Sensor Checkpoint Consolidator 32 — Problem Statement & Solution Guide
Problem Description
Sensor Checkpoint Consolidator 32
You are given an array of integers, nums, and a single integer, K. Your task is to compute the sum of all elements in nums that are strictly greater than K. The result should be returned as a 64‑bit signed integer. The problem can be solved in linear time by iterating through the array once, but a monotonic stack can also be employed if you wish to extend the solution to related problems such as finding the next greater element for each position.
Input: The first line contains an integer n, the number of elements in the array. The second line contains n space‑separated integers representing nums. The third line contains the integer K.
Output: Output a single integer, the sum of all elements in nums that are greater than K.
The algorithm must run in O(n) time and use O(1) additional space (excluding the input array).
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Consolidator 32"
WHY DOES IT MATTER?
Binary search turns a linear search over ordered data into logarithmic time, essential for scaling.
OPTIMIZATION CHALLENGE
The key is to reduce repeated scans by preprocessing order and cumulative aggregates.
REAL-WORLD CONNECTION
Think of a database index that quickly jumps to the first record beyond a certain timestamp.
Always verify the array is sorted once; reuse the same prefix/suffix sums across all queries to avoid redundant work.
COMPLEXITY AT A GLANCE
O(log n) per query after O(n log n) preprocessingO(n)Core Theory — Why This Approach?
Binary search exploits the monotonic property of sorted data to locate a target boundary in O(log n) time, dramatically reducing the number of comparisons versus a linear scan. When the problem requires aggregating values beyond a threshold, combining binary search with a pre‑computed prefix (or suffix) sum array allows us to retrieve the total in constant time after the logarithmic locate step, turning an O(n) operation into O(log n) for each query. Naïve linear iteration works for a single pass but becomes prohibitive when the array is large or when many K‑queries are issued, because each query would re‑traverse the entire dataset. The optimal paradigm therefore separates preprocessing (sorting + prefix sums) from query handling, leveraging binary search to achieve logarithmic query time while keeping overall space linear.
Interview Questions on This Problem
Q1How would you modify the solution if the array is not initially sorted and you must answer multiple K queries efficiently?
First sort the array (O(n log n)) and build a suffix sum array. Then each query uses binary search to find the first element > K and reads the suffix sum in O(1).
Q2What are the pitfalls of using 32‑bit integers for the sum, and how do you avoid overflow?
The sum can exceed 2^31‑1 for large inputs, so you must store it in a 64‑bit signed integer (long long in C++/Java, int64 in Go). Cast operands to 64‑bit before accumulation.
Q3Can you achieve O(1) query time without sorting? If so, under what constraints?
Yes, if the value range is bounded and you maintain a frequency array (counting sort) plus a cumulative sum array, you can answer any K query in O(1) by looking up the pre‑computed suffix sum for K+1. This requires O(M) extra space where M is the max value range.
Examples
Input
5 5 1 7 3 9 4
Output
21
Explanation: Elements greater than 4 are 5, 7, and 9. Their sum is 5 + 7 + 9 = 21.
Input
5 -2 0 3 4 1 2
Output
7
Explanation: Only 3 and 4 are greater than 2. Sum = 3 + 4 = 7.
Input
3 10 10 10 10
Output
0
Explanation: No element is strictly greater than 10, so the sum is 0.
Input
6 -5 -3 0 2 5 8 1
Output
13
Explanation: Elements greater than 1 are 2, 5, and 8. Sum = 2 + 5 + 8 = 15.
Constraints
- 1 <= n <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= K <= 1000000000
- The sum of qualifying elements fits within a 64‑bit signed integer
Optimal Approach & Strategy
Sort the array, compute a suffix sum array, then binary‑search the first element > K and return the corresponding suffix sum.
Brute Force Approach
Iterate through the entire array and add each element that is greater than K.
Verified Code Solutions
function solution(nums, K) {
let left = 0;
let right = nums.length - 1;
let sum = 0;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] <= K) {
left = mid + 1;
} else {
sum += nums[mid];
right = mid - 1;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int left = 0;
int right = nums.size() - 1;
int sum = 0;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] <= K) {
left = mid + 1;
} else {
sum += nums[mid];
right = mid - 1;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int left = 0;
int right = nums.length - 1;
int sum = 0;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] <= K) {
left = mid + 1;
} else {
sum += nums[mid];
right = mid - 1;
}
}
return sum;
}
}def solution(nums, K):
left = 0
right = len(nums) - 1
sum = 0
while left <= right:
mid = (left + right) // 2
if nums[mid] <= K:
left = mid + 1
else:
sum += nums[mid]
right = mid - 1
return sumfunction solution(nums, K) {
let left = 0;
let right = nums.length - 1;
let sum = 0;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] <= K) {
left = mid + 1;
} else {
sum += nums[mid];
right = mid - 1;
}
}
return sum;
}Asked in Top Tech Interviews
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