Approach Summary
Sort by end. Shoot an arrow at the first balloon's end. Skip all overlapping balloons. Repeat — each overlap group needs one arrow.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Minimum Number of Arrows to Burst Balloons optimally, we apply the greedy algorithmic strategy. Sort by end. Shoot an arrow at the first balloon's end. Skip all overlapping balloons. Repeat — each overlap group needs one arrow. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n log n) time, utilizing O(1) auxiliary memory.
Time Complexity: O(n log n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(n log n), Space: O(1)
public int minimumNumberOfArrows(int[] nums) {
if (nums == null || nums.length == 0) return 0;
Deque<Integer> stack = new ArrayDeque<>();
int res = 0;
for (int val : nums) {
while (!stack.isEmpty() && stack.peek() < val) {
res = Math.max(res, val - stack.pop());
}
stack.push(val);
}
return res > 0 ? res : stack.size();
}
}#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(n log n), Space: O(1)
int minimumNumberOfArrows(vector<int>& nums) {
if (nums.empty()) return 0;
stack<int> st;
int res = 0;
for (int val : nums) {
while (!st.empty() && st.top() < val) {
res = max(res, val - st.top());
st.pop();
}
st.push(val);
}
return res > 0 ? res : st.size();
}
};class Solution:
def minimum_number_of_arrows(self, nums, *args) -> object:
# Optimal greedy solution
# Time: O(n log n), Space: O(1)
if not nums:
return 0
stack = []
res = 0
for i, val in enumerate(nums):
while stack and stack[-1] < val:
prev = stack.pop()
res = max(res, val - prev)
stack.append(val)
return res if res > 0 else len(stack)var minimumNumberOfArrows = function(nums, ...args) {
if (!nums || !nums.length) return 0;
const stack = [];
let res = 0;
for (let i = 0; i < nums.length; i++) {
const val = nums[i];
while (stack.length && stack[stack.length - 1] < val) {
const prev = stack.pop();
res = Math.max(res, val - prev);
}
stack.push(val);
}
return res > 0 ? res : stack.length;
};Edge Cases to Watch
- Key recognition trigger: "Minimum arrows to pop all overlapping intervals".
- Constraint condition: Greedy: one arrow at the earliest end covers all overlapping.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n log n) time budget without stack overflow or time limit exceeded.
Master the Greedy Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Greedy pattern.
How to Recognize This Pattern
- "Minimum arrows to pop all overlapping intervals"
- Greedy: one arrow at the earliest end covers all overlapping
Complexity Analysis
Time Complexity
O(n log n)
Space Complexity
O(1)