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.
This approach uses the concept of a prefix sum to efficiently monitor the cumulative effects of the differences array. By calculating the prefix sum, we can track how the potential hidden sequence evolves from a starting point. Subsequently, determining the feasible range for the starting point of the hidden sequence allows us to count how many valid hidden sequences exist within the given bounds.
In this C solution, we iterate through the differences array to calculate the prefix sum. This allows us to determine how much the hidden sequence could diverge from its initial value. By tracking the minimum and maximum prefix sums, we can calculate the minimum and maximum initial values of the hidden sequence that would fit within the specified bounds. The difference between these two bounds plus one gives us the count of valid hidden sequences.
Time Complexity: O(n), where n is the number of elements in the differences array. Space Complexity: O(1), constant space used for tracking prefix sums.
An alternative method involves dynamically computing the range of possible values for each element in the hidden sequence. Starting with a base value, each element's value is recursively determined based on previous elements and the differences array. Finally, the overall range of valid starting points is deduced by verifying constraints imposed by lower and upper bounds.
This C code uses an array to track cumulative values derived from the differences array by incrementally adjusting the base value. Upon determining the minimum and maximum values in the range, it then calculates how many starting values fit the specified bounds, optimized via selective tracking and array manipulation.
Time Complexity: O(n), resulting from the need to iterate through the differences twice (one for accumulation and one for range computation). Space Complexity: O(n), owing to the use of additional storage for cumulative values.
Since the array differences is already determined, the difference between the maximum and minimum values of the elements in the array hidden is also fixed. We just need to ensure that this difference does not exceed upper - lower.
Let's assume the first element of the array hidden is 0. Then, hidden[i] = hidden[i - 1] + differences[i - 1], where 1 leq i leq n. Let the maximum value of the array hidden be mx and the minimum value be mi. If mx - mi leq upper - lower, then we can construct a valid hidden array. The number of possible constructions is upper - lower - (mx - mi) + 1. Otherwise, it is impossible to construct a valid hidden array, and we return 0.
The time complexity is O(n), where n is the length of the array differences. The space complexity is O(1).
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| Approach | Complexity |
|---|---|
| Prefix Sum and Range Validation | Time Complexity: O(n), where n is the number of elements in the differences array. Space Complexity: O(1), constant space used for tracking prefix sums. |
| Dynamic Range Computation | Time Complexity: O(n), resulting from the need to iterate through the differences twice (one for accumulation and one for range computation). Space Complexity: O(n), owing to the use of additional storage for cumulative values. |
| Prefix Sum | — |
| 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. |
Count the Hidden Sequences | Detailed Explanation | Maths | Leetcode 2145 | codestorywithMIK • codestorywithMIK • 11,482 views views
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