Sum Elements Greater Than K — Problem Statement & Solution Guide

ArraysEasyLinear Scan

TimeO(n)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Sum Elements Greater Than K problem optimally.

TopicArrays

PatternLinear Scan

TimeO(n)

SpaceO(1)

Problem Description

You are provided with an integer array nums and a threshold integer K. Your task is to compute the aggregate sum of all elements within the array that strictly exceed the value of K. Elements equal to or less than K must be excluded from the calculation.

If no elements in the array satisfy the condition of being greater than K, the function should return 0. The solution should efficiently process the array to determine the final sum.

DSA Pattern Breakdown

"Sum Elements Greater Than K"

easy

WHY DOES IT MATTER?

The filter‑then‑aggregate pattern appears in countless real‑world analytics tasks—summing sales above a target, counting high‑frequency events, or computing risk exposure beyond a threshold. Mastering this pattern builds a foundation for more complex streaming and big‑data pipelines.

OPTIMIZATION CHALLENGE

The key insight is that the predicate (> K) is independent of other elements, allowing us to avoid any sorting, auxiliary containers, or nested loops. By maintaining a single accumulator, we reduce both time to O(n) and space to O(1).

REAL-WORLD CONNECTION

Think of a financial trading system that streams price ticks; it must continuously sum the value of trades that exceed a risk limit K. The same linear scan with a running total is used to enforce limits in near‑real time without storing the entire history.

During an interview, write the loop first, then immediately add the conditional check and accumulator. Resist the urge to create extra arrays or use built‑in filter functions unless the language guarantees O(1) extra space.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to a single linear scan of the input array, accumulating only those values that satisfy the predicate > K. In algorithmic terms, this is a classic example of a filter‑then‑aggregate pattern, which can be expressed as Σ_{i=0}^{n-1} [nums[i] > K] * nums[i]. The naive approach of nested loops or repeated passes would inflate the time complexity to O(n²) and quickly become infeasible for large n (e.g., n > 10⁶) because each element would be examined multiple times. By recognizing that the predicate is stateless and does not depend on other elements, we can collapse the operation into a single pass, achieving optimal O(n) time.

The optimal paradigm leverages the fact that addition is associative and commutative, allowing us to maintain a running total while iterating. No auxiliary data structures are required beyond a scalar accumulator, which keeps the auxiliary space constant. This approach also aligns with cache‑friendly sequential memory access, minimizing branch mispredictions and ensuring the solution scales linearly with input size. In environments where the array may be streamed or stored in external memory, the same linear‑time, O(1)-space algorithm can be applied incrementally, making it robust for both in‑memory and out‑of‑core scenarios.

Interview Questions on This Problem

Q1How would you modify the solution if the array is sorted in descending order and you need to stop processing as soon as you encounter a value ≤ K?

Because the array is sorted descending, once you hit an element ≤ K, all subsequent elements will also be ≤ K. You can break out of the loop at that point, which still yields O(n) worst‑case but can be O(m) where m is the count of elements > K, offering early‑exit optimization.

Q2What changes are required if the input can contain 64‑bit integers and the sum may overflow a 32‑bit integer?

Use a 64‑bit integer type (e.g., long long in C++, long in Java, or Python's arbitrary‑precision int) for the accumulator. Additionally, consider checking for overflow if the language does not handle it automatically, or use built‑in big‑integer libraries.

Q3Explain how you would parallelize the computation on a multi‑core system while preserving correctness.

Divide the array into chunks, let each core compute a local sum of elements > K for its chunk, then perform a final reduction (sum) of the local results. This map‑reduce style maintains O(n/p) work per core plus O(p) reduction overhead, where p is the number of cores.

Examples

Example 1

Input

nums = [12, 4, 8, 15, 2], K = 10

Output

Explanation: Iterate through the array: 12 > 10 (add 12), 4 <= 10 (skip), 8 <= 10 (skip), 15 > 10 (add 15), 2 <= 10 (skip). Total sum = 12 + 15 = 27.

Example 2

Input

nums = [5, 5, 5], K = 5

Output

Explanation: All elements are equal to K (5). Since the condition requires elements strictly greater than K, no elements are included. Total sum = 0.

Example 3

Input

nums = [-3, 0, 7, 12, 12], K = 6

Output

Explanation: Check each element: -3 <= 6 (skip), 0 <= 6 (skip), 7 > 6 (add 7), 12 > 6 (add 12), 12 > 6 (add 12). Total sum = 7 + 12 + 12 = 31. Wait, 7+12+12 is 31. Let me re-calculate. 7+12=19, 19+12=31. Correct output is 31.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
-10^9 <= K <= 10^9

Optimal Approach & Strategy

Perform one linear scan, adding each element to a sum only when it exceeds K, achieving O(n) time and O(1) space.

Brute Force Approach

Iterate over the array for each element, checking all other elements to decide if it should be added, leading to O(n²) time.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var sumElementsGreaterThanK = function(nums, K) {
    let sum = 0;
    for (let num of nums) {
        if (num > K) {
            sum += num;
        }
    }
    return sum;
};

class Solution {
public:
    int sumElementsGreaterThanK(vector<int>& nums, int K) {
        int sum = 0;
        for (int num : nums) {
            if (num > K) {
                sum += num;
            }
        }
        return sum;
    }
};

class Solution {
    public int sumElementsGreaterThanK(int[] nums, int K) {
        int sum = 0;
        for (int num : nums) {
            if (num > K) {
                sum += num;
            }
        }
        return sum;
    }
}

class Solution:
    def sumElementsGreaterThanK(self, nums: List[int], K: int) -> int:
        return sum(num for num in nums if num > K)

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var sumElementsGreaterThanK = function(nums, K) {
    let sum = 0;
    for (let num of nums) {
        if (num > K) {
            sum += num;
        }
    }
    return sum;
};

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