Problem Description

In this problem, we have an array representing the range of positions occupied by cars on a number line. Each car's position is defined by a pair [start_i, end_i], indicating that the car covers every integer coordinate from start_i to end_i inclusively. We need to determine the total number of unique integer points covered by any part of a car on this number line.

To clarify with an example, if we have a car occupying positions from 2 to 4, it covers three points on the number line: 2, 3, and 4.

Intuition

We can approach this problem by visualizing the number line and cars as intervals which can overlap. It may initially seem that we need to count every position each car occupies, but that would be inefficient, especially if the number line and the cars' range are large.

The provided solution uses a clever approach generally known as "difference array" or "prefix sum array" technique. The key insight here is to track changes at the start and just past the end positions of the cars' ranges. We increment at the start position, indicating the beginning of a car, and decrement right after the end position, indicating the end of a car's coverage.

This way, when we traverse the modified array and accumulate these changes, we can determine the coverage at each point on the number line. A positive number indicates that we are within a car's range at that point.

Finally, we sum up all the positions on the number line where the accumulated sum is greater than zero, which signifies a point being covered by at least one car.

Here are the steps in detail:

1. Initialize a difference array d with zeros. Its size should be large enough to cover the maximum possible end point of any car.

2. Iterate through each car's range in nums. For each range [a, b], increment d[a] by 1 (indicating a car starts) and decrement d[b + 1] by 1 (indicating the point after a car ends).

3. Utilize accumulate from Python's itertools to turn the difference array into a prefix sum array. This will tell us the coverage at each point on the line.

4. Count and return how many points on the number line have a positive coverage.

Learn more about Math and Prefix Sum patterns.

Solution Approach

The solution adopts the difference array technique, a classical computational algorithm for efficiently implementing range update queries. We can describe the steps in the process as follows:

1. Initialize the Difference Array: A difference array d is initialized with zeros and is large enough to cover all the potential end points of the cars. This array will help us record the change at the start and just past the end of each car's interval.

2. Update the Difference Array: We iterate over each car's range in nums, for each car defined by a range [a, b]. We increment d[a] by 1 to denote the start of the car's range and decrement d[b + 1] by 1 to denote the point just after the end of the car's range. Thus, these increments and decrements represent the change at those specific points.

3. Cumulative Sum (Accumulate): After updating the difference array, we use accumulate from Python's itertools to calculate the cumulative sum. This results in an array where each index now represents how many cars are covering that specific point on the number line.

4. Count Positive Coverages: We then count all the points with a positive value in the accumulated array. A positive value indicates that the point is covered by at least one car. So, the final count gives us the total number of unique integer points covered by cars.

Code Explanation:

class Solution:
    def numberOfPoints(self, nums: List[List[int]]) -> int:
        # Step 1: Initialize the difference array with zeros.
        d = [0] * 110
      
        # Step 2: Update the difference array based on car ranges.
        for a, b in nums:
            d[a] += 1       # Increment at the start of the car's range.
            d[b + 1] -= 1   # Decrement right after the end of the car's range.
          
        # Step 3: Apply accumulate to get the [prefix sum](/problems/subarray_sum), which gives us coverage per point.
        # Step 4: Sum up the count of points where the accumulated value is greater than 0.
        return sum(s > 0 for s in accumulate(d))

The choice of 110 as the array size is arbitrary and should cover the problem's given constraints. This number would need to be adjusted according to the specific constraints of the problem, such as the maximum value of the end of the cars' ranges.

This solution takes O(N+R) time, where N is the number of cars and R is the range of the number line we are interested in. The space complexity is O(R), which is required for the difference array.

Example Walkthrough

To help illustrate the solution approach, let's walk through a small example. Suppose our input is the list of car ranges nums = [[1, 3], [2, 5], [6, 8]]. This means we have three cars covering the following ranges on the number line:

• Car 1 covers from point 1 to point 3.
• Car 2 covers from point 2 to point 5.
• Car 3 covers from point 6 to point 8.

Let's follow the steps of the solution:

1. Initialize the Difference Array: We initialize a difference array, d, with zeros. For this example, we can choose a length that covers the maximum end point in our input, which is 8. But to be certain we cover the end points plus one, we'll extend it to 10: d = [0] * 10.

2. Update the Difference Array: We iterate over each car's range and update d accordingly.

   • For range [1, 3], we increment d[1] by 1 and decrement d[4] by 1 (since we increment one past the end of the interval).
   • For range [2, 5], we increment d[2] by 1 and decrement d[6] by 1.
   • For range [6, 8], we increment d[6] by 1 and decrement d[9] by 1.

   After these updates, d looks like this: [0, 1, 1, 0, -1, 0, 1, 0, -1, -1].

3. Cumulative Sum (Accumulate): Now, we apply accumulate to d to get the prefix sum array. This reflects the number of cars covering each point. After calculating, the array looks like this: [0, 1, 2, 2, 1, 1, 2, 2, 1, 0].

4. Count Positive Coverages: We count all the positive values in the accumulated array, which gives us the total number of unique integer points covered by cars. There are 8 positive values, so 8 points on the number line are covered by at least one car.

Following the implementation described in the content, our resulting numberOfPoints would be 8 for the given example. This demonstrates how the difference array method efficiently computes the required coverage, even with overlapping intervals.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is determined by several key operations:

1. Initializing the difference array d with 110 elements set to 0. This is an O(1) operation since the size of the array is constant and does not depend on the input nums.
2. Iterating over the input nums, where each element is a list [a, b]. For each element, the code performs constant-time updates to the difference array d. If the length of nums is n, this operation takes O(n) time.
3. Accumulating the differences in d to compute the prefix sums. The use of the accumulate function from Python's itertools module has a time complexity of O(m), where m is the size of the difference array, which is a constant 110 in this case.
4. Summing up the positive values in the accumulated list with a list comprehension. This operation is again O(m) because it iterates over every element in the accumulated list.

Since both m and the size of the difference array are constant, the dominant term in the time complexity is the iteration over nums, giving an overall time complexity of O(n), where n is the length of nums.

Space Complexity

The space complexity is determined by the additional memory used by the program, which is primarily the difference array d. The array has a constant size of 110, which gives us an O(1) space complexity. The use of the accumulate function generates a temporary list with the same size as d, but since the size of d is constant, this does not impact the asymptotic space complexity. Therefore, the total space complexity remains O(1).

Learn more about how to find time and space complexity quickly using problem constraints.