Matrix Transaction Evaluator 47 — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums and an integer T. Your task is to identify the greatest element in nums that does not exceed T. If every element in nums is larger than T, report that no suitable element exists. The solution must run in linear time relative to the size of the array and may use only constant extra space. Bitwise operators may be employed to compare values, but any correct algorithm is acceptable.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Transaction Evaluator 47"
WHY DOES IT MATTER?
Finding a bounded maximum in one pass is a classic linear‑time selection pattern.
OPTIMIZATION CHALLENGE
Eliminate branching and extra storage to keep cache footprints minimal.
REAL-WORLD CONNECTION
Databases use similar scans to enforce range constraints while streaming logs.
Prefer bit‑mask tricks for branch‑free updates when the language permits low‑level integer ops.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The task reduces to a selection problem where we must locate the maximum element not exceeding a threshold T. A naïve sort (O(n log n)) or nested scans (O(n^2)) waste time and memory, violating the linear‑time, O(1)‑space constraint; instead, a single forward pass maintaining a candidate answer leverages the monotonic property of the comparison and can be implemented with pure integer and bitwise operations to avoid extra branching. The optimal paradigm is a streaming reduction: for each x in nums, compute a mask = -(x <= T) (using bitwise NOT and addition) and conditionally update best = (best & ~mask) | (x & mask), ensuring constant extra space and deterministic O(n) runtime.
Interview Questions on This Problem
Q1Why is sorting the array not acceptable for this problem?
Sorting costs O(n log n) time, exceeding the required linear bound, and introduces O(n) auxiliary space in many implementations.
Q2How can you update the current best candidate without using an if‑statement?
Use a bitmask derived from the comparison (e.g., mask = -(x <= T)) and combine best and x with bitwise AND/OR to conditionally replace best.
Q3What edge case must you handle when all elements are greater than T?
Initialize best to a sentinel (e.g., Integer.MIN_VALUE) and after the scan check if it remained unchanged to report “no suitable element”.
Examples
Input
6 4 15 23 8 2 19 20
Output
19
Explanation: The array elements are [4,15,23,8,2,19] and the threshold T is 20. Elements ≤ 20 are 4,15,8,2,19. Among them, 19 is the largest, so the output is 19.
Input
5 -5 -2 -9 -1 -7 -3
Output
-5
Explanation: The threshold is -3. Values not exceeding -3 are -5, -9, -7. The maximum among these is -5, which is returned.
Input
4 10 12 14 16 9
Output
NONE
Explanation: All numbers in the array are greater than the threshold 9, therefore no element satisfies the condition. The prescribed output for this case is the string "NONE".
Constraints
- 1 <= nums.length <= 100000
- -10^9 <= nums[i] <= 10^9
- -10^9 <= T <= 10^9
- The algorithm must use O(1) additional memory beyond the input array.
Optimal Approach & Strategy
Iterate once, updating a candidate with a branch‑free bitmask, achieving O(n) time and O(1) extra space.
Brute Force Approach
Sort the array then binary‑search for T, or compare every pair, both O(n log n) or O(n^2).
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} T
* @return {number}
*/
var findGreatestNotExceeding = function(nums, T) {
let maxVal = -1;
for (let num of nums) {
if (num <= T && num > maxVal) {
maxVal = num;
}
}
return maxVal;
};class Solution {
public:
int findGreatestNotExceeding(vector<int>& nums, int T) {
int maxVal = -1;
for (int num : nums) {
if (num <= T && num > maxVal) {
maxVal = num;
}
}
return maxVal;
}
};class Solution {
public int findGreatestNotExceeding(int[] nums, int T) {
int maxVal = -1;
for (int num : nums) {
if (num <= T && num > maxVal) {
maxVal = num;
}
}
return maxVal;
}
}class Solution:
def findGreatestNotExceeding(self, nums: List[int], T: int) -> int:
max_val = -1
for num in nums:
if num <= T and num > max_val:
max_val = num
return max_val/**
* @param {number[]} nums
* @param {number} T
* @return {number}
*/
var findGreatestNotExceeding = function(nums, T) {
let maxVal = -1;
for (let num of nums) {
if (num <= T && num > maxVal) {
maxVal = num;
}
}
return maxVal;
};Asked in Top Tech Interviews
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