Approach Summary
Sort intervals and queries. Sweep queries in order; add intervals whose start ≤ query to a min-heap by size. Remove stale entries (end < query).
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 Interval to Include Each Query optimally, we apply the intervals algorithmic strategy. Sort intervals and queries. Sweep queries in order; add intervals whose start ≤ query to a min-heap by size. Remove stale entries (end < query). 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 + q) log n) time, utilizing O(n + q) auxiliary memory.
Time Complexity: O((n + q) log n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(n + q) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O((n + q) log n), Space: O(n + q)
public int minimumIntervalToIncludeQuery(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 + q) log n), Space: O(n + q)
int minimumIntervalToIncludeQuery(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_interval_to_include_query(self, nums, *args) -> object:
# Optimal intervals solution
# Time: O((n + q) log n), Space: O(n + q)
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 minimumIntervalToIncludeQuery = 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: Offline query + sweep line + min-heap.
- Constraint condition: Process queries in sorted order, lazy-evict heap.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O((n + q) log n) time budget without stack overflow or time limit exceeded.
Master the Intervals Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Intervals pattern.
How to Recognize This Pattern
- Offline query + sweep line + min-heap
- Process queries in sorted order, lazy-evict heap
Complexity Analysis
Time Complexity
O((n + q) log n)
Space Complexity
O(n + q)