Vault Buffer Validator 45 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers and a positive integer K. Your task is to determine the sum of the K largest elements in the array. The array may contain negative numbers, zeros, and positives. K is always at least 1 and at most the length of the array. The input consists of two lines: the first line contains two integers N and K, where N is the number of elements in the array; the second line contains N space‑separated integers representing the array. The output should be a single integer: the sum of the K greatest values found in the array.
The problem is a classic greedy selection: to maximize the sum of K elements, one must always pick the largest available numbers. Efficient solutions typically involve sorting the array or using a selection algorithm to isolate the top K values, ensuring that the overall time complexity remains acceptable for large inputs.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Validator 45"
WHY DOES IT MATTER?
Selecting top‑K elements is a fundamental building block for ranking, recommendation, and resource allocation problems.
OPTIMIZATION CHALLENGE
The key is to reduce work from sorting all N items to handling only K, cutting time from O(N log N) to O(N log K) or O(N).
REAL-WORLD CONNECTION
It mirrors maintaining a leaderboard where only the highest scores are kept in memory.
Prefer a min‑heap when K is small relative to N; switch to QuickSelect when memory is tight or you need deterministic linear time.
COMPLEXITY AT A GLANCE
O(N log K) or O(N) averageO(K)Core Theory — Why This Approach?
The naive solution sorts the entire array or scans it K times, leading to O(N log N) or O(N·K) time, which becomes prohibitive for massive N (e.g., 10^7). The optimal greedy paradigm leverages a min‑heap of size K or the QuickSelect partition algorithm to isolate the K largest values in linear or near‑linear time, guaranteeing the sum can be computed without full sorting. By maintaining only the top K candidates, we exploit the monotonic property that any element smaller than the current K‑th largest can be discarded, dramatically reducing work. This approach aligns with the selection problem, where the goal is to find the K‑th order statistic and all elements above it, enabling O(N) average‑case or O(N log K) worst‑case performance.
Interview Questions on This Problem
Q1How would you compute the sum of the K largest numbers without sorting the entire array?
Use a min‑heap of size K, pushing each element and popping when the heap exceeds K; the heap then contains the K largest values whose sum is computed.
Q2What is the time complexity of QuickSelect for finding the K‑th largest element, and why is it suitable here?
Average O(N) time; it partitions the array around a pivot, recursively processing only the side that contains the K‑th largest, then sums the top K elements.
Q3Why might a naïve O(N·K) solution time out on large inputs?
Because it repeatedly scans the array K times, leading to up to 10^14 operations for N=10^7 and K≈10^7, far exceeding typical time limits.
Examples
Input
5 3 1 3 5 7 9
Output
21
Explanation: The array is [1,3,5,7,9]. The three largest numbers are 9, 7, and 5. Their sum is 9+7+5 = 21.
Input
6 1 -2 -5 0 4 3 1
Output
4
Explanation: Only the single largest element is required. The maximum value in the array is 4, so the sum is 4.
Input
4 4 10 20 30 40
Output
100
Explanation: K equals the array length, so all elements are summed: 10+20+30+40 = 100.
Input
7 5 5 1 9 3 8 2 7
Output
32
Explanation: The five largest numbers are 9, 8, 7, 5, and 3. Their sum is 9+8+7+5+3 = 32.
Constraints
- 1 <= N <= 100000
- 1 <= K <= N
- -1000000000 <= nums[i] <= 1000000000
- The sum of the K selected elements fits within a 64‑bit signed integer.
Optimal Approach & Strategy
Use a min‑heap of size K (O(N log K)) or QuickSelect (average O(N)) to isolate the K largest without full sorting.
Brute Force Approach
Sort the entire array descending and sum the first K elements, which costs O(N log N) time.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
if (k > nums.length) {
for (let num of nums) {
sum += num;
}
} else {
for (let i = 0; i < k; i++) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
if (k > nums.size()) {
for (int num : nums) {
sum += num;
}
} else {
for (int i = 0; i < k; i++) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
if (k > nums.length) {
for (int num : nums) {
sum += num;
}
} else {
for (int i = 0; i < k; i++) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
total_sum = 0
if k > len(nums):
for num in nums:
total_sum += num
else:
for i in range(k):
total_sum += nums[i]
return total_sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
if (k > nums.length) {
for (let num of nums) {
sum += num;
}
} else {
for (let i = 0; i < k; i++) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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