Node Matrix Validator — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Conditional sum using a running accumulator
O(n)O(1)Problem Description
Given an integer array nums and an integer K, return the sum of all elements strictly greater than K. Include each qualifying occurrence, including negative values greater than a negative threshold. Return 0 if no element qualifies. Do not modify the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Validator"
WHY DOES IT MATTER?
A conditional sum combines filtering and aggregation in one pass.
OPTIMIZATION CHALLENGE
Accumulate qualifying values directly to avoid allocating a filtered list.
REAL-WORLD CONNECTION
Add measurements that exceed a chosen threshold without copying them into another collection.
Use strict greater-than, include qualifying negative numbers, and choose a sum type wide enough for the constraints.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The inclusion rule depends only on each value and K. Maintain the invariant that total is the sum of all qualifying values already visited. Initially no values have been visited, so total is 0. Adding a value exactly when it exceeds K preserves the invariant. After the last value, total is the required sum. Every value may affect the result, so a single query on arbitrary input requires examining all n values in the worst case. One running sum uses constant auxiliary space. The largest possible sum is 100000000000000, which requires a 64-bit integer in C++ and Java and is exactly representable by JavaScript Number. Python integers support this range.
Interview Questions on This Problem
Q1What happens when a value equals K?
It is excluded because the comparison is strictly greater than K.
Q2Can the result be negative?
Yes. With a negative threshold, qualifying negative values can produce a negative sum.
Q3Why use a 64-bit sum in C++ and Java?
Individual values fit in 32 bits, but their combined sum can exceed the signed 32-bit range.
Examples
Input
nums = [1, 2, 3, 4, 5], K = 3
Output
9
Explanation: Only 4 and 5 are strictly greater than 3: 4 + 5 = 9.
Input
nums = [10, 20, 30], K = 50
Output
0
Explanation: No value exceeds 50.
Input
nums = [-5, -1, 0, 1, 5], K = -2
Output
5
Explanation: Include -1, 0, 1 and 5: -1 + 0 + 1 + 5 = 5.
Input
nums = [100], K = 99
Output
100
Explanation: The only value exceeds 99, so return 100.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Initialize total to 0. Visit each array element; add the value only when it is strictly greater than K. After each visit, total equals the sum of qualifying values seen so far. Return total after the scan. This takes O(n) time and O(1) auxiliary space without modifying the input. Use a 64-bit integer for the sum in C++ and Java.
Brute Force Approach
Collect all values greater than K in a temporary list, then sum that list. This takes O(n) time and up to O(n) extra space. Sorting or checking combinations is unnecessary.
Step-by-Step Dry Run
For nums = [-5, -1, 0, 1, 5] and K = -2: start total = 0. Skip -5. Add -1, giving -1; add 0, leaving -1; add 1, giving 0; add 5, giving 5. Return 5.
Code Solutions
function solution(nums, K) {
let total = 0;
for (const value of nums) {
if (value > K) total += value;
}
return total;
}
// Input driver: length/dimensions, values, then K when applicable.
if (require.main === module) {
const input = require("fs").readFileSync(0, "utf8").trim();
if (input) {
const tokens = input.split(/\s+/).map(Number); let pos = 0;
const n = tokens[pos++];
const nums = tokens.slice(pos, pos + n); pos += n;
const K = tokens[pos++];
console.log(solution(nums, K));
}
}#include <iostream>
#include <vector>
class Solution {
public:
long long solution(const std::vector<int>& nums, int K) {
long long total = 0;
for (int value : nums) {
if (value > K) total += value;
}
return total;
}
};
// Input driver: length/dimensions, values, then K when applicable.
int main() {
int n; if (!(std::cin >> n)) return 0;
std::vector<int> nums(n);
for (int& value : nums) std::cin >> value;
int K; std::cin >> K;
std::cout << Solution().solution(nums, K) << "\n";
}
class Solution {
public long solution(int[] nums, int K) {
long total = 0L;
for (int value : nums) {
if (value > K) total += value;
}
return total;
}
}
// Input driver: length/dimensions, values, then K when applicable.
public class Main {
public static void main(String[] args) {
java.util.Scanner in = new java.util.Scanner(System.in);
if (!in.hasNextInt()) return;
int n = in.nextInt();
int[] nums = new int[n];
for (int i = 0; i < n; i++) nums[i] = in.nextInt();
int K = in.nextInt();
System.out.println(new Solution().solution(nums, K));
}
}
def solution(nums, K):
total = 0
for value in nums:
if value > K:
total += value
return total
# Input driver: length/dimensions, values, then K when applicable.
if __name__ == "__main__":
import sys
tokens = iter(map(int, sys.stdin.buffer.read().split()))
n = next(tokens)
nums = [next(tokens) for _ in range(n)]
K = next(tokens)
print(solution(nums, K))
function solution(nums, K) {
let total = 0;
for (const value of nums) {
if (value > K) total += value;
}
return total;
}
// Input driver: length/dimensions, values, then K when applicable.
if (require.main === module) {
const input = require("fs").readFileSync(0, "utf8").trim();
if (input) {
const tokens = input.split(/\s+/).map(Number); let pos = 0;
const n = tokens[pos++];
const nums = tokens.slice(pos, pos + n); pos += n;
const K = tokens[pos++];
console.log(solution(nums, K));
}
}
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