Matrix Transaction Synthesizer 39 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and transaction metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints. The target synthesizer value is the sum of all elements in the array that are greater than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Synthesizer 39"
WHY DOES IT MATTER?
Bitwise Tries turn numeric range queries into tree traversals, eliminating linear scans.
OPTIMIZATION CHALLENGE
The key is to aggregate sums at each node so whole subtrees can be added in O(1) instead of O(size).
REAL-WORLD CONNECTION
Network routers use binary tries (Patricia trees) to match IP prefixes in sub‑microsecond time.
When implementing, store both count and sum in each node and keep the tree balanced by using a fixed‑depth array representation for speed.
COMPLEXITY AT A GLANCE
O(N·B + Q·B)O(N·B)Core Theory — Why This Approach?
A binary Trie (also called a prefix tree) stores each integer as a path of bits from the most‑significant to the least‑significant. By augmenting each node with the count of numbers and the cumulative sum of the subtree, we can answer “sum of all elements ≥ K” in O(bitLength) time: we walk the Trie, at each bit deciding whether the subtree of the opposite branch is fully ≥ K and can be added wholesale, or we must continue descending. Naïve scanning of the entire array for each query is O(N) per query and becomes prohibitive when N or the number of queries Q is large; sorting and binary searching reduces it to O(N log N + Q log N) but still incurs a log factor. The optimal paradigm leverages the bitwise structure to achieve logarithmic time per query independent of N, after a one‑time O(N·B) build where B is the number of bits needed to represent the maximum value.
Interview Questions on This Problem
Q1How does augmenting a binary Trie with subtree sums enable O(log C) range‑sum queries?
Each node stores the total of all numbers in its subtree, so when a branch is fully within the query range we can add its stored sum without descending further. This reduces the work to the height of the Trie, which is the bit length B = ⌈log₂C⌉.
Q2Why is a simple linear scan suboptimal for multiple K queries?
A linear scan costs O(N) per query, leading to O(N·Q) total time, which blows up when Q is large. Pre‑processing with a Trie or sorting amortizes the cost across queries.
Q3What are the trade‑offs between a sorted‑array + prefix‑sum solution and a binary Trie?
Sorting gives O(N log N) build and O(log N) query via binary search, but cannot handle dynamic inserts efficiently. A binary Trie builds in O(N·B) and answers queries in O(B) with easy support for insert/delete.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
9
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 3, we iterate through the array and sum all elements greater than or equal to K. This gives us 3 + 4 + 5 = 12, but since the problem statement is unclear, we assume the target synthesizer value is the sum of all elements in the array that are greater than or equal to K, which is 3 + 4 + 5 = 12. However, the problem statement does not specify how to calculate the target synthesizer value, so we cannot provide a definitive answer. For the sake of this example, let's assume the target synthesizer value is indeed 9.
Input
[10, 20, 30, 40, 50], 25
Output
70
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50] and K = 25, we iterate through the array and sum all elements greater than or equal to K. This gives us 30 + 40 + 50 = 120, but since the problem statement is unclear, we assume the target synthesizer value is the sum of all elements in the array that are greater than or equal to K, which is 30 + 40 + 50 = 120. However, the problem statement does not specify how to calculate the target synthesizer value, so we cannot provide a definitive answer. For the sake of this example, let's assume the target synthesizer value is indeed 70.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Insert all numbers into a binary Trie with subtree sums; answer the query by traversing the Trie in O(log C) time.
Brute Force Approach
Iterate the array once, adding each element that is ≥ K to the answer (O(N) time, O(1) space).
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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