int meetRequirement(int n, vector<vector<int>>& lights, vector<int>& requirement) {
        vector<int> d(100010);
        for (auto& e : lights) {
            int i = max(0, e[0] - e[1]), j = min(n - 1, e[0] + e[1]);
            ++d[i];
            --d[j + 1];
        }
        int s = 0, ans = 0;
        for (int i = 0; i < n; ++i) {
            s += d[i];
            if (s >= requirement[i]) ++ans;
        }
        return ans;
    }
};

  • class Solution:
    def meetRequirement(
        self, n: int, lights: List[List[int]], requirement: List[int]
    ) -> int:
        d = [0] * 100010
        for p, r in lights:
            i, j = max(0, p - r), min(n - 1, p + r)
            d[i] += 1
            d[j + 1] -= 1
        return sum(s >= r for s, r in zip(accumulate(d), requirement))


  • func meetRequirement(n int, lights [][]int, requirement []int) int {
	d := make([]int, 100010)
	for _, e := range lights {
		i, j := max(0, e[0]-e[1]), min(n-1, e[0]+e[1])
		d[i]++
		d[j+1]--
	}
	var s, ans int
	for i, r := range requirement {
		s += d[i]
		if s >= r {
			ans++
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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