Minimum Number of Arrows to Burst Balloons

class Solution {
public:
    int minArrowsToBurstBalloons(vector<vector<int>>& balloons) {
        if (balloons.empty()) return 0;

        // sorting on the end position of balloons
        sort(balloons.begin(), balloons.end(), [](const vector<int> &a, const vector<int> &b) {
            return a[1] < b[1];
        });

        int arrowCount = 1;
        int start, end, lastEnd = balloons[0][1];
        for (const auto &balloon : balloons) {
            start = balloon[0];
            end = balloon[1];

            // New arrow is needed when a balloon starts after the last one ends
            if (lastEnd < start) {
                arrowCount++;
                lastEnd = end;
            }
        }
        return arrowCount;
    }
};

This solution implements an algorithm to find the minimum number of arrows required to burst all the given balloons using C++ programming language. The code operates by first determining if the balloon list is empty. If it is, you return 0, meaning no arrows are needed.

The key operational steps are as follows:

1. Sort the balloons vector based on the end coordinates of the balloons. This allows for an efficient approach to determine the minimum number of arrows by focusing on where balloons end.

2. Initialize arrowCount to 1 since at least one arrow will be needed if there are any balloons.

3. Loop through the sorted balloon intervals and detect the requirement for new arrows. For each balloon, it checks whether the starting point of the current balloon is beyond the end point of the last encountered balloon.

   • If true, increment the arrowCount, indicating a need for a new arrow, and update lastEnd to the current balloon's end.

4. The function returns the count of arrows required.

This algorithm leverages sorting and a simple iteration, ensuring that the time complexity is dominated by the sorting process, hence the overall time complexity is O(n log n), where n is the number of balloons.

class Solver {
    public int countArrows(int[][] coordinates) {
        if (coordinates.length == 0) return 0;

        // Sort by balloon endpoint
        Arrays.sort(coordinates, (first, second) -> Integer.compare(first[1], second[1]));

        int requiredArrows = 1;
        int start, end, lastEnd = coordinates[0][1];
        for (int[] balloon: coordinates) {
            start = balloon[0];
            end = balloon[1];

            // Check if a new arrow is needed
            if (lastEnd < start) {
                requiredArrows++;
                lastEnd = end;
            }
        }

        return requiredArrows;
    }
}

The provided Java program aims to solve the problem of determining the minimum number of arrows required to burst all balloons represented by intervals. Each balloon is given as a range where it can be burst.

The approach taken includes the following steps:

1. Check if the input array of balloon coordinates is empty. If it is, return 0 since no arrows are needed.
2. Sort the balloon coordinates by their end points. This facilitates the process of finding overlapping intervals which can be burst by a single arrow.
3. Initialize the required number of arrows to one. Assume at least one arrow is needed for the first balloon.
4. Track the end point of the last burst balloon in lastEnd.
5. Traverse each balloon, comparing the start point of the current balloon with lastEnd.
6. If the start point is greater than lastEnd, this indicates that the current balloon doesn't overlap with the previous one, hence an additional arrow is necessary.
7. Update lastEnd to the end point of the current balloon if a new arrow is used.

By efficiently checking for overlapping intervals and counting the necessary instances where a new arrow must be used, the program successfully calculates the minimum number of arrows required. The use of sorting by end coordinates simplifies the logic needed to determine overlapping balloons. This solution is both effective in terms of time complexity, due to sorting taking O(n log n) time, and in terms of space complexity, as it operates in O(1) space, not counting the input.

class Solution:
    def minArrowsNeeded(self, intervals: List[List[int]]) -> int:
        if not intervals:
            return 0

        # Sort by ending coordinate
        intervals.sort(key=lambda point: point[1])

        num_arrows = 1
        end_of_last_burst = intervals[0][1]
        for start, end in intervals:
            # Check for non-overlapping balloons
            if end_of_last_burst < start:
                num_arrows += 1
                end_of_last_burst = end

        return num_arrows

This solution deals with finding the minimum number of arrows required to burst all balloons characterized by intervals where each interval represents a balloon's horizontal relative position of start and end points. The solution, implemented in Python 3, follows these steps:

1. Check if there are no balloons. If true, return 0 as no arrows are needed.

2. Sort the list of intervals by each balloon's ending point. This step is crucial as it simplifies the process of finding overlapping balloons.

3. Initialize a counter for arrows required (num_arrows) starting with one arrow, assuming at least one balloon exists.

4. Keep track of the ending point of the last burst balloon (end_of_last_burst). This is initially set to the end point of the first balloon in the sorted list.

5. Iterate over each balloon's interval. If the starting point of the current balloon is greater than end_of_last_burst, it indicates that this balloon doesn't overlap with the previously considered ones, hence an additional arrow is needed. Update end_of_last_burst to the current balloon's end point.

6. Finally, return the total number of arrows needed (num_arrows).

This approach ensures that the minimum number of arrows is used by making use of sorting based on end positions of balloons, which allows for an efficient check of overlaps. The provided Python 3 code effectively applies logical decisions to ascertain when additional arrows are needed and successfully calculates the result based on interval management.