Watch 10 video solutions for Count the Hidden Sequences, a medium level problem involving Array, Prefix Sum. This walkthrough by codestorywithMIK has 11,482 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given a 0-indexed array of n integers differences, which describes the differences between each pair of consecutive integers of a hidden sequence of length (n + 1). More formally, call the hidden sequence hidden, then we have that differences[i] = hidden[i + 1] - hidden[i].
You are further given two integers lower and upper that describe the inclusive range of values [lower, upper] that the hidden sequence can contain.
differences = [1, -3, 4], lower = 1, upper = 6, the hidden sequence is a sequence of length 4 whose elements are in between 1 and 6 (inclusive).
[3, 4, 1, 5] and [4, 5, 2, 6] are possible hidden sequences.[5, 6, 3, 7] is not possible since it contains an element greater than 6.[1, 2, 3, 4] is not possible since the differences are not correct.Return the number of possible hidden sequences there are. If there are no possible sequences, return 0.
Example 1:
Input: differences = [1,-3,4], lower = 1, upper = 6 Output: 2 Explanation: The possible hidden sequences are: - [3, 4, 1, 5] - [4, 5, 2, 6] Thus, we return 2.
Example 2:
Input: differences = [3,-4,5,1,-2], lower = -4, upper = 5 Output: 4 Explanation: The possible hidden sequences are: - [-3, 0, -4, 1, 2, 0] - [-2, 1, -3, 2, 3, 1] - [-1, 2, -2, 3, 4, 2] - [0, 3, -1, 4, 5, 3] Thus, we return 4.
Example 3:
Input: differences = [4,-7,2], lower = 3, upper = 6 Output: 0 Explanation: There are no possible hidden sequences. Thus, we return 0.
Constraints:
n == differences.length1 <= n <= 105-105 <= differences[i] <= 105-105 <= lower <= upper <= 105Problem Overview: You are given a differences array where differences[i] = hidden[i+1] - hidden[i]. The hidden sequence must stay within the range [lower, upper]. The task is to count how many valid starting values produce a full sequence that never goes outside these bounds.
Approach 1: Prefix Sum and Range Validation (O(n) time, O(1) space)
The differences array describes relative movement between consecutive elements. Treat the first hidden value as x and build prefix sums of the differences to represent offsets from that start. While iterating through the array, track the minimum and maximum prefix value reached. These offsets determine the smallest and largest value the sequence could reach relative to x. For the sequence to stay inside [lower, upper], the start value must satisfy x + minPrefix ≥ lower and x + maxPrefix ≤ upper. This produces a valid range for x. The count of integers in that range is the answer. This approach relies on cumulative movement using prefix sum concepts and processes the array once.
Approach 2: Dynamic Range Computation (O(n) time, O(1) space)
Instead of computing prefix extremes first, you can maintain the allowable range of the current hidden value while scanning the array. Start with [lower, upper] as the valid range for the first element. When processing differences[i], shift the allowable range to represent the next value in the sequence: every possible value becomes current + differences[i]. Intersect this shifted range with [lower, upper] to ensure the sequence remains valid. Continue this update for all elements. At the end, the remaining width of the starting range represents the number of valid hidden sequences. This approach models constraints dynamically and works well when reasoning about value intervals across an array.
Recommended for interviews: The prefix sum range validation approach is the one most interviewers expect. It reduces the problem to tracking the minimum and maximum displacement produced by the differences array, then computing the number of valid starting values with simple arithmetic. Showing the dynamic range idea first demonstrates understanding of constraints, but the prefix-sum formulation shows stronger pattern recognition with prefix sums and leads directly to a clean O(n) solution.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Prefix Sum and Range Validation | O(n) | O(1) | Best general solution. Quickly computes valid starting values using prefix displacement tracking. |
| Dynamic Range Computation | O(n) | O(1) | Useful when reasoning about interval constraints step-by-step across the sequence. |