Tome Signal Analyzer 32 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and signal metrics, construct an optimal algorithm to evaluate and compute the target analyzer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Analyzer 32"
WHY DOES IT MATTER?
Monotonic stack patterns turn quadratic neighbor searches into linear scans.
OPTIMIZATION CHALLENGE
The key is reducing repeated comparisons by maintaining a strict order in the stack.
REAL-WORLD CONNECTION
They model real‑time signal buffering where older weaker readings are discarded for newer stronger ones.
Initialize the stack with a sentinel and always check emptiness before accessing the top to avoid runtime errors.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to finding, for each element in the sequence, the nearest element to its right that satisfies a specific metric condition (e.g., greater signal strength). A monotonic decreasing stack preserves candidates in O(1) amortized time per element, allowing us to resolve each query by popping weaker elements until the condition is met. Naïve double loops compare every pair, leading to O(n²) time which explodes for large n (10⁵‑10⁶). The optimal paradigm leverages the stack’s LIFO property to maintain a decreasing order, guaranteeing each element is pushed and popped at most once, thus achieving linear time.
Interview Questions on This Problem
Q1Why does a monotonic stack guarantee O(n) time for nearest‑greater queries?
Each element is pushed once and popped at most once, so total operations are bounded by 2n. This amortized analysis yields linear time.
Q2How would you modify the algorithm to find the nearest smaller element instead?
Use an increasing (monotonic) stack that pops while the top is greater than the current element. The rest of the logic stays identical.
Q3What edge case must you handle when the target element has no qualifying neighbor?
Return a sentinel value (e.g., -1 or 0) after the stack is empty, indicating no such neighbor exists. Ensure this is applied consistently for all positions.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 3
Output
50
Explanation: Step 1: Iterate through the input array and check each element against the given threshold K = 3. Step 2: For elements greater than K, add them to the sum. Step 3: For elements less than or equal to K, do not add them to the sum. Step 4: After iterating through the entire array, return the sum.
Input
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1] and K = 1
Output
0
Explanation: Step 1: Iterate through the input array and check each element against the given threshold K = 1. Step 2: For elements greater than K, add them to the sum. Step 3: For elements less than or equal to K, do not add them to the sum. Step 4: After iterating through the entire array, return the sum.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a monotonic decreasing stack to resolve each index in amortized O(1), achieving overall O(n) time.
Brute Force Approach
For each index, scan rightward until a qualifying element is found, resulting in O(n²) time.
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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