Network Network Aligner 41 — Problem Statement & Solution Guide
Problem Description
You are given a sequence of integers nums and a single integer K. Your task is to compute the sum of all elements in nums that are less than or equal to K. The input consists of three lines: the first line contains an integer N, the length of the array; the second line contains N space‑separated integers representing nums; the third line contains the integer K. Output a single integer – the required sum. The solution must run in linear time relative to N and use only O(1) additional memory beyond the input storage.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Network Aligner 41"
WHY DOES IT MATTER?
Aggregating condition‑based values is a fundamental pattern for data summarization.
OPTIMIZATION CHALLENGE
Avoiding sorting or extra data structures reduces time from O(N log N) to O(N).
REAL-WORLD CONNECTION
Think of filtering transaction amounts below a risk threshold in a financial monitoring system.
Keep the accumulator in a primitive type and early‑exit if possible to save cycles.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a linear scan where each element is compared against the threshold K and, if it satisfies the condition, added to an accumulator. This is an instance of a prefix‑sum style aggregation that can be solved in O(N) time without auxiliary data structures. Naïve approaches might attempt sorting or building a binary search tree to answer multiple queries, which incurs O(N log N) overhead and unnecessary memory for a single‑pass requirement. The optimal paradigm leverages the fact that the condition is monotonic per element, allowing a single traversal with constant extra space, which scales linearly with input size.
Interview Questions on This Problem
Q1How would you handle the case where K is smaller than all array elements?
The loop will never add any value, so the accumulator remains zero. This edge case is naturally covered by the same linear algorithm.
Q2Can you extend this solution to answer multiple K queries efficiently?
Yes, by sorting the array and building a prefix sum array, each query can be answered in O(log N) via binary search. This trades preprocessing time for faster query handling.
Q3What is the impact of integer overflow in the sum and how do you mitigate it?
If the sum exceeds the language's integer limits, it may wrap around or throw an error. Use a larger numeric type (e.g., long long) or check for overflow during accumulation.
Examples
Input
5 3 -1 7 4 2 4
Output
8
Explanation: The elements not exceeding K=4 are 3, -1, 4, and 2. Their sum is 3 + (-1) + 4 + 2 = 8.
Input
3 10 20 30 5
Output
0
Explanation: No element in the array is ≤ 5, therefore the sum of qualifying elements is 0.
Input
5 -5 -2 0 5 10 0
Output
-7
Explanation: Elements ≤ 0 are -5, -2, and 0. Adding them yields -5 + (-2) + 0 = -7.
Constraints
- 1 <= N <= 2*10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
- The resulting sum fits in a signed 64‑bit integer.
Optimal Approach & Strategy
Perform a single pass, adding elements ≤ K directly to a running sum, achieving O(N) time and O(1) space.
Brute Force Approach
Sort the array then iterate until elements exceed K, summing as you go, which costs O(N log N).
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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