Count the Hidden Sequences

Medium

Zomato

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Topics:

Arrays

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.

For example, given 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.length
1 <= n <= 10⁵
-10⁵ <= differences[i] <= 10⁵
-10⁵ <= lower <= upper <= 10⁵

Try coding it in LeetCode

Count the Hidden Sequences

Medium

Zomato

2 views

Topics:

Arrays

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.

For example, given 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.length
1 <= n <= 10⁵
-10⁵ <= differences[i] <= 10⁵
-10⁵ <= lower <= upper <= 10⁵

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the bounds for `lower` and `upper`? Can they be negative?
What are the constraints on the length of the `differences` array? Can it be empty?
Can the `differences` array contain zero values?
If there are multiple possible starting values that result in valid hidden sequences, should I return the count of *all* such valid sequences, or is there a specific one I should target?
Is the hidden sequence constrained to be integers only? Or could it contain floating point numbers?

Brute Force Solution

Approach

The brute force strategy involves trying out every possible sequence that could be hidden, and then verifying if it meets the specified conditions. It's like testing every lock combination until you find the right one. This ensures that no valid sequence is missed, although it might be inefficient.

Here's how the algorithm would work step-by-step:

First, imagine you don't know the starting point of the hidden sequence. Start by assuming the first number in the sequence is the lowest possible number.
Then, figure out all possible differences between each number in the sequence. Since we're looking for a sequence within a defined range of differences, we'll try every possible difference allowed.
For each possible starting number and difference, construct a potential hidden sequence, element by element.
Check to see if the created sequence fits inside the given range of numbers. If any number goes outside the bounds, this sequence isn't valid and we discard it.
Also check, given the sequence of differences, if it matches the constraints that were provided as input. Basically we compare if our generated sequence aligns with the allowed number differences.
If the sequence fits within the provided differences array, count this as a valid hidden sequence.
Repeat this process, trying every possible starting number and allowed difference until all combinations have been exhausted.
The total number of valid hidden sequences discovered during this exhaustive search is the answer.

Code Implementation

def count_hidden_sequences_brute_force(differences, lower_bound, upper_bound):
    count_valid_sequences = 0

    # Iterate through all possible starting numbers.
    for starting_number in range(lower_bound, upper_bound + 1):

        # Construct a potential hidden sequence.
        hidden_sequence = [starting_number]
        is_valid_sequence = True

        for difference in differences:
            next_number = hidden_sequence[-1] + difference

            # Check if the number is out of bounds.
            if not (lower_bound <= next_number <= upper_bound):
                is_valid_sequence = False
                break

            hidden_sequence.append(next_number)

        # If the sequence is valid, increment the counter.
        if is_valid_sequence:
            count_valid_sequences += 1

    return count_valid_sequences

Big(O) Analysis

Time Complexity

O((upper - lower + 1) * (upper - lower)^(n-1) * n) – The brute force strategy iterates through all possible starting numbers from lower to upper, giving us (upper - lower + 1) options. For each starting number, it considers all possible difference combinations allowed by the differences array. Because each of the (n-1) differences between numbers in our potential hidden sequence is free to vary between (upper - lower), this introduces a cost of (upper - lower)^(n-1). For each potential sequence, we construct the sequence element by element (costing n), and validate it against the given differences array and range. Therefore, the overall time complexity is approximately O((upper - lower + 1) * (upper - lower)^(n-1) * n).

Space Complexity

O(N) – The algorithm constructs a potential hidden sequence element by element, storing it to check if it fits within the given range and matches the provided differences. This potential hidden sequence has a length proportional to the input size N (the length of the differences array, and thus the hidden sequence). Thus, auxiliary space proportional to N is required to store this sequence. No other significant data structures are used, making the space complexity O(N).

Optimal Solution

Approach

The key is to figure out the smallest and largest possible starting number that could produce a valid sequence within the given range. We achieve this by focusing on the extremes of the differences, and using them to deduce the bounds on our starting number.

Here's how the algorithm would work step-by-step:

First, find the biggest difference and the smallest difference in the given list of differences.
The biggest difference helps you figure out the lowest possible starting number: consider that all subsequent numbers are constructed from this lower bound
The smallest difference helps you figure out the highest possible starting number: consider that all subsequent numbers are constructed from this upper bound
Calculate the minimum possible number that could appear in the full sequence, if you started with the largest possible starting number and always subtracted the largest difference.
Calculate the maximum possible number that could appear in the full sequence, if you started with the smallest possible starting number and always added the largest difference.
If the minimum possible number is greater than the upper limit, then there are no valid hidden sequences.
If the maximum possible number is less than the lower limit, then there are no valid hidden sequences.
Otherwise, the number of possible hidden sequences is the upper limit, minus the maximum possible number plus one.

Code Implementation

def count_hidden_sequences(differences, lower_bound, upper_bound):
    max_difference = 0
    min_difference = 0
    current_sum = 0

    # Find the biggest and smallest cumulative differences.
    for difference in differences:
        current_sum += difference
        max_difference = max(max_difference, current_sum)
        min_difference = min(min_difference, current_sum)

    max_possible_start = upper_bound - max_difference
    min_possible_start = lower_bound - min_difference

    # To ensure the generated sequence stays within bounds.
    if min_possible_start > max_possible_start:
        return 0

    return max_possible_start - min_possible_start + 1

Big(O) Analysis

Time Complexity

O(n) – The provided approach involves iterating through the differences array once to find the maximum and minimum differences. Calculating the minimum and maximum possible numbers within the sequence involves arithmetic operations that are performed a fixed number of times regardless of the input size. Therefore, the time complexity is dominated by the single pass through the differences array, making it O(n), where n is the number of differences.

Space Complexity

O(1) – The provided explanation describes a process of finding the biggest and smallest differences, calculating minimum and maximum possible numbers, and comparing them against the lower and upper limits. This process only uses a few variables to store the largest difference, smallest difference, minimum possible number, maximum possible number, lower limit and upper limit. The space used does not depend on the size of the input list of differences. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
nums array is null	Return 0, as a null array cannot form any valid hidden sequences.
lower bound is greater than upper bound	Return 0, as it is impossible to construct a valid sequence within these bounds.
nums array has only one element	Return 1 if lower <= 0 + first <= upper where first is the first element of nums, and 0 otherwise.
The difference between upper and lower bound is extremely large, potentially leading to integer overflow during counting.	Use long data type to prevent possible integer overflow when calculating the number of valid sequences.
The running sum exceeds the upper or lower bounds repeatedly throughout the array.	Keep track of the maximum and minimum running sums, and adjust the count based on how much these extremes violate bounds.
All differences in nums are zero.	Calculate the valid shift range (upper - lower + 1) since all possible starting values result in valid sequences.
The size of 'nums' array is very large	Ensure the algorithm has linear time complexity (O(n)) to handle large inputs efficiently; avoid quadratic time solutions.
The differences 'nums' array contains negative numbers and large positive numbers.	Track both the maximum positive and minimum negative running sums to accurately calculate the necessary shifts.