Frequency Window Constraint Resolver — Problem Statement & Solution Guide

Sliding WindowHardMinimum Window Substring

TimeO(N + M)

SpaceO(M)

Problem Description

You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the frequency window constraint using the **Minimum Window Substring** methodology.

Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.

DSA Pattern Breakdown

"Frequency Window Constraint Resolver"

hard

WHY DOES IT MATTER?

The sliding window pattern is essential for problems where a contiguous segment must satisfy a dynamic constraint, allowing linear-time solutions where brute force would be quadratic. It reduces redundant work by reusing previously computed information as the window slides.

OPTIMIZATION CHALLENGE

The core optimization is to maintain a counter of missing required characters, enabling constant-time validity checks and preventing repeated scans of the window.

REAL-WORLD CONNECTION

In real-time monitoring, you might need to detect the shortest period where all critical metrics exceed thresholds. The sliding window algorithm can process streaming data in O(1) per event, enabling near-instant alerts.

During interviews, emphasize the two-pointer technique, the use of a frequency map, and the invariant that the counter reflects missing characters. Show how you update the counter when expanding or contracting the window.

COMPLEXITY AT A GLANCE

⏱ Time:O(N + M)

💾 Space:O(M)

Core Theory — Why This Approach?

The Minimum Window Substring problem is a classic example of the sliding window paradigm, where we maintain a dynamic window over the input string and adjust its boundaries to satisfy a frequency constraint. A naive approach would enumerate all substrings and check if they contain the required characters, leading to O(N^2) time and O(1) space, which quickly becomes infeasible for large N. The optimal solution uses two pointers (left and right) to expand the window until it satisfies the constraint, then contracts from the left to find the minimal window, all while maintaining a frequency map of characters. This yields an O(N) time complexity because each character is processed at most twice, and O(M) space where M is the number of distinct characters in the target string.

The key insight is that the window’s validity can be checked in constant time using a counter that tracks how many required characters are still missing. When the counter reaches zero, the window is valid; we then try to shrink it from the left, updating the counter as characters are removed. This two-phase expansion and contraction ensures we never revisit the same substring multiple times, which is why the algorithm is linear.

Because the problem is essentially a frequency constraint over a contiguous segment, it maps directly to many real-world scenarios such as finding the shortest time window that covers all required events in log streams or locating the minimal subarray that satisfies a set of resource quotas in distributed systems. Understanding this pattern equips engineers to solve a wide range of sliding window problems efficiently.

Interview Questions on This Problem

Q1How would you modify the algorithm if the target string contains duplicate characters?

The frequency map must count each occurrence of the target characters. The counter should represent the total number of required characters, not just distinct ones, and we decrement it only when a character’s count in the window matches its required count. This ensures the window contains the exact multiplicity needed.

Q2What is the time complexity if the alphabet size is constant, e.g., ASCII?

With a constant alphabet, the frequency map can be an array of fixed size, so the space complexity becomes O(1). The time complexity remains O(N) because the algorithm still processes each character at most twice, independent of alphabet size.

Q3In a distributed log aggregation system, how would you apply this algorithm to find the minimal time window covering all event types?

Treat the log entries as the input string and the set of event types as the target. Use a sliding window over timestamps, maintaining a count of each event type seen. When all types are present, record the window and try to shrink it by moving the left pointer, updating counts accordingly. This yields the shortest time interval containing all required events.

Examples

Example 1

Input

[10, -5, 3, 2, 7]

Output

Explanation: Step-by-step: Given the array [10, -5, 3, 2, 7], we need to find the smallest window that contains all non-negative values. The window [10, 3, 2, 7] contains all non-negative values and has a size of 4. However, the window [10, -5, 3, 2, 7] has a size of 5 and also contains all non-negative values. Therefore, the correct output is 5.

Example 2

Input

[1, 2, 3, 4, 5]

Output

Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5], we need to find the smallest window that contains all non-negative values. The entire array contains all non-negative values and has a size of 5. Therefore, the correct output is 5.

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N) or O(N log N)
Space Complexity: O(N) or O(1)

Optimal Approach & Strategy

Use two pointers to maintain a sliding window, a frequency map for target characters, and a counter for missing characters. Expand right until valid, then contract left to minimize, achieving O(N) time.

Brute Force Approach

Check every possible substring and count characters to see if it satisfies the requirement. This takes O(N^2) time and is impractical for large inputs.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N + M)

function solution(nums) {
   let left = 0, right = 0, minLen = Infinity, minSum = 0;
   let sum = 0;
   while (right < nums.length) {
       sum += nums[right];
       while (sum < 0) {
           sum -= nums[left];
           left++;
       }
       if (right - left + 1 < minLen) {
           minLen = right - left + 1;
           minSum = sum;
       }
       right++;
   }
   return minLen;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           int left = 0, right = 0, minLen = INT_MAX, minSum = 0;
           int sum = 0;
           while (right < nums.size()) {
               sum += nums[right];
               while (sum < 0) {
                   sum -= nums[left];
                   left++;
               }
               if (right - left + 1 < minLen) {
                   minLen = right - left + 1;
                   minSum = sum;
               }
               right++;
           }
           return minLen;
       }
   };

class Solution {
   public int solution(int[] nums) {
       int left = 0, right = 0, minLen = Integer.MAX_VALUE, minSum = 0;
       int sum = 0;
       while (right < nums.length) {
           sum += nums[right];
           while (sum < 0) {
               sum -= nums[left];
               left++;
           }
           if (right - left + 1 < minLen) {
               minLen = right - left + 1;
               minSum = sum;
           }
           right++;
       }
       return minLen;
   }
}

def solution(nums):
   left = 0
   right = 0
   minLen = float('inf')
   minSum = 0
   sum = 0
   while right < len(nums):
       sum += nums[right]
       while sum < 0:
           sum -= nums[left]
           left += 1
       if right - left + 1 < minLen:
           minLen = right - left + 1
           minSum = sum
       right += 1
   return minLen

function solution(nums) {
   let left = 0, right = 0, minLen = Infinity, minSum = 0;
   let sum = 0;
   while (right < nums.length) {
       sum += nums[right];
       while (sum < 0) {
           sum -= nums[left];
           left++;
       }
       if (right - left + 1 < minLen) {
           minLen = right - left + 1;
           minSum = sum;
       }
       right++;
   }
   return minLen;
}

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Frequency Window Constraint Resolver Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Sliding Window

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Frequency Window Constraint Resolver — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor