Tome Voyage Optimizer 17 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the processing pipeline for a sequence of data packets, where each packet is represented by an integer value. The system requires calculating the 'Net Throughput Score' for each packet based on its relationship with subsequent packets in the stream. For a given packet at index i, its Net Throughput Score is defined as the value of the first subsequent packet (at index j > i) that is strictly greater than the current packet's value. If no such subsequent packet exists, the score is defined as -1. Your objective is to compute an array of these scores for every packet in the input sequence.
This problem models a scenario where each data element's utility is determined by the next significant improvement in the stream. The challenge lies in efficiently determining these next greater elements for all positions without resorting to a brute-force O(n^2) approach, which would be computationally prohibitive for large datasets. You must design an algorithm that processes the sequence in linear time to produce the final score array.
Input: An array of integers representing the data packet values.
Output: An array of integers of the same length, where each element at index i contains the value of the next greater element to the right of index i, or -1 if no such element exists.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Optimizer 17"
WHY DOES IT MATTER?
Monotonic stacks turn a seemingly quadratic relationship (searching rightward for a greater element) into a linear scan by preserving only the necessary candidates, dramatically reducing runtime for massive data streams.
OPTIMIZATION CHALLENGE
The key insight is that any element smaller than the current one can never be the answer for any element to its left, so it can be discarded immediately, ensuring each element is processed a constant number of times.
REAL-WORLD CONNECTION
In network routers, packets are prioritized based on upcoming higher‑priority traffic; a stack of pending priorities can quickly reveal the next higher priority without scanning the entire queue, mirroring the NGE logic.
When coding, initialize the stack empty and iterate from the end; always remember to push the current element after determining its answer – this order prevents self‑comparison and keeps the stack state valid.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem asks for the first element to the right of each index that is strictly greater than the current element – a classic Next Greater Element (NGE) scenario. A naive double‑loop scans every pair (i, j) with j > i, leading to O(n²) time, which quickly becomes infeasible for large streams (n can be up to 10⁶ in typical interview constraints). The optimal paradigm leverages a monotonic stack: while traversing the array from right to left, the stack maintains a decreasing sequence of candidate values. For each element, we pop all smaller or equal values because they can never serve as a next greater for any earlier element, then the top of the stack (if any) is the required answer. This yields a linear O(n) solution because each element is pushed and popped at most once. The approach also naturally handles edge cases such as duplicate values and strictly decreasing sequences by returning a sentinel (e.g., -1) when the stack is empty.
Interview Questions on This Problem
Q1How would you modify the Next Greater Element algorithm to return the index of the next greater element instead of its value?
Store indices on the stack instead of values. While processing element i, pop indices whose corresponding values are ≤ arr[i]; the top of the stack then holds the index of the next greater element. Push i onto the stack after processing.
Q2Can the Next Greater Element problem be solved in O(1) extra space?
Yes, by reusing the input array to store results and using it as a simulated stack via in‑place overwriting, but this sacrifices readability and may require careful handling of original values; typically O(n) auxiliary space is acceptable and clearer.
Q3Explain how the monotonic stack technique applies to the "Largest Rectangle in Histogram" problem.
Both problems rely on a stack that maintains a monotonic order (increasing heights for the histogram). For each bar, we pop until we find a bar shorter than the current, using the popped bar's height to compute area with the current index as the right boundary. This reuse of the monotonic stack pattern demonstrates its versatility for range‑based queries.
Examples
Input
[4, 5, 2, 10, 8]
Output
[5, 10, 10, -1, -1]
Explanation: For index 0 (value 4), the first greater value to the right is 5 at index 1. For index 1 (value 5), the first greater value is 10 at index 3. For index 2 (value 2), the first greater value is 10 at index 3. For index 3 (value 10), there is no greater value to the right, so the result is -1. For index 4 (value 8), there is no greater value to the right, so the result is -1.
Input
[3, 1, 4, 1, 5, 9, 2, 6]
Output
[4, 4, 5, 5, 9, -1, 6, -1]
Explanation: Index 0 (3) -> next greater is 4. Index 1 (1) -> next greater is 4. Index 2 (4) -> next greater is 5. Index 3 (1) -> next greater is 5. Index 4 (5) -> next greater is 9. Index 5 (9) -> no greater value, result -1. Index 6 (2) -> next greater is 6. Index 7 (6) -> no greater value, result -1.
Input
[10, 10, 10, 10]
Output
[-1, -1, -1, -1]
Explanation: All elements are equal. Since the condition requires a strictly greater value, no element has a next greater element. Thus, all results are -1.
Input
[1, 2, 3, 4, 5]
Output
[2, 3, 4, 5, -1]
Explanation: The array is strictly increasing. Each element's next greater element is the immediate next element. The last element (5) has no subsequent elements, so its result is -1.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The input array may contain duplicate values.
- The solution must run in O(n) time complexity.
- The solution must use O(n) auxiliary space.
Optimal Approach & Strategy
Traverse the array from right to left using a monotonic decreasing stack; pop smaller elements, the stack top becomes the next greater, then push the current element. This runs in O(n) time with O(n) auxiliary space.
Brute Force Approach
For each index i, scan forward j = i+1 … n‑1 until you find an element greater than arr[i]; record it or -1 if none exists. This double loop costs O(n²) time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int K) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
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 i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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