Tome Voyage Aligner 23 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and voyage metrics, construct an optimal algorithm to evaluate and compute the target aligner value under given operational constraints. The input array contains integers representing tome and voyage metrics, and the target value K is an integer representing the operational constraint.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Aligner 23"
WHY DOES IT MATTER?
Binary search reduces exponential brute‑force possibilities to logarithmic steps, enabling real‑time responses.
OPTIMIZATION CHALLENGE
The key is formulating a monotonic predicate that can be evaluated in O(1) per check.
REAL-WORLD CONNECTION
It mirrors how databases use indexed lookups to locate records instantly.
Always verify edge conditions (empty array, all true/false) before entering the loop to avoid infinite loops.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
Binary search leverages the monotonic property of sorted data to eliminate half of the remaining search space with each comparison, yielding logarithmic time complexity. Naïve linear scans examine every element, leading to O(n) time which becomes prohibitive for large n and tight time constraints.
The optimal paradigm transforms the problem into a decision function: given an index, can we achieve the target aligner value K? By applying binary search on the index or value domain, we converge to the minimal or maximal feasible solution in O(log n) iterations, while maintaining O(1) auxiliary space.
Interview Questions on This Problem
Q1Why does binary search require a sorted (or monotonic) input?
Because each comparison must reliably discard one half of the remaining candidates. Without monotonicity the discarded half could contain the answer, breaking correctness.
Q2How can you adapt binary search to find the first index that satisfies a predicate?
Use a left‑biased loop that moves the high pointer to mid when the predicate is true, otherwise move low to mid+1. The loop terminates with low pointing to the minimal satisfying index.
Q3What is the time complexity of binary search on an array of size n and why?
O(log n) because the search interval halves each iteration. After k steps the interval size is n/2^k, reaching 1 when k ≈ log₂ n.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
55
Explanation: Step 1: Initialize sum to 0. Step 2: Iterate through the array from left to right. Step 3: For each element, check if it is greater than K. Step 4: If it is, add it to the sum. Step 5: Return the sum.
Input
[60, 60, 60, 60, 60, 60, 60, 60, 60, 60]
Output
600
Explanation: Step 1: Initialize sum to 0. Step 2: Iterate through the array from left to right. Step 3: For each element, check if it is greater than K. Step 4: If it is, add it to the sum. Step 5: Return the sum.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Apply binary search on the index/value space using a monotonic predicate, achieving O(log n) time with constant extra space.
Brute Force Approach
Iterate through the array and evaluate the aligner condition for each element until the target is met, resulting in O(n) 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
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.