Subarray Sum Frequency — Problem Statement & Solution Guide
Problem Description
Given an integer array and an integer k, return the number of continuous subarrays where the sum of the elements equals k. The function should handle empty arrays and return 0 in such cases.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Subarray Sum Frequency"
WHY DOES IT MATTER?
The prefix-sum-with-hashmap pattern turns an O(n^2) problem into O(n), which is critical for large datasets and real-time systems. It demonstrates mastery of cumulative data structures and hash-based counting, skills highly valued in algorithmic interviews.
OPTIMIZATION CHALLENGE
The bottleneck in the naive solution is recomputing sums for overlapping subarrays. By storing prefix sums and using a hash map to count occurrences, we eliminate redundant calculations, reducing time from quadratic to linear.
REAL-WORLD CONNECTION
Consider a bank’s transaction ledger: each day’s balance is a prefix sum. To find periods where the net change equals a target, you simply look up previous balances. Similarly, in network traffic analysis, cumulative packet counts help identify bursts of activity matching a threshold.
Always initialize the hash map with {0:1} to account for subarrays that start at index 0. Forgetting this leads to off-by-one errors and missed counts.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The Subarray Sum Frequency problem is a classic example of leveraging prefix sums and hash maps to transform a quadratic-time brute force into a linear-time solution. In the naive approach, we iterate over all possible start and end indices, summing elements on the fly, which leads to O(n^2) time and is infeasible for large arrays (e.g., n > 10^5). The optimal paradigm uses the observation that if the cumulative sum up to index j is S_j and the cumulative sum up to index i-1 is S_{i-1}, then the subarray [i, j] sums to k iff S_j - S_{i-1} = k. By maintaining a hash map that counts how many times each prefix sum has appeared, we can, for each new prefix sum S_j, immediately determine how many previous indices satisfy S_{i-1} = S_j - k, thus counting all qualifying subarrays ending at j in constant time.
This technique reduces the problem to a single pass over the array, achieving O(n) time while using O(n) auxiliary space for the hash map. It also gracefully handles negative numbers and zero-length arrays, as the initial prefix sum of 0 is stored in the map with a count of 1. The algorithm’s elegance lies in its ability to convert a seemingly combinatorial counting problem into a simple lookup operation, a pattern that appears in many interview questions involving subarray sums, submatrix sums, and even string pattern matching.
The key insight is that the difference between two prefix sums equals the sum of the subarray between them. By precomputing and storing prefix sums, we avoid recomputing sums for overlapping subarrays, thereby eliminating redundant work. This pattern is a cornerstone of efficient algorithm design and is frequently tested in technical interviews at top-tier companies.
Interview Questions on This Problem
Q1How would you modify the algorithm to count subarrays whose product equals k instead of the sum?
For products, you can take logarithms to convert multiplication into addition, but this only works for positive numbers. A more robust approach is to use a hashmap of prefix products, but you must handle zeros separately, as division by zero is undefined. The algorithm then checks if current_product / k exists in the map, similar to the sum case.
Q2A fintech platform needs to detect fraudulent transactions that sum to a suspicious amount within any 24-hour window. How does the subarray sum algorithm help, and what constraints must you consider?
Treat each transaction amount as an array element and the 24-hour window as a sliding window. By maintaining a running prefix sum and a hash map of sums within the window, you can detect if any contiguous set of transactions equals the suspicious amount in O(n) time. Constraints include handling large transaction volumes, ensuring the window slides efficiently, and dealing with negative adjustments (refunds).
Q3In a high-growth startup, you need to process real-time streams of sensor data and count subarrays summing to a threshold k. What data structure would you use to support continuous updates and queries?
A balanced binary search tree or a Fenwick tree can maintain prefix sums dynamically, but the simplest is a hash map of prefix sums updated as new data arrives. For each new element, update the cumulative sum and query the map for cumulative_sum - k. This gives O(1) amortized per update and query, suitable for real-time streaming.
Examples
Input
[1, 1, 1], 2
Output
2
Explanation: Step-by-step: with input [1, 1, 1] and target sum 2, we find two subarrays [1, 1] at indices (0, 1) and (1, 2) that sum up to 2.
Input
[1, 2, 3], 3
Output
2
Explanation: Step-by-step: with input [1, 2, 3] and target sum 3, we find two subarrays [1, 2] at indices (0, 1) and [3] at index (2) that sum up to 3.
Constraints
- 1 <= nums.length <= 20000
- -10^4 <= nums[i] <= 10^4
- -10^5 <= k <= 10^5
Optimal Approach & Strategy
Maintain a cumulative sum and a hash map of its frequencies. For each new element, update the sum and add the count of (sum - k) from the map to the answer. This runs in O(n) time and O(n) space.
Brute Force Approach
Loop over all start indices, accumulate sums for each end index, and increment a counter when the sum equals k. This takes O(n^2) time and is impractical for large arrays.
Verified Code Solutions
function subarraySum(nums, k) { let count = 0; for (let i = 0; i < nums.length; i++) { let sum = 0; for (let j = i; j < nums.length; j++) { sum += nums[j]; if (sum === k) { count++; } } } return count; }class Solution { public: int subarraySum(vector<int>& nums, int k) { int count = 0; for (int i = 0; i < nums.size(); i++) { int sum = 0; for (int j = i; j < nums.size(); j++) { sum += nums[j]; if (sum == k) { count++; } } } return count; } }class Solution { public int subarraySum(int[] nums, int k) { int count = 0; for (int i = 0; i < nums.length; i++) { int sum = 0; for (int j = i; j < nums.length; j++) { sum += nums[j]; if (sum == k) { count++; } } } return count; } }def subarraySum(nums, k): count = 0; for i in range(len(nums)): sum = 0; for j in range(i, len(nums)): sum += nums[j]; if sum == k: count += 1; return countfunction subarraySum(nums, k) { let count = 0; for (let i = 0; i < nums.length; i++) { let sum = 0; for (let j = i; j < nums.length; j++) { sum += nums[j]; if (sum === k) { count++; } } } return count; }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.