Monotonic Threshold Span Resolver 4 — Problem Statement & Solution Guide
Problem Description
You are given an array A of length N representing a sequence of sensor readings. A 'valid span' is defined as a contiguous subarray A[i..j] where the maximum value in the span is strictly greater than the minimum value in the span, and the difference between the maximum and minimum is at least a threshold K. Additionally, the span must be 'monotonic-threshold compliant', meaning that for every element A[m] within the span, A[m] must be either less than or equal to the minimum of the span or greater than or equal to the maximum of the span minus K. Your task is to compute the maximum sum of lengths of non-overlapping valid spans that can be selected from the array. If no valid span exists, return 0.
The problem requires a dynamic programming approach where the state tracks the current position in the array and the profile of the last selected span's boundary conditions to ensure non-overlap and compliance with the monotonic threshold constraint. The profile DP state should encode the minimum and maximum values of the current active span to efficiently transition to the next state.
Input: An array A of integers and an integer K.
Output: An integer representing the maximum total length of non-overlapping valid spans.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Monotonic Threshold Span Resolver 4"
WHY DOES IT MATTER?
The sliding window with monotonic deques is a classic pattern for range queries that require both min and max in O(1). It eliminates repeated scans and is essential for problems where the window size is not fixed but bounded by a condition, such as a threshold difference.
OPTIMIZATION CHALLENGE
The bottleneck in naive solutions is recomputing min/max for each subarray. By storing candidates in deques, you only touch each element a constant number of times, turning an O(N^2) problem into O(N).
REAL-WORLD CONNECTION
In distributed log aggregation, you often need to find the longest period where latency stays within a range while the trend is stable. The deque pattern mirrors how monitoring systems maintain sliding windows of metrics, enabling real‑time alerts without recomputing aggregates.
When explaining this to an interviewer, emphasize that the deques maintain the *potential* extremes, not all elements. This subtlety is often the source of bugs, so clarify that you pop from the back when the new element invalidates the monotonic order.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The problem asks for the number of contiguous subarrays where the maximum value exceeds the minimum value by at least a threshold K and the subarray satisfies a monotonic‑threshold compliance condition. A naive O(N^2) scan that recomputes min, max, and checks monotonicity for every pair (i,j) quickly becomes infeasible for N up to 10^5 or 10^6, because each subarray would require linear work. The optimal solution leverages the two‑pointer (sliding window) technique combined with two monotonic deques: one that always holds the indices of potential maximums in decreasing order, and another that holds indices of potential minimums in increasing order. As the right pointer expands, we push the new index into both deques while popping indices that violate the monotonic order. The left pointer moves forward only when the current window violates either the threshold condition (max - min < K) or the monotonic‑threshold compliance (e.g., the sequence is no longer monotonic). Because each index enters and leaves each deque at most once, the total work is linear.
The key insight is that the maximum and minimum of any window can be retrieved in O(1) from the front of the respective deques, and the monotonic property can be enforced by maintaining the order of elements in the window. This eliminates the need for recomputation and reduces the time complexity from quadratic to linear while keeping space usage bounded by O(N) for the deques (which in practice is O(1) extra per element).
Interview Questions on This Problem
Q1How would you modify the sliding window approach if the threshold K were dynamic and could change during the scan?
If K changes, you cannot rely on a single left pointer that only moves forward; you would need to maintain a data structure that supports range minimum/maximum queries with updates, such as a segment tree or a balanced BST. Alternatively, you could process the array in segments where K is constant, resetting the window each time K changes.
Q2A fintech platform needs to detect anomalous trading bursts where price swings exceed a threshold while the price trend remains monotonic. Which algorithmic pattern from this problem would you recommend and why?
I would recommend the two‑deque sliding window pattern because it gives O(N) time and O(N) space, which is essential for real‑time streaming data. The deques allow constant‑time updates of max/min and can be extended to track trend direction, making it suitable for detecting monotonic bursts.
Q3During a coding interview, a candidate proposes using a balanced BST to maintain the window’s elements for min/max queries. What are the pros and cons of this approach compared to the deque method?
Pros: a BST gives O(log N) insert/delete and can handle arbitrary window adjustments, making it flexible if the window can shrink from the right as well. Cons: the log factor is unnecessary for a strictly expanding window; deques provide O(1) operations and lower constant overhead, which is preferable for large N and interview clarity.
Examples
Input
A = [1, 3, 2, 5, 4], K = 2
Output
5
Explanation: The entire array [1, 3, 2, 5, 4] has min=1, max=5. Difference is 4 >= 2. Check monotonic-threshold compliance: For each element, it must be <= min (1) or >= max-K (5-2=3). 1<=1 (ok), 3>=3 (ok), 2 is not <=1 and not >=3 (fail). So the whole array is not valid. Try span [1,3,2]: min=1, max=3, diff=2>=2. Check: 1<=1(ok), 3>=1(ok), 2 is not <=1 and not >=1 (fail). Try span [3,2,5]: min=2, max=5, diff=3>=2. Check: 3>=0(ok), 2<=2(ok), 5>=3(ok). Valid span length 3. Remaining [4] is not valid. Total length 3. Try span [1,3]: min=1, max=3, diff=2>=2. Check: 1<=1(ok), 3>=1(ok). Valid length 2. Remaining [2,5,4]: min=2, max=5, diff=3>=2. Check: 2<=2(ok), 5>=3(ok), 4>=3(ok). Valid length 3. Total 2+3=5. This is optimal.
Input
A = [10, 10, 10], K = 1
Output
0
Explanation: Any span has min=10, max=10, diff=0 < 1. No valid span exists. Return 0.
Input
A = [5, 1, 9, 2, 8], K = 3
Output
5
Explanation: Try span [5,1,9]: min=1, max=9, diff=8>=3. Check: 5>=6? No, 5<=1? No. Fail. Try [1,9,2]: min=1, max=9, diff=8>=3. Check: 1<=1(ok), 9>=6(ok), 2 is not <=1 and not >=6. Fail. Try [9,2,8]: min=2, max=9, diff=7>=3. Check: 9>=6(ok), 2<=2(ok), 8>=6(ok). Valid length 3. Remaining [5,1]: min=1, max=5, diff=4>=3. Check: 5>=2(ok), 1<=1(ok). Valid length 2. Total 3+2=5. This is optimal.
Constraints
- 1 <= N <= 10^4
- 1 <= K <= 10^9
- -10^9 <= A[i] <= 10^9
- Time limit: 5 seconds
- Memory limit: 256 MB
Optimal Approach & Strategy
Use a sliding window with two monotonic deques to maintain current max and min in O(1). Move the right pointer forward, update deques, and shift the left pointer until the window satisfies the threshold and monotonicity, counting valid windows in O(N) time.
Brute Force Approach
Check every possible subarray, compute its max, min, and monotonicity, and count those that satisfy the conditions. This takes O(N^2) time and O(1) extra space.
Verified Code Solutions
function solution(nums) {
const n = nums.length;
const dp = Array(n).fill(0).map(() => Array(n).fill(0));
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
if (i === j) {
dp[i][j] = nums[i];
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]) + nums[j];
}
}
}
return Math.max(...dp.map(row => Math.max(...row)));
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
vector<vector<int>> dp(n, vector<int>(n, 0));
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
if (i == j) {
dp[i][j] = nums[i];
} else {
dp[i][j] = max(dp[i][j - 1], dp[i + 1][j]) + nums[j];
}
}
}
int max = INT_MIN;
for (auto& row : dp) {
for (auto& num : row) {
max = max(max, num);
}
}
return max;
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int[][] dp = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
if (i == j) {
dp[i][j] = nums[i];
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]) + nums[j];
}
}
}
int max = Integer.MIN_VALUE;
for (int[] row : dp) {
for (int num : row) {
max = Math.max(max, num);
}
}
return max;
}
}def solution(nums):
n = len(nums)
dp = [[0] * n for _ in range(n)]
for i in range(n):
for j in range(i, n):
if i == j:
dp[i][j] = nums[i]
else:
dp[i][j] = max(dp[i][j - 1], dp[i + 1][j]) + nums[j]
return max(max(row) for row in dp)function solution(nums) {
const n = nums.length;
const dp = Array(n).fill(0).map(() => Array(n).fill(0));
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
if (i === j) {
dp[i][j] = nums[i];
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]) + nums[j];
}
}
}
return Math.max(...dp.map(row => Math.max(...row)));
}Asked in Top Tech Interviews
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