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Prefix Hash & Frequency Counting

Solve Subarray Sum Equals K, Longest Substring Without Repeating Characters, and Prefix Hash problems.

Anuj Kumar Singh Written by Anuj Kumar Singh (Lead Engineer, 13+ yrs exp) 5 min read Verified Spring Boot 3+ Guide

Real-World Analogy

Frequency counting is keeping a tally sheet of votes cast during an election—incrementing the vote count for a candidate instantly upon reading each ballot.

Frequency Counting & Subarray Sums

Using a HashMap to store cumulative prefix sums allows finding subarrays summing to $K$ in $O(N)$ time.

Production Code Example:

SubarraySumK.java
import java.util.HashMap;
public class SubarraySumK {
    public int subarraySum(int[] nums, int k) {
        HashMap<Integer, Integer> map = new HashMap<>();
        map.put(0, 1);
        int count = 0, sum = 0;
        for (int x : nums) {
            sum += x;
            if (map.containsKey(sum - k)) count += map.get(sum - k);
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
        return count;
    }
}

Key Complexity & Algorithmic Takeaways:

When implementing Prefix Hash & Frequency Counting in coding interviews and production applications, keep these core guidelines in mind:

  • Time Complexity Analysis: Always evaluate best-case, average-case, and worst-case time complexities ($O(1)$, $O(\log n)$, $O(n)$, $O(n \log n)$, $O(n^2)$).
  • Space Complexity & Memory Bounds: Account for auxiliary memory usage, call stack frame recursion overhead, and heap allocations.
  • Edge Cases & Validation: Test empty inputs, null pointers, single-element collections, duplicate values, and integer overflow bounds.
  • Optimal vs Naive Solutions: Start with a clear brute-force solution, then optimize using techniques like Hashing, Two Pointers, Windowing, or Dynamic Programming.

Summary Takeaway:

Mastering Prefix Hash & Frequency Counting provides the foundational problem-solving skills needed to pass technical coding interviews at top tech companies and write ultra-performant software systems.

Key Formula

If $PrefixSum[R] - PrefixSum[L-1] = K$, then $PrefixSum[L-1] = PrefixSum[R] - K$.