Optimal Recipe Component — Problem Statement & Solution Guide
Problem Description
You are given an array ingredients representing the sequence of items available in a supply chain, and an integer k representing the number of distinct critical components required for a specific recipe. Your task is to determine the minimum length of a contiguous subarray within ingredients that contains at least k distinct elements. If no such subarray exists, return -1.
The input consists of a single integer array ingredients where each element represents a unique item identifier, and an integer k indicating the target count of distinct items. The output should be the smallest window size that satisfies the distinctness condition. This problem requires efficient sliding window techniques to achieve optimal time complexity.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Recipe Component"
WHY DOES IT MATTER?
The Sliding Window pattern is essential for optimizing subarray problems where the property of interest (like distinct count) is monotonic or can be maintained incrementally. It transforms O(N^2) or O(N^3) brute-force solutions into O(N) linear time, which is crucial for handling large datasets in modern applications.
OPTIMIZATION CHALLENGE
The key insight is that once the window satisfies the condition (distinct >= K), expanding the right pointer further is useless for minimizing length; instead, we must shrink the left pointer to find the smallest valid window ending at the current right pointer. This 'expand until valid, shrink while valid' strategy ensures linear complexity.
REAL-WORLD CONNECTION
This is analogous to monitoring network traffic for a specific set of IP addresses. You want to find the shortest time window where all K critical servers were accessed. You don't need to re-scan the entire log for every possible start time; you just slide a window forward, adding new events and dropping old ones, to find the tightest cluster of activity.
In interviews, explicitly state that you are using a 'variable-sized sliding window' and explain the invariant: 'The window is always the smallest valid window ending at the current right pointer.' This demonstrates a deep understanding of the algorithm's state management.
COMPLEXITY AT A GLANCE
O(N)O(K)Core Theory — Why This Approach?
The problem of finding the minimum length subarray containing at least K distinct elements is a classic application of the Sliding Window technique, specifically the 'variable-sized' or 'expanding-then-shrinking' window. Unlike fixed-size windows, this approach dynamically adjusts the window boundaries to maintain a specific invariant: the count of distinct elements within the window must be greater than or equal to K. The core theoretical underpinning relies on the monotonicity of the window size relative to the right pointer. As the right pointer expands, the number of distinct elements is non-decreasing. Once the condition (distinct >= K) is met, we can attempt to shrink the window from the left to minimize its length, as any further expansion of the right pointer without shrinking will only increase the window size, potentially missing a smaller valid subarray that starts later.
Interview Questions on This Problem
Q1At a fintech platform, we need to identify the shortest time window in a transaction log where at least 5 different currency types were traded. How would you design an efficient algorithm to find this minimum window size in a stream of transactions?
I would use a sliding window approach with two pointers, left and right. I'd maintain a frequency map to track distinct currencies. I expand the right pointer until I have 5 distinct currencies. Then, I shrink the left pointer as much as possible while keeping the distinct count >= 5, updating the minimum window size. This ensures O(N) time complexity, which is critical for real-time transaction processing.
Q2In a distributed system, we have a log of service calls. We need to find the minimum number of consecutive calls that include at least K unique service endpoints. Why is a brute-force O(N^2) approach unacceptable, and how does the sliding window technique optimize this?
Brute force is O(N^2) or O(N^2 * K) due to checking every subarray, which is too slow for high-throughput logs. The sliding window technique leverages the fact that if a window [L, R] is valid, any window [L', R] where L' > L is either valid or invalid, but we only need to check the boundary where it becomes invalid. This reduces the problem to O(N) because each element is added and removed from the window at most once.
Q3You are building a recommendation engine that needs to find the shortest sequence of user interactions containing at least K different content categories. How do you handle the case where no such sequence exists?
I initialize the minimum length to infinity. I use a sliding window to find all valid windows. If the minimum length remains infinity after processing the entire array, I return -1. This handles the edge case where the total distinct elements in the array are less than K, ensuring robustness in the recommendation pipeline.
Examples
Input
ingredients = [1, 2, 3, 1, 2, 4], k = 3
Output
3
Explanation: The subarray [1, 2, 3] has length 3 and contains 3 distinct elements. The subarray [2, 3, 1] also has length 3. No subarray of length 2 contains 3 distinct elements. Thus, the minimum length is 3.
Input
ingredients = [1, 1, 1, 2, 2, 3], k = 2
Output
2
Explanation: The subarray [1, 2] at indices 2-3 has length 2 and contains 2 distinct elements. The subarray [2, 3] at indices 4-5 also has length 2. No subarray of length 1 contains 2 distinct elements. Thus, the minimum length is 2.
Input
ingredients = [5, 5, 5, 5], k = 2
Output
-1
Explanation: All elements are identical (5). No subarray contains 2 distinct elements. Thus, the answer is -1.
Input
ingredients = [1, 2, 3, 4, 5], k = 5
Output
5
Explanation: The entire array [1, 2, 3, 4, 5] has length 5 and contains 5 distinct elements. No smaller subarray can contain 5 distinct elements. Thus, the minimum length is 5.
Constraints
- 1 <= ingredients.length <= 10^5
- 1 <= ingredients[i] <= 10^9
- 1 <= k <= 10^5
- k <= ingredients.length
Optimal Approach & Strategy
Use a sliding window with two pointers and a frequency map. Expand the right pointer to include elements until K distinct elements are found, then shrink the left pointer to minimize the window size while maintaining the K distinct constraint.
Brute Force Approach
Iterate through all possible start and end indices of subarrays, counting distinct elements in each using a set. This results in O(N^2) time complexity, which is inefficient for large inputs.
Verified Code Solutions
function minSubarrayLength(ingredients, k) {
if (k === 0) return 0;
if (ingredients.length < k) return -1;
const freq = new Map();
let distinct = 0;
let left = 0;
let minLen = Infinity;
for (let right = 0; right < ingredients.length; right++) {
const val = ingredients[right];
freq.set(val, (freq.get(val) || 0) + 1);
if (freq.get(val) === 1) {
distinct++;
}
while (distinct >= k) {
const currentLen = right - left + 1;
if (currentLen < minLen) {
minLen = currentLen;
}
const leftVal = ingredients[left];
freq.set(leftVal, freq.get(leftVal) - 1);
if (freq.get(leftVal) === 0) {
distinct--;
}
left++;
}
}
return minLen === Infinity ? -1 : minLen;
}
const ingredients = [1, 2, 3, 1, 2, 4];
const k = 3;
console.log(minSubarrayLength(ingredients, k));#include <iostream>
#include <vector>
#include <unordered_map>
#include <climits>
using namespace std;
int minSubarrayLength(vector<int>& ingredients, int k) {
if (k == 0) return 0;
if (ingredients.size() < k) return -1;
unordered_map<int, int> freq;
int distinct = 0;
int left = 0;
int minLen = INT_MAX;
for (int right = 0; right < ingredients.size(); right++) {
int val = ingredients[right];
freq[val]++;
if (freq[val] == 1) {
distinct++;
}
while (distinct >= k) {
int currentLen = right - left + 1;
if (currentLen < minLen) {
minLen = currentLen;
}
int leftVal = ingredients[left];
freq[leftVal]--;
if (freq[leftVal] == 0) {
distinct--;
}
left++;
}
}
return minLen == INT_MAX ? -1 : minLen;
}
int main() {
vector<int> ingredients = {1, 2, 3, 1, 2, 4};
int k = 3;
int result = minSubarrayLength(ingredients, k);
cout << result << endl;
return 0;
}import java.util.*;
public class Solution {
public static int minSubarrayLength(int[] ingredients, int k) {
if (k == 0) return 0;
if (ingredients.length < k) return -1;
Map<Integer, Integer> freq = new HashMap<>();
int distinct = 0;
int left = 0;
int minLen = Integer.MAX_VALUE;
for (int right = 0; right < ingredients.length; right++) {
int val = ingredients[right];
freq.put(val, freq.getOrDefault(val, 0) + 1);
if (freq.get(val) == 1) {
distinct++;
}
while (distinct >= k) {
int currentLen = right - left + 1;
if (currentLen < minLen) {
minLen = currentLen;
}
int leftVal = ingredients[left];
freq.put(leftVal, freq.get(leftVal) - 1);
if (freq.get(leftVal) == 0) {
distinct--;
}
left++;
}
}
return minLen == Integer.MAX_VALUE ? -1 : minLen;
}
public static void main(String[] args) {
int[] ingredients = {1, 2, 3, 1, 2, 4};
int k = 3;
int result = minSubarrayLength(ingredients, k);
System.out.println(result);
}
}from typing import List
def min_subarray_length(ingredients: List[int], k: int) -> int:
if k == 0:
return 0
if len(ingredients) < k:
return -1
freq = {}
distinct = 0
left = 0
min_len = float('inf')
for right in range(len(ingredients)):
val = ingredients[right]
freq[val] = freq.get(val, 0) + 1
if freq[val] == 1:
distinct += 1
while distinct >= k:
current_len = right - left + 1
if current_len < min_len:
min_len = current_len
left_val = ingredients[left]
freq[left_val] -= 1
if freq[left_val] == 0:
distinct -= 1
left += 1
return -1 if min_len == float('inf') else min_len
if __name__ == "__main__":
ingredients = [1, 2, 3, 1, 2, 4]
k = 3
print(min_subarray_length(ingredients, k))function minSubarrayLength(ingredients, k) {
if (k === 0) return 0;
if (ingredients.length < k) return -1;
const freq = new Map();
let distinct = 0;
let left = 0;
let minLen = Infinity;
for (let right = 0; right < ingredients.length; right++) {
const val = ingredients[right];
freq.set(val, (freq.get(val) || 0) + 1);
if (freq.get(val) === 1) {
distinct++;
}
while (distinct >= k) {
const currentLen = right - left + 1;
if (currentLen < minLen) {
minLen = currentLen;
}
const leftVal = ingredients[left];
freq.set(leftVal, freq.get(leftVal) - 1);
if (freq.get(leftVal) === 0) {
distinct--;
}
left++;
}
}
return minLen === Infinity ? -1 : minLen;
}
const ingredients = [1, 2, 3, 1, 2, 4];
const k = 3;
console.log(minSubarrayLength(ingredients, k));Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.