Tome Cache Aligner 46 — 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 aligner value under given operational constraints. The input array contains integers, and K is a non-negative integer.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Aligner 46"
WHY DOES IT MATTER?
Monotonic stack patterns turn quadratic range queries into linear passes.
OPTIMIZATION CHALLENGE
The key is reducing repeated comparisons by ensuring each element is examined only twice (push and pop).
REAL-WORLD CONNECTION
They model cache eviction policies where the next larger timestamp determines when data is flushed.
Initialize the stack with a sentinel and always store indices, not values, to compute distances without extra passes.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The problem reduces to finding, for each element, the span of consecutive elements where it remains the maximum (or minimum) under the given K constraint. A monotonic stack maintains a decreasing (or increasing) sequence of indices, allowing us to compute the nearest element that breaks the constraint in O(1) amortized time. A naïve double‑loop would compare each pair, leading to O(N²) time which explodes for N > 10⁵, especially when K is large and the constraint must be re‑checked for every sub‑range. By processing the array once and using the stack to pop elements that no longer satisfy the constraint, we achieve a linear‑time solution that respects the stack‑based pattern of “next greater/less element”.
Interview Questions on This Problem
Q1How does a monotonic stack help compute nearest greater elements in linear time?
It stores indices in a monotonic order so that each element is pushed and popped at most once, guaranteeing O(N) operations. The top of the stack always represents the closest candidate that satisfies the monotonic condition.
Q2Why does the naïve O(N²) approach fail for large N even if K is small?
Because it still examines every pair of indices, leading to ~10¹⁰ operations for N = 10⁵, which exceeds time limits regardless of K. Stack‑based linear scans avoid redundant comparisons.
Q3What edge case must be handled when the stack becomes empty while processing?
An empty stack means the current element is the new global extremum, so its span extends to the array boundary. You must treat the boundary index (‑1 or N) accordingly.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
6
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K=3, we iterate through the array and sum numbers less than or equal to K. The numbers 1, 2, and 3 are less than or equal to K, so we sum them up to get 6.
Input
[1, 2, 3, 4, 5], 45
Output
0
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K=45, we iterate through the array and sum numbers less than or equal to K. However, all numbers in the array are greater than K, so we return 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a monotonic stack to find left/right bounds for each element in a single pass – O(N) time.
Brute Force Approach
Check every possible sub‑array, compute the aligner value, and keep the maximum – O(N²) time.
Verified Code Solutions
function solution(nums, K) {
if (K < 0) {
throw new Error('K cannot be negative');
}
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array contains non-numeric values');
}
if (num <= K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int K) {
if (K < 0) {
throw invalid_argument('K cannot be negative');
}
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (K < 0) {
throw new IllegalArgumentException('K cannot be negative');
}
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
if K < 0:
raise ValueError('K cannot be negative')
sum = 0
for num in nums:
if not isinstance(num, (int, float)):
raise ValueError('Input array contains non-numeric values')
if num <= K:
sum += num
return sumfunction solution(nums, K) {
if (K < 0) {
throw new Error('K cannot be negative');
}
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array contains non-numeric values');
}
if (num <= K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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