You are given an integer array nums.
Define a frequency balance subarray as follows:
f such that every distinct value in the subarray occurs either f or 2 * f times, and both frequencies occur among the distinct values.Return an integer denoting the length of the longest frequency balance subarray.
Example 1:
Input: nums = [1,2,2,1,2,3,3,3]
Output: 5
Explanation:
[2, 1, 2, 3, 3].Example 2:
Input: nums = [5,5,5,5]
Output: 4
Explanation:
[5, 5, 5, 5].Example 3:
Input: nums = [1,2,3,4]
Output: 1
Explanation:
Since all elements appear only once, the length of the longest frequency balance subarray is 1.
Constraints:
1 <= nums.length <= 1031 <= nums[i] <= 109Problem Overview: Given an array, identify subarrays where the frequencies of the elements are balanced. A balanced subarray means every distinct value appears the same number of times inside that subarray. The task is to efficiently detect or count such segments.
Approach 1: Brute Force Enumeration (O(n^3) time, O(k) space)
Generate every possible subarray using two nested loops and recompute frequencies from scratch for each segment. For each candidate range [i, j], build a frequency map and check whether all values have the same count. The check requires iterating over the map to confirm that the minimum and maximum frequencies match. This approach is straightforward and demonstrates the definition of a balanced subarray clearly, but recomputing frequencies repeatedly makes it too slow for large inputs.
Approach 2: Incremental Frequency Map (O(n^2) time, O(k) space)
Fix the starting index and expand the right boundary one element at a time. Maintain a running hash map that tracks frequencies of elements in the current subarray. After each expansion, update the map and check if all frequencies are equal by comparing the smallest and largest counts. Because the map updates happen in constant time per step, the expensive recomputation disappears. This approach reduces the complexity significantly while keeping the implementation simple using hash map frequency tracking.
Approach 3: Frequency Pattern Hashing (Near O(n) average, O(n) space)
A more optimized strategy represents the frequency distribution as a normalized pattern. Maintain counts for each value and track a "frequency of frequencies" structure so you can determine when all active elements share the same count. By storing normalized prefix states in a hash map, repeated patterns indicate that the segment between them maintains balanced frequency growth. This idea is similar to prefix-difference techniques used in prefix sum problems and leverages constant-time hash lookups to detect matches quickly.
Recommended for interviews: The incremental hash map approach is the most practical explanation during interviews. Starting with brute force shows understanding of the definition, but moving to the O(n^2) expansion technique demonstrates control over frequency counting and hash-based optimization. Discussing prefix-state hashing shows deeper algorithmic insight and familiarity with advanced counting techniques.
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Try solving it yourself| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Enumeration | O(n^3) | O(k) | Useful for understanding the definition or verifying small inputs |
| Incremental Frequency Map | O(n^2) | O(k) | General solution when constraints allow quadratic expansion |
| Frequency Pattern Hashing | O(n) average | O(n) | Large inputs where repeated normalized frequency states can be hashed |
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