How do you solve Minimum Number of Arrows to Burst Balloons on LeetCode?

Minimum Number of Arrows to Burst Balloons (LeetCode #452) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Shooting an arrow at the end of a balloon can potentially burst multiple overlapping balloons.

What is the time complexity of Minimum Number of Arrows to Burst Balloons?

The optimal solution for Minimum Number of Arrows to Burst Balloons runs in O(n log n) time and uses O(1) space. The complexity is dominated by the sorting step, which is O(n log n). The subsequent iteration through the points is O(n).

What is the brute force solution for Minimum Number of Arrows to Burst Balloons?

The brute force approach for Minimum Number of Arrows to Burst Balloons is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each balloon and determining if an arrow shot at a specific position can burst it. This means we would try shooting arrows at every possible position and count how many arrows are needed to burst all balloons. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Number of Arrows to Burst Balloons?

Minimum Number of Arrows to Burst Balloons uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting, Interval Scheduling.

What are the interview tips for Minimum Number of Arrows to Burst Balloons?

Always consider sorting as a first step when dealing with interval problems. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice identifying overlapping intervals in different contexts.

What are common mistakes when solving Minimum Number of Arrows to Burst Balloons?

Not sorting the balloons before processing them, which leads to incorrect results. Failing to update the current_end correctly after shooting an arrow.

#452

Minimum Number of Arrows to Burst Balloons

Medium

ArrayGreedySortingGreedySortingInterval Scheduling

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

The optimal approach uses a greedy algorithm. By sorting the balloons by their end positions, we can ensure that we always shoot the arrow at the end of the current balloon that overlaps with others, minimizing the number of arrows needed.

⚙️

Algorithm

3 steps

1Step 1: Sort the balloons by their end positions.
2Step 2: Initialize arrows to 0 and set the current_end to the end of the first balloon.
3Step 3: Iterate through the sorted balloons, and if the start of the current balloon is greater than current_end, increment arrows and update current_end.

solution.py11 lines

1def findMinArrowShots(points):
2    if not points:
3        return 0
4    points.sort(key=lambda x: x[1])
5    arrows = 1
6    current_end = points[0][1]
7    for start, end in points[1:]:
8        if start > current_end:
9            arrows += 1
10            current_end = end
11    return arrows

ℹ

Complexity note: The complexity is dominated by the sorting step, which is O(n log n). The subsequent iteration through the points is O(n).

1Shooting an arrow at the end of a balloon can potentially burst multiple overlapping balloons.
2Sorting the balloons by their end positions allows us to minimize the number of arrows needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.