Solution Explanation: Count Positions on Street With Required Brightness

This problem involves determining the number of positions on a street that meet a minimum brightness requirement, given the positions and ranges of streetlights. The optimal approach leverages the concept of a difference array for efficient calculation.

Approach: Difference Array

A difference array allows for efficient updates to ranges of values. Instead of directly updating each element within a range, we only modify the beginning and ending points of the range. This significantly reduces the time complexity compared to iterating through each element in the range.

1. Building the Difference Array:

• Initialize a difference array d of size n + 1 (where n is the length of the street) with zeros. The extra element simplifies boundary handling.
• For each streetlight [position, range]:
  • Calculate the start index i = max(0, position - range) and end index j = min(n - 1, position + range) of the lit area.
  • Increment d[i] by 1 (representing the start of the lit area).
  • Decrement d[j + 1] by 1 (representing the end of the lit area). This is crucial; the prefix sum will correctly account for the lighting.

2. Prefix Sum Calculation:

• Calculate the prefix sum of the difference array d. This prefix sum at index k represents the brightness at position k on the street. The decrement in d[j+1] ensures that the brightness doesn't overcount beyond the lamp's range.

3. Checking Requirements:

• Iterate through the prefix sum and the requirement array simultaneously.
• For each position i, if the prefix sum d[i] is greater than or equal to requirement[i], increment the count of positions meeting the requirement.

4. Returning the Result:

• Return the final count of positions that meet the brightness requirement.

Time and Space Complexity

• Time Complexity: O(n + m), where n is the length of the street and m is the number of streetlights. The dominant operations are iterating through the streetlights (O(m)) to build the difference array and calculating the prefix sum (O(n)).
• Space Complexity: O(n), due to the difference array d which stores brightness information for each position.

Code Implementation (Python)

from itertools import accumulate
 
class Solution:
    def meetRequirement(self, n: int, lights: List[List[int]], requirement: List[int]) -> int:
        d = [0] * (n + 1)
        for p, r in lights:
            i, j = max(0, p - r), min(n - 1, p + r)
            d[i] += 1
            d[j + 1] -= 1
        
        prefix_sum = list(accumulate(d))  #Efficient prefix sum calculation
        
        count = 0
        for i in range(n):
            if prefix_sum[i] >= requirement[i]:
                count += 1
        return count
 

The code in other languages (Java, C++, Go, TypeScript) follows the same logic, only differing in syntax. The use of accumulate (Python) or equivalent functions in other languages provides an efficient way to compute the prefix sum. The difference array approach avoids redundant calculations making it far more efficient than brute force methods.