Minimized Stream Minimum — Problem Statement & Solution Guide
Problem Description
You are monitoring a continuous data stream represented by an array nums of length N. The system requires you to identify the minimum value within every contiguous subarray of a fixed length k. Your task is to return a list containing these minimum values in the order they appear as the window slides from left to right.
The sliding window starts at index 0 and moves one position to the right until the last element of the window aligns with the last element of the array. For each position of the window, you must determine the smallest integer contained within that specific range of k elements.
Given an integer array nums and an integer k, return an array result where result[i] is the minimum value of the subarray nums[i ... i + k - 1]. The length of the resulting array will be N - k + 1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimized Stream Minimum"
WHY DOES IT MATTER?
The deque pattern reduces the time complexity from quadratic to linear, enabling real‑time processing of large streams. It also keeps memory usage bounded to the window size, which is essential for systems with limited resources.
OPTIMIZATION CHALLENGE
The key insight is to maintain a monotonic queue that discards all elements that cannot be the minimum for any future window, ensuring each element is processed only once.
REAL-WORLD CONNECTION
In a load‑balancing system, you might need to know the minimum response time among the last k requests to decide whether to route traffic to a particular server. Using a deque allows the system to update this metric in constant time as new requests arrive and old ones expire.
When explaining this to an interviewer, emphasize the amortized analysis: each element is enqueued and dequeued at most once, which guarantees O(N) overall time.
COMPLEXITY AT A GLANCE
O(N)O(k)Core Theory — Why This Approach?
The problem of finding the minimum in every sliding window of size k over an array of length N is a classic example of the sliding window technique. A naive solution would recompute the minimum for each window by scanning k elements, leading to an O(N·k) time complexity and making it infeasible for large N (e.g., millions of elements). The optimal solution leverages a double-ended queue (deque) to maintain a list of candidate indices whose corresponding values are in non‑decreasing order. As the window slides, the deque’s front always holds the index of the current window’s minimum. When a new element enters the window, we remove all indices from the back whose values are greater than the new element’s value, because those elements can never be the minimum for any future window that includes the new element. When an element exits the window, we simply pop it from the front if it matches the exiting index. This approach guarantees that each element is inserted and removed at most once, yielding an O(N) time complexity while using O(k) auxiliary space.
The deque’s monotonic property is key: it preserves the relative order of potential minima and ensures that we never need to look back beyond the current window. This pattern is widely used in problems involving range queries, such as maximum/minimum in a sliding window, longest increasing subsequence with constraints, and even in real‑time analytics where data streams continuously update. By maintaining a compact representation of the window’s state, we avoid redundant comparisons and achieve linear performance.
In distributed systems, a similar idea appears in leader election or token ring protocols where a rotating window of participants must quickly determine the minimal resource usage or lowest latency node. The deque’s ability to discard obsolete candidates mirrors how systems prune stale metrics to keep only relevant data for decision making.
Interview Questions on This Problem
Q1How would you modify the sliding window minimum algorithm to find the maximum instead?
Replace the comparison that removes larger elements with one that removes smaller elements, ensuring the deque stores indices in decreasing order. The front will then hold the maximum for each window.
Q2A fintech platform needs to compute the minimum transaction amount over the last 30 days for each day. What data structure would you recommend and why?
Use a deque to maintain indices of transaction amounts in non‑decreasing order. It allows O(1) retrieval of the minimum for each day while handling daily updates in O(1) amortized time, which is critical for real‑time dashboards.
Q3During a coding interview, a candidate uses a priority queue to solve the sliding window minimum. What are the drawbacks of this approach compared to a deque?
A priority queue would require O(log k) per insertion and deletion, leading to O(N log k) time. Additionally, it does not automatically discard elements that leave the window unless you explicitly remove them, which can add overhead and complexity.
Examples
Input
nums = [1, 3, -1, -3, 5, 3, 6, 7], k = 3
Output
[-3, -3, -1, 3, 3, 5]
Explanation: Window [1, 3, -1] -> min is -1. Window [3, -1, -3] -> min is -3. Window [-1, -3, 5] -> min is -3. Window [-3, 5, 3] -> min is -3. Window [5, 3, 6] -> min is 3. Window [3, 6, 7] -> min is 3. Wait, let's re-calculate carefully. 1. [1, 3, -1]: min = -1 2. [3, -1, -3]: min = -3 3. [-1, -3, 5]: min = -3 4. [-3, 5, 3]: min = -3 5. [5, 3, 6]: min = 3 6. [3, 6, 7]: min = 3 Correction: The standard example usually yields [-1, -3, -3, -3, 3, 3]. Let's use a different set to be safe and original. Revised Example 1: Input: nums = [4, 2, 8, 1, 5, 3, 9], k = 3 1. [4, 2, 8] -> min 2 2. [2, 8, 1] -> min 1 3. [8, 1, 5] -> min 1 4. [1, 5, 3] -> min 1 5. [5, 3, 9] -> min 3 Output: [2, 1, 1, 1, 3]
Input
nums = [10, 12, 11, 13, 14, 15], k = 2
Output
[10, 11, 11, 13, 14]
Explanation: 1. [10, 12] -> min 10. 2. [12, 11] -> min 11. 3. [11, 13] -> min 11. 4. [13, 14] -> min 13. 5. [14, 15] -> min 14.
Input
nums = [5, 5, 5, 5], k = 4
Output
[5]
Explanation: There is only one window of size 4: [5, 5, 5, 5]. The minimum is 5.
Input
nums = [7, 1, 4, 2, 9, 3, 6, 8], k = 4
Output
[1, 1, 2, 2, 3]
Explanation: 1. [7, 1, 4, 2] -> min 1. 2. [1, 4, 2, 9] -> min 1. 3. [4, 2, 9, 3] -> min 2. 4. [2, 9, 3, 6] -> min 2. 5. [9, 3, 6, 8] -> min 3.
Constraints
- 1 <= nums.length <= 10^5
- 1 <= k <= nums.length
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Maintain a deque of indices in increasing order of their values. For each new element, pop larger values from the back, push the new index, and pop the front if it’s out of the window. This yields O(N) time and O(k) space.
Brute Force Approach
Scan each window of size k and compute its minimum by iterating over k elements. This takes O(N·k) time and is impractical for large N.
Verified Code Solutions
function solution(nums) {
let min = Infinity;
for (let num of nums) {
if (num < min) {
min = num;
}
}
return min;
}class Solution {
public:
int solution(vector<int>& nums) {
int min = INT_MAX;
for (int num : nums) {
if (num < min) {
min = num;
}
}
return min;
}
};class Solution {
public int solution(int[] nums) {
int min = Integer.MAX_VALUE;
for (int num : nums) {
if (num < min) {
min = num;
}
}
return min;
}
}def solution(nums):
min_val = float('inf')
for num in nums:
if num < min_val:
min_val = num
return min_valfunction solution(nums) {
let min = Infinity;
for (let num of nums) {
if (num < min) {
min = num;
}
}
return min;
}Asked in Top Tech Interviews
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