Tome Cache Tracker 17 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and cache metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints. The input array nums represents the tome and cache metrics, and the integer K represents the target tracker value. The solution should return the sum of all elements in the array that are greater than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Tracker 17"
WHY DOES IT MATTER?
Counting subarrays with a target sum is a classic sliding‑window‑compatible pattern that appears in many analytics pipelines.
OPTIMIZATION CHALLENGE
The key is reducing the quadratic enumeration to linear by reusing cumulative information.
REAL-WORLD CONNECTION
It mirrors detecting a specific transaction total in a stream of financial logs.
Initialize the hashmap with a zero‑sum entry and update counts after processing each element to avoid off‑by‑one errors.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to counting subarrays whose elements sum to a target K. By leveraging prefix sums, each subarray sum can be expressed as the difference between two prefix sums, enabling O(1) lookup of complementary sums. A naive double‑loop enumerates all O(n²) subarrays and accumulates their sums, which quickly exceeds time limits for large n. The optimal paradigm uses a hash map to store frequencies of encountered prefix sums, allowing each element to be processed in constant amortized time while maintaining linear overall complexity.
Interview Questions on This Problem
Q1How does the prefix‑sum + hashmap technique achieve O(n) time for subarray‑sum‑equals‑K?
It records the count of each cumulative sum seen so far; for each new prefix sum, the needed complement (current‑sum‑K) is looked up instantly. This transforms the subarray search into a constant‑time frequency query.
Q2Why can we safely use a single pass without resetting the hashmap?
Because each prefix sum represents all subarrays ending at the current index, and earlier frequencies remain valid for future complements. Resetting would discard needed historical information.
Q3What edge case must be handled when K equals 0?
Subarrays with a net sum of zero must be counted, which includes cases where the same prefix sum appears multiple times. Initializing the hashmap with {0:1} ensures these are captured.
Examples
Input
[3, 4, 5, 2, 2, 2, 3]
Output
12
Explanation: Step-by-step: Given the input [3, 4, 5, 2, 2, 2, 3] and K = 3, we iterate through the array and sum all elements greater than or equal to K, which are 3, 4, and 5. Therefore, the output is 12.
Input
[2, 2, 2, 3]
Output
9
Explanation: Step-by-step: Given the input [2, 2, 2, 3] and K = 2, we iterate through the array and sum all elements greater than or equal to K, which are 2, 2, 2, and 3. Therefore, the output is 9.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a hash map to store prefix‑sum frequencies and compute complements on the fly in one linear scan.
Brute Force Approach
Iterate over every possible start index and accumulate sums to every end index, checking if each equals K.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num >= K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int K) {
int sum = 0;
for (int num : nums) {
if (num >= K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num >= K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num >= K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num >= K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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