Node Matrix Synthesizer 16 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a sequence of integer values representing sensor readings from a distributed node matrix. The goal is to determine the 'synthesizer value' for each element in the sequence. For every element at index i, the synthesizer value is defined as the count of preceding elements (indices j where j < i) that are strictly greater than the current element nums[i]. If no such preceding elements exist, the value is 0.
Given an array of integers, return a new array of the same length where each position contains the computed synthesizer value for that specific element. This problem requires an efficient approach to handle large input sizes, leveraging the properties of monotonic sequences or binary search on a sorted structure to achieve optimal performance.
The input will be a single array of integers. The output must be an array of integers where the i-th element corresponds to the count of previous elements strictly greater than the i-th element of the input array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Synthesizer 16"
WHY DOES IT MATTER?
Counting greater‑than‑left elements is a classic order‑statistics pattern used in inversion counting and ranking problems.
OPTIMIZATION CHALLENGE
The key is reducing the naïve O(n²) scan to O(n log n) by using a data structure that supports fast prefix queries.
REAL-WORLD CONNECTION
It mirrors real‑time alert systems where you need to know how many prior sensor spikes exceed the current reading.
Always compress values first; it prevents overflow and keeps the BIT size linear to the input length.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
The synthesizer value is the count of earlier readings that are strictly greater than the current reading, which can be modeled as a dynamic prefix‑sum query over a sorted value domain. A naive O(n²) scan fails for large n because each element would re‑examine all previous elements, leading to timeouts on typical interview constraints. By compressing the sensor values and maintaining a Fenwick Tree (or Balanced BST) that stores frequencies of seen values, we can query the number of greater elements in O(log n) and update the structure in O(log n), achieving an overall O(n log n) solution. This approach leverages the optimal paradigm of offline coordinate compression combined with a binary indexed tree to turn a quadratic counting problem into a logarithmic one per element.
Interview Questions on This Problem
Q1How does coordinate compression enable binary indexed trees to handle arbitrary integer ranges?
It maps each distinct sensor reading to a dense index in [1, m] where m ≤ n, preserving order. The BIT then operates on this compact index space, allowing O(log n) updates and queries.
Q2Why is a Fenwick Tree preferred over a segment tree for this problem?
Fenwick Trees have lower constant factors and simpler code for prefix‑sum operations. Both provide O(log n) time, but the BIT uses less memory and is easier to implement.
Q3Can this problem be solved with a balanced BST, and what would be the trade‑off?
Yes, an order‑statistic tree can insert and query rank in O(log n). However, it incurs higher overhead and requires custom augmentation compared to the BIT.
Examples
Input
nums = [5, 3, 8, 1, 4]
Output
[0, 1, 0, 3, 2]
Explanation: For index 0 (5): No preceding elements, count = 0. For index 1 (3): Preceding [5]. 5 > 3, count = 1. For index 2 (8): Preceding [5, 3]. None > 8, count = 0. For index 3 (1): Preceding [5, 3, 8]. 5 > 1, 3 > 1, 8 > 1, count = 3. For index 4 (4): Preceding [5, 3, 8, 1]. 5 > 4, 8 > 4, count = 2.
Input
nums = [10, 10, 10, 10]
Output
[0, 0, 0, 0]
Explanation: For index 0 (10): No preceding elements, count = 0. For index 1 (10): Preceding [10]. 10 is not strictly greater than 10, count = 0. For index 2 (10): Preceding [10, 10]. None strictly greater, count = 0. For index 3 (10): Preceding [10, 10, 10]. None strictly greater, count = 0.
Input
nums = [1, 2, 3, 4, 5]
Output
[0, 0, 0, 0, 0]
Explanation: The array is strictly increasing. For any element at index i, all preceding elements are smaller. Therefore, the count of preceding elements strictly greater than the current element is 0 for all indices.
Input
nums = [5, 4, 3, 2, 1]
Output
[0, 1, 2, 3, 4]
Explanation: The array is strictly decreasing. For index 1 (4): 5 > 4, count = 1. For index 2 (3): 5 > 3, 4 > 3, count = 2. For index 3 (2): 5 > 2, 4 > 2, 3 > 2, count = 3. For index 4 (1): 5 > 1, 4 > 1, 3 > 1, 2 > 1, count = 4.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The output array must have the same length as the input array.
Optimal Approach & Strategy
Compress values, then iterate left‑to‑right updating a Fenwick Tree; query the sum of frequencies of values greater than the current compressed index.
Brute Force Approach
Loop over each element and scan all prior elements, incrementing a counter when a prior value is greater.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0 || nums.length === 1) return 0;
nums.sort((a, b) => a - b);
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] > K) {
right = mid - 1;
} else if (nums[mid] < K) {
left = mid + 1;
} else {
return 0;
}
}
return 0;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() == 0 || nums.size() == 1) return 0;
sort(nums.begin(), nums.end());
int left = 0;
int right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] > K) {
right = mid - 1;
} else if (nums[mid] < K) {
left = mid + 1;
} else {
return 0;
}
}
return 0;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0 || nums.length == 1) return 0;
Arrays.sort(nums);
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] > K) {
right = mid - 1;
} else if (nums[mid] < K) {
left = mid + 1;
} else {
return 0;
}
}
return 0;
}
}def solution(nums, K):
if len(nums) == 0 or len(nums) == 1:
return 0
nums.sort()
left = 0
right = len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] > K:
right = mid - 1
elif nums[mid] < K:
left = mid + 1
else:
return 0
return 0function solution(nums, K) {
if (nums.length === 0 || nums.length === 1) return 0;
nums.sort((a, b) => a - b);
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] > K) {
right = mid - 1;
} else if (nums[mid] < K) {
left = mid + 1;
} else {
return 0;
}
}
return 0;
}Asked in Top Tech Interviews
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