Minimum Number Of Arrows To Burst Balloons

                                            
class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& points) {
        if (points.empty()) {
            return 0;
        }
        sort(points.begin(), points.end(), [](vector<int>& a, vector<int>& b) {
            return a[1] < b[1];
        });
        int arrows = 1;
        int end = points[0][1];
        for (int i = 1; i < points.size(); i++) {
            if (points[i][0] > end) {
                arrows++;
                end = points[i][1];
            }
        }
        return arrows;
    }
};

                                            
class Solution {
    public int findMinArrowShots(int[][] points) {
        if (points == null || points.length == 0) {
            return 0;
        }
        
        Arrays.sort(points, (a, b) -> Integer.compare(a[1], b[1]));
        
        int arrows = 1;
        int end = points[0][1];
        
        for (int i = 1; i < points.length; i++) {
            if (points[i][0] > end) {
                arrows++;
                end = points[i][1];
            }
        }
        
        return arrows;
    }
}

                                            
class Solution(object):
    def findMinArrowShots(self, points):
        if not points:
            return 0
        
        points.sort(key=lambda x: x[1])
        arrows = 1
        end = points[0][1]
        
        for i in range(1, len(points)):
            if points[i][0] > end:
                arrows += 1
                end = points[i][1]
        
        return arrows

                                            
public class Solution {
    public int FindMinArrowShots(int[][] points) {
        if (points == null || points.Length == 0) return 0;
        
        Array.Sort(points, (a, b) => a[1].CompareTo(b[1]));
        
        int arrows = 1;
        int end = points[0][1];
        
        for (int i = 1; i < points.Length; i++) {
            if (points[i][0] > end) {
                arrows++;
                end = points[i][1];
            }
        }
        
        return arrows;
    }
}

                                            
/**
 * @param {number[][]} points
 * @return {number}
 */
var findMinArrowShots = function(points) {
    if (points.length === 0) return 0;
    
    points.sort((a, b) => a[1] - b[1]);
    
    let arrows = 1;
    let currEnd = points[0][1];
    
    for (let i = 1; i < points.length; i++) {
        if (points[i][0] > currEnd) {
            arrows++;
            currEnd = points[i][1];
        }
    }
    
    return arrows;
};

                                            
function findMinArrowShots(points: number[][]): number {
    if (points.length === 0) return 0;
    
    points.sort((a, b) => a[1] - b[1]);
    
    let arrows = 1;
    let end = points[0][1];
    
    for (let i = 1; i < points.length; i++) {
        if (points[i][0] > end) {
            arrows++;
            end = points[i][1];
        }
    }
    
    return arrows;
};

                                            
class Solution {
    /**
     * @param Integer[][] $points
     */
    function findMinArrowShots($points) {
        if (empty($points)) return 0;
        
        usort($points, function($a, $b) {
            return $a[1] <=> $b[1];
        });
        
        $arrows = 1;
        $end = $points[0][1];
        
        foreach ($points as $point) {
            if ($point[0] > $end) {
                $arrows++;
                $end = $point[1];
            }
        }
        
        return $arrows;
    }
}

                                            
class Solution {
    func findMinArrowShots(_ points: [[Int]]) -> Int {
        guard points.count > 0 else {
            return 0
        }
        
        let sortedPoints = points.sorted { $0[1] < $1[1] }
        var arrows = 1
        var end = sortedPoints[0][1]
        
        for i in 1..<sortedPoints.count {
            if sortedPoints[i][0] > end {
                arrows += 1
                end = sortedPoints[i][1]
            }
        }
        
        return arrows
    }
}

                                            
class Solution {
    fun findMinArrowShots(points: Array<IntArray>): Int {
        if (points.isEmpty()) return 0
        points.sortBy { it[1] }
        
        var arrows = 1
        var end = points[0][1]
        
        for (i in 1 until points.size) {
            if (points[i][0] > end) {
                arrows++
                end = points[i][1]
            }
        }
        
        return arrows
    }
}

                                            
class Solution {
  int findMinArrowShots(List<List<int>> points) {
    if (points.isEmpty) return 0;
    
    points.sort((a, b) => a[1].compareTo(b[1]));
    
    int arrows = 1;
    int end = points[0][1];
    
    for (int i = 1; i < points.length; i++) {
      if (points[i][0] > end) {
        arrows++;
        end = points[i][1];
      }
    }
    
    return arrows;
  }
}

                                            
func findMinArrowShots(points [][]int) int {
    if len(points) == 0 {
        return 0
    }

    sort.Slice(points, func(i, j int) bool {
        return points[i][1] < points[j][1]
    })

    arrows := 1
    end := points[0][1]

    for i := 1; i < len(points); i++ {
        if points[i][0] > end {
            arrows++
            end = points[i][1]
        }
    }

    return arrows
}

                                            
# @param {Integer[][]} points
# @return {Integer}
def find_min_arrow_shots(points)
    return 0 if points.empty?
    points.sort_by!{ |p| p[1] }
    arrows = 1
    end_pos = points[0][1]
    
    (1...points.length).each do |i|
        if points[i][0] > end_pos
            arrows += 1
            end_pos = points[i][1]
        end
    end
    
    return arrows
end

                                            
object Solution {
    def findMinArrowShots(points: Array[Array[Int]]): Int = {
        if (points.isEmpty) return 0
        
        val sortedPoints = points.sortBy(_.last)
        
        var arrows = 1
        var end = sortedPoints(0)(1)
        
        for (i <- 1 until sortedPoints.length) {
            if (sortedPoints(i)(0) > end) {
                arrows += 1
                end = sortedPoints(i)(1)
            }
        }
        
        arrows
    }
}

                                            
impl Solution {
    pub fn find_min_arrow_shots(points: Vec<Vec<i32>>) -> i32 {
        if points.is_empty() {
            return 0;
        }
        
        let mut points = points;
        points.sort_by_key(|x| x[1]);
        
        let mut arrows = 1;
        let mut end = points[0][1];
        
        for i in 1..points.len() {
            if points[i][0] > end {
                arrows += 1;
                end = points[i][1];
            }
        }
        
        arrows
    }
}

Explanation

To solve this coding challenge, you need to determine the minimum number of arrows required to burst all the balloons represented as intervals on the x-axis. The key is to take advantage of overlapping intervals. By sorting the intervals based on their end points, you can effectively cover more balloons with each arrow shot. The method can be broken down into the following steps:

Edge Case Handling : If there are no balloons, then return 0 since no arrows are needed.
Sorting : Sort the intervals based on their end points. This helps in deciding the minimum number of intervals that can be covered by a single arrow.
Counting Arrows : Start with one arrow pointing to the end of the first interval. Loop through the rest of the intervals.
Overlap Check : For each interval, if the starting point of the current interval is greater than the end point tracked, increment the arrow count and update the end point.
Result : The final count of arrows gives the minimum number needed to burst all balloons.

Detailed Steps in Pseudocode

Input Validation :

If the
points

array is empty, then return 0 right away because no balloons exist to burst.

Sorting :

Sort the
points

array based on the end position of each balloon. This helps in maximizing the number of balloons that can be burst by a single arrow.

Initialize Variables :

Begin with a count of one arrow.
Set the
end

variable to the end position of the first balloon in the sorted list.

Iterate Through Balloons :

For each balloon, check if the start position of the balloon is beyond the current
end

.
If it is, increment the arrow count and update the
end

position to the end of the current balloon. This means a new arrow is needed.
If not, the current balloon can be burst with the existing arrow.

Return Result :

The final arrow count is returned as the result.

Pseudocode

                                            
# Function to find the minimum number of arrows to burst balloons
function find_minimum_arrows(points):
    # Edge case: If there are no balloons, return 0
    if points is empty:
        return 0
    
    # Sort balloons by their end positions
    sort points by points[i][1]
    
    # Initialize the number of arrows with 1
    arrows = 1
    
    # Set the end of the first balloon as the current endpoint
    current_end = points[0][1]
    
    # Iterate through the rest of the balloons
    for i from 1 to length of points - 1:
        # If the start of the current balloon is greater than current_end
        if points[i][0] > current_end:
            # Increment the arrow count
            arrows += 1
            # Update the current_end to the end of the current balloon
            current_end = points[i][1]
    
    # Return the total number of arrows needed
    return arrows

Further Elucidation

Sorting Rationale : By sorting the balloons by their end positions, each group of overlapping balloons can be easily identified because the balloons with an earlier end must be part of the earliest shot arrow.
Tracking Overlaps : The
current_end

helps in tracking where the current arrow shot is supposed to end. Whenever a new balloon's start is beyond this
current_end

, it indicates that a new arrow is needed because the current balloon cannot be burst by the previously considered arrow.
Time Complexity : Sorting the balloons takes \(O(n \log n)\), and iterating through the sorted list takes \(O(n)\), making the solution efficient even for larger inputs.

By following these steps, you should be able to effectively determine the minimum number of arrows required to burst all the balloons in the given problem.