Longest Constrained Subsequence — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Longest Constrained Subsequence problem optimally.
O(n)O(1)Problem Description
You are provided with a non-empty array of integers representing a sequence of resource allocations and a single integer limit. Your task is to determine the maximum length of a contiguous subarray such that the cumulative sum of its elements does not exceed the given limit.
The solution must identify the longest window where the constraint is satisfied. If no such subarray exists (which is impossible given the non-empty constraint and typical positive value assumptions in this context, but logically handled), return 0. However, assuming standard non-negative or mixed values where a single element might exceed the limit, the logic must handle edge cases where the minimum possible sum (single element) is already greater than the limit.
Input consists of the array resource_values and the integer threshold. Output is a single integer representing the length of the longest valid contiguous subarray.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Longest Constrained Subsequence"
WHY DOES IT MATTER?
The sliding‑window pattern is a cornerstone for any problem that asks for optimal sub‑structures over contiguous ranges under a monotonic constraint, enabling linear‑time solutions where brute force would be quadratic.
OPTIMIZATION CHALLENGE
Realizing that the sum can be updated incrementally and that the left boundary never needs to move backward eliminates the need for recomputing sums, collapsing O(n^2) work into O(n).
REAL-WORLD CONNECTION
Think of a network bandwidth monitor that continuously tracks data usage over a rolling time window; you need to know the longest period where usage stayed under a quota, which is exactly a sliding‑window sum check.
During an interview, write the two‑pointer skeleton first, then add the while‑shrink loop; this structure makes it easy to reason about correctness and edge cases.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the longest contiguous subarray whose sum does not exceed a given limit. A naïve solution would enumerate every possible subarray, compute its sum, and keep the maximum length that satisfies the constraint – this incurs O(n^2) time because each of the O(n^2) subarrays requires O(1) or O(n) work to compute the sum. For large inputs (n up to 10^5 or more) this quickly becomes infeasible. The optimal paradigm leverages the monotonic nature of the sum when extending a window: as we move the right pointer forward, the window sum either stays within the limit or exceeds it, in which case we can shrink the window from the left until the constraint is restored. This two‑pointer or sliding‑window technique processes each element at most twice, yielding linear time. The key insight is that the sum of a contiguous segment can be updated incrementally, avoiding recomputation, and the window’s left boundary only moves forward, guaranteeing O(n) overall complexity.
The sliding‑window algorithm works as follows: maintain two indices, left and right, and a running sum of elements between them. Expand right, adding arr[right] to the sum. If the sum exceeds the limit, increment left while subtracting arr[left] from the sum until the sum is ≤ limit again. After each adjustment, record the window length if it is larger than any previously seen. Because each element is added once and removed at most once, the total work is linear. This approach also uses O(1) extra space beyond the input array, making it optimal for both time and memory.
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution if the array could contain negative numbers?
With negatives the sum can increase after shrinking the window, breaking the monotonic property; you would need a prefix‑sum + binary search or a deque to maintain a monotonic queue of prefix sums, resulting in O(n log n) or O(n) with more complex data structures.
Q2Can you solve the problem in O(n) time using prefix sums without explicit two‑pointer logic?
Yes, compute prefix sums P[i] and maintain a deque of candidate start indices with non‑decreasing prefix values; for each i, pop from the front while P[i]-P[deque[0]]>limit, then the window length is i-deque[0]; this also runs in O(n).
Q3What is the worst‑case scenario for the sliding‑window algorithm and why does it still stay O(n)?
The worst case occurs when every new element forces the left pointer to move forward (e.g., all elements are larger than the limit). Even then each element is visited a constant number of times (once by right, once by left), so total operations remain proportional to n.
Examples
Input
resource_values = [2, 3, 1, 4, 5], threshold = 7
Output
3
Explanation: Start with window [2], sum=2. Add 3 -> [2,3], sum=5. Add 1 -> [2,3,1], sum=6. Add 4 -> [2,3,1,4], sum=10 > 7. Shrink from left: remove 2 -> [3,1,4], sum=8 > 7. Remove 3 -> [1,4], sum=5. Add 5 -> [1,4,5], sum=10 > 7. Remove 1 -> [4,5], sum=9 > 7. Remove 4 -> [5], sum=5. The maximum length observed was 3 (from [2,3,1]).
Input
resource_values = [1, 2, 3, 4, 5], threshold = 15
Output
5
Explanation: The sum of the entire array is 1+2+3+4+5 = 15, which is equal to the threshold. Since the sum does not exceed the threshold, the entire array is a valid subarray. The length is 5.
Input
resource_values = [10, 1, 1, 1, 1], threshold = 5
Output
4
Explanation: Start with [10], sum=10 > 5. Shrink: window becomes empty, then add 1 -> [1], sum=1. Add 1 -> [1,1], sum=2. Add 1 -> [1,1,1], sum=3. Add 1 -> [1,1,1,1], sum=4. The next element is out of bounds. The maximum length is 4.
Input
resource_values = [5, 5, 5], threshold = 4
Output
0
Explanation: The first element is 5, which is greater than the threshold 4. The window shrinks to empty. The next element is 5, which is also greater than 4. The window shrinks to empty. The last element is 5, which is greater than 4. The window shrinks to empty. No valid subarray exists, so the length is 0.
Constraints
- 1 <= resource_values.length <= 10^5
- 1 <= resource_values[i] <= 10^9
- 1 <= threshold <= 10^14
Optimal Approach & Strategy
Use two pointers (sliding window) with a running sum, expanding right and shrinking left only when needed – O(n) time, O(1) extra space.
Brute Force Approach
Check every possible subarray, compute its sum, and keep the longest that stays under the limit – O(n^2) time.
Code Solutions
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = input[idx++];
const arr = input.slice(idx, idx + n); idx += n;
const limit = input[idx];
function longestConstrainedSubsequence(arr, limit) {
let sum = 0, left = 0, best = 0;
for(let right = 0; right < arr.length; ++right) {
sum += arr[right];
while(left <= right && sum > limit) {
sum -= arr[left];
++left;
}
best = Math.max(best, right - left + 1);
}
return best;
}
console.log(longestConstrainedSubsequence(arr, limit));#include <bits/stdc++.h>
using namespace std;
int longestConstrainedSubsequence(const vector<int>& arr, int limit) {
long long sum = 0;
int left = 0, best = 0;
for(int right = 0; right < (int)arr.size(); ++right) {
sum += arr[right];
while(left <= right && sum > limit) {
sum -= arr[left];
++left;
}
best = max(best, right - left + 1);
}
return best;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<int> arr(n);
for(int i=0;i<n;++i) cin>>arr[i];
int limit; cin>>limit;
cout<<longestConstrainedSubsequence(arr, limit);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
public static int longestConstrainedSubsequence(int[] arr, int limit) {
long sum = 0;
int left = 0, best = 0;
for(int right = 0; right < arr.length; ++right) {
sum += arr[right];
while(left <= right && sum > limit) {
sum -= arr[left];
left++;
}
best = Math.max(best, right - left + 1);
}
return best;
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int[] arr = new int[n];
st = new StringTokenizer(br.readLine());
for(int i = 0; i < n; i++) arr[i] = Integer.parseInt(st.nextToken());
int limit = Integer.parseInt(br.readLine().trim());
System.out.print(longestConstrainedSubsequence(arr, limit));
}
}import sys
def longest_constrained_subsequence(arr, limit):
sum_ = 0
left = 0
best = 0
for right, val in enumerate(arr):
sum_ += val
while left <= right and sum_ > limit:
sum_ -= arr[left]
left += 1
best = max(best, right - left + 1)
return best
def main():
data = list(map(int, sys.stdin.read().strip().split()))
if not data:
return
n = data[0]
arr = data[1:1+n]
limit = data[1+n]
print(longest_constrained_subsequence(arr, limit))
if __name__ == "__main__":
main()const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = input[idx++];
const arr = input.slice(idx, idx + n); idx += n;
const limit = input[idx];
function longestConstrainedSubsequence(arr, limit) {
let sum = 0, left = 0, best = 0;
for(let right = 0; right < arr.length; ++right) {
sum += arr[right];
while(left <= right && sum > limit) {
sum -= arr[left];
++left;
}
best = Math.max(best, right - left + 1);
}
return best;
}
console.log(longestConstrainedSubsequence(arr, limit));Asked in Top Tech Interviews
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