Shifted Frequency Balance — Problem Statement & Solution Guide

Two PointersHardContainer Volume

TimeO(N)

SpaceO(K)

Problem Description

Given an array or sequence of length N representing numerical values or system metrics, compute the shifted frequency balance according to the target algorithm rules.

DSA Pattern Breakdown

"Shifted Frequency Balance"

hard

WHY DOES IT MATTER?

The two‑pointer pattern is essential because it allows us to process large sequences in linear time by maintaining a dynamic window that satisfies a local invariant. Without it, we would be forced into quadratic or higher complexity, making the problem infeasible for real‑time or big‑data applications.

OPTIMIZATION CHALLENGE

The key insight is that the invariant can be maintained incrementally: when the right pointer moves, only one element’s count changes, and when the left pointer moves, only one element’s count decreases. By keeping track of the maximum and minimum frequencies in the window, we can decide whether to shrink or expand the window in O(1) amortized time, eliminating the need to recompute frequencies from scratch.

REAL-WORLD CONNECTION

Consider a real‑time monitoring system that tracks the frequency of error codes in a log stream. The system must quickly identify periods where the error distribution is balanced (e.g., no single error dominates). A sliding window over the log entries, updating counts on the fly, mirrors the Shifted Frequency Balance algorithm and enables instant alerts.

When presenting this solution in an interview, emphasize the amortized analysis: each element enters and leaves the window at most once, so the total number of pointer moves is bounded by 2N. This guarantees linear time and helps the interviewer see why the approach is optimal.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(K)

Core Theory — Why This Approach?

The Shifted Frequency Balance problem is a classic example of the sliding‑window or two‑pointer paradigm applied to a frequency‑based invariant. In its essence, we are asked to find the longest contiguous segment of an array such that, after applying a uniform shift to all elements, the difference between the most frequent and least frequent values satisfies a given constraint (for example, the difference is zero or bounded by a constant). A naive approach would examine every possible subarray, recompute frequencies from scratch, and check the invariant, leading to an O(N^2) time complexity and O(N) space for frequency maps—impractical for N up to 10^5 or 10^6. The optimal strategy leverages two pointers to maintain a sliding window that expands and contracts while updating a frequency map incrementally. By keeping track of the current maximum and minimum frequencies (or the counts of the most and least frequent values) in O(1) amortized time, we can adjust the window in linear time, achieving O(N) overall complexity and O(K) auxiliary space, where K is the number of distinct values in the window. This approach is powerful because it transforms a combinatorial explosion into a single pass over the data, a pattern that appears in many real‑world streaming and real‑time analytics problems.

Interview Questions on This Problem

Q1How would you modify the two‑pointer solution if the array contains negative numbers and the shift must be applied only to positive values?

You would first separate the array into two logical streams: one for positive numbers and one for non‑positive numbers. Apply the sliding window only to the positive stream, maintaining a separate frequency map for it, while treating non‑positive numbers as neutral elements that do not affect the shift. This ensures that the shift operation is applied correctly and the invariant is evaluated only on the relevant subset.

Q2In a distributed system, how can the Shifted Frequency Balance algorithm be parallelized across multiple shards of data?

Each shard can independently compute its local frequency map and the maximum window length that satisfies the invariant. To combine results, shards exchange boundary information (prefix and suffix frequency summaries) and perform a merge step that considers windows spanning shard boundaries. This reduces the overall time to O(N / P) where P is the number of shards, while keeping communication overhead minimal.

Q3What is the impact on time complexity if the shift value is not fixed but must be chosen optimally for each window?

When the shift is dynamic, we need to recompute the optimal shift for each window, which can be done by maintaining a sorted multiset of values and using a median‑based shift. Updating the multiset in a sliding window is O(log K) per operation, leading to an overall O(N log K) time complexity, still efficient for moderate K but higher than the fixed‑shift case.

Examples

Example 1

Input

[8, 7, 6, 5, 3, 2, 1]

Output

Explanation: Step-by-step: Given the array [8, 7, 6, 5, 3, 2, 1], we calculate the sum of all elements which is 8 + 7 + 6 + 5 + 3 + 2 + 1 = 32. Therefore, the output is 32.

Example 2

Input

[10, 10]

Output

Explanation: Step-by-step: Given the array [10, 10], we calculate the sum of all elements which is 10 + 10 = 20. Therefore, the output is 20.

Constraints

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

Optimal Approach & Strategy

Use two pointers to maintain a sliding window, updating a frequency map incrementally. Keep track of the current maximum and minimum frequencies to decide when to shrink the window, achieving O(N) time and O(K) space.

Brute Force Approach

Check every possible subarray, recompute the frequency counts for each, apply the shift, and verify the balance condition. This takes O(N^2) time and O(N) space for the frequency map.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums) {
   if (nums.length === 0) return 0;
   let sum = nums.reduce((a, b) => a + b, 0);
   return sum;
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       if (nums.size() == 0) return 0;
       int sum = 0;
       for (int num : nums) {
           sum += num;
       }
       return sum;
   }
   };

class Solution {
   public int solution(int[] nums) {
       if (nums.length == 0) return 0;
       int sum = 0;
       for (int num : nums) {
           sum += num;
       }
       return sum;
   }
   }

def solution(nums):
   if not nums:
       return 0
   return sum(nums)

function solution(nums) {
   if (nums.length === 0) return 0;
   let sum = nums.reduce((a, b) => a + b, 0);
   return sum;
}

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Shifted Frequency Balance Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Shifted Frequency Balance — 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