2145. Count the Hidden Sequences

MediumArrayPrefix Sum
Leetcode Link

Problem Description

You are presented with an array differences which is used to calculate the values of a hidden sequence hidden. The differences array gives you the difference between each pair of consecutive elements in the hidden sequence. That is, for each i in the differences array, differences[i] = hidden[i + 1] - hidden[i].

The hidden sequence itself is not known, but it does have length n + 1, where n is the length of the differences array. The problem also stipulates that any number in the hidden sequence must be within an inclusive range denoted by two given integers, lower and upper.

Your task is to find out how many different hidden sequences can be possibly made that adhere to the differences provided and also stay within the boundaries given by lower and upper. If no such sequences can be made, your answer should be 0.

Intuition

To solve this problem, we think about the constraints that the lower and upper bounds put on the possible values of the hidden sequence. Since we know the differences between each pair of consecutive values, we can simulate the sequence to find the minimum and maximum values that occur if we start the sequence at zero. These minimum and maximum values are offset versions of what the real sequence could look like.

Knowing the most extreme (minimum and maximum) values that our sequence reaches, we can calculate how many different starting points for the hidden sequence are possible that would keep the sequence within the bounds of lower and upper. Essentially, we are sliding the window of the extreme values of our simulated sequence along the scale of lower to upper and checking for overlaps.

Since we are given that the hidden sequence values must be in the range [lower, upper] (inclusive), we subtract the maximum value we found from upper and the result is the span of numbers that could potentially be the start of a hidden sequence. We also subtract the range span of the hidden sequence (max - min), so we get the number of sequences that can be formed. Adding 1 accounts for the inclusive boundaries. If the span is negative, that means there are no possible sequences, so we use max function to set the count to 0 in such cases.

The one-liner calculation in the provided solution performs this sliding window computation to find the count of valid starting numbers, and thus, by extension, the count of valid sequences.

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Solution Approach

To implement the solution, we make use of a simple linear scan algorithm to iterate through the differences array and accumulate the changes to calculate the possible minimum and maximum values that the hidden sequence can take.

  1. We start by initializing a variable num to 0, which represents the current value of the hidden sequence, assuming it starts at 0. Alongside, we initialize two other variables, mi and mx, which stand for the minimum and the maximum values that we encounter as we construct the hidden sequence. Both are initially set to 0.

  2. Then, we iterate over the differences array, and for each difference d, we add d to num. As we simulate the sequence, we update mi and mx with these two lines:

    1mi = min(mi, num)
2mx = max(mx, num)

    If num is lesser than our current minimum, we update mi to reflect num, and similarly, if num is greater than our current maximum, we update mx to be num.

  3. After we finish iterating through all the elements in differences, we will have the most extreme values that the hidden sequence could attain if it started at 0. Now, we must consider the given bounds lower and upper.

  4. The formula to calculate the number of valid starting points for the hidden sequence, and hence the number of possible sequences, is:

    1return max(0, upper - lower - (mx - mi) + 1)

    We subtract mx - mi from upper - lower to get the number of positions between lower and upper that could be the start of the hidden sequence, while still ensuring all of its values do not breach the bounds. We add 1 because both lower and upper are inclusive.

  5. If the result of the subtraction is less than 0, it implies that there's no possible starting point that keeps the sequence within the bounds, therefore there are 0 possible sequences. We use the max function to handle this scenario, which helps to ensure that we do not return a negative number of possible sequences.

This approach is efficient, with a time complexity of O(n) where n is the length of the differences array, since we only need to scan through the differences array once to arrive at the solution.

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Example Walkthrough

Let's consider the differences array as [-2, -1, 2, 1], with the bounds lower equal to 1, and upper equal to 6.

  1. We start by initializing our variables num, mi, and mx to 0. These will keep track of our current sequence value (assuming we start at 0), and the minimum and maximum values we encounter.

  2. We iterate through the differences array:

    • For the first difference, -2, we update num to 0 - 2 = -2. We also update mi to -2 (since -2 is less than the old minimum 0) and mx remains as 0.
    • For the second difference, -1, num becomes -2 - 1 = -3. mi is updated to -3 and mx remains as 0.
    • For the third difference, 2, num becomes -3 + 2 = -1. mi remains as -3 and mx remains as 0.
    • For the last difference, 1, num is now -1 + 1 = 0. Again mi and mx remain as -3 and 0, respectively.

  3. After iterating through the array, we have mi = -3 and mx = 0. Our hidden sequence, if starting at 0, would range in values between -3 and 0.

  4. We now compare the span of our possible hidden sequence values with the given bounds:

    • We first calculate the span of numbers between lower and upper: upper - lower which is 6 - 1 = 5.
    • Next, we find the span of our hidden sequence by computing mx - mi: 0 - (-3) = 3.
    • To find how many sequences can fit, we subtract the hidden sequence span from the bounds span and add 1: (5 - 3) + 1 = 3.

  5. The final result is 3. Meaning we can have 3 different possible starting points for the hidden sequence that would keep the sequence within the bounds given by lower and upper. The valid sequences would start at 1, 2, and 3, leading to the following possible sequences within the bounds [1, 6]:

    • Starting at 1: [1, -1, -2, 0, 1]
    • Starting at 2: [2, 0, -1, 1, 2]
    • Starting at 3: [3, 1, 0, 2, 3]

Each starting value leads to a sequence that, when applying the differences, remains within the bounds of 1 to 6. Hence, there are 3 valid hidden sequences.

Solution Implementation

1from typing import List
2
3class Solution:
4    def numberOfArrays(self, differences: List[int], lower: int, upper: int) -> int:
5        # Initialize the variables: current_sum to track the running sum,
6        # min_value to keep the minimum value encountered, and max_value
7        # for the maximum value encountered.
8        current_sum = min_value = max_value = 0
9      
10        # Iterate through each difference in the array
11        for diff in differences:
12            # Add the current difference to the running sum
13            current_sum += diff
14            # Update the minimum value if the new current_sum is lower
15            min_value = min(min_value, current_sum)
16            # Update the maximum value if the new current_sum is higher
17            max_value = max(max_value, current_sum)
18      
19        # Calculate the width of the range spanned by the differences
20        range_width = max_value - min_value
21        # Calculate the total number of distinct arrays that can be formed
22        # within the given upper and lower bounds. Here we also include the
23        # '+ 1' offset to account for inclusive bounds.
24        # If the resulting number is negative, we use max(0, ...) to default to 0.
25        # This represents cases where no valid arrays can be formulated.
26        num_of_arrays = max(0, (upper - lower) - range_width + 1)
27      
28        return num_of_arrays
29
1class Solution {
2
3    /**
4     * Calculate the number of valid arrays that can be constructed with the given conditions.
5     * 
6     * @param differences An array of integers representing the difference between consecutive elements in the target array.
7     * @param lower The lower bound for the elements of the target array.
8     * @param upper The upper bound for the elements of the target array.
9     * @return The number of valid arrays that can be constructed.
10     */
11    public int numberOfArrays(int[] differences, int lower, int upper) {
12        // Initialize running sum, minimum and maximum values observed while simulating the array creation
13        long runningSum = 0;
14        long minObserved = 0;
15        long maxObserved = 0;
16
17        // Iterate over the array of differences
18        for (int difference : differences) {
19            // Update the running sum with the current difference
20            runningSum += difference;
21          
22            // Update the minimum observed sum, if necessary
23            minObserved = Math.min(minObserved, runningSum);
24          
25            // Update the maximum observed sum, if necessary
26            maxObserved = Math.max(maxObserved, runningSum);
27        }
28
29        // Compute the number of possible starting values that satisfy the bounds
30        int totalValidArrays = Math.max(0, (int) (upper - lower - (maxObserved - minObserved) + 1));
31
32        // Return the computed total number of valid arrays
33        return totalValidArrays;
34    }
35}
36
1#include <vector> // Include necessary header for using vectors
2#include <algorithm> // Include necessary header for using min and max functions
3
4class Solution {
5public:
6    int numberOfArrays(vector<int>& differences, int lower, int upper) {
7        long long runningSum = 0; // This will keep track of the accumulated sum of differences
8        long long minSum = 0; // This will keep the minimum sum encountered
9        long long maxSum = 0; // This will keep the maximum sum encountered
10
11        // Iterate over the differences array
12        for (int &difference : differences) {
13            runningSum += difference; // Accumulate the sum of differences
14            minSum = std::min(minSum, runningSum); // Update the minimum sum if necessary
15            maxSum = std::max(maxSum, runningSum); // Update the maximum sum if necessary
16        }
17      
18        // Calculate the range of the final array values
19        long long validRange = upper - lower - (maxSum - minSum) + 1;
20
21        // If the range is negative, set it to zero
22        return std::max(0LL, validRange);
23    }
24};
25
1// TypeScript does not have a standard header system like C++, 
2// so you don't 'include' modules. Instead, you import them if necessary.
3// In this case, no import is needed since arrays and Math functions are built-in.
4
5// A global variable to keep track of the accumulated sum of differences
6let runningSum: number = 0;
7
8// A global variable to keep track of the minimum sum encountered
9let minSum: number = 0;
10
11// A global variable to keep track of the maximum sum encountered
12let maxSum: number = 0;
13
14// Function to calculate the number of valid arrays from the given differences
15// and the bounds provided by lower and upper limits
16function numberOfArrays(differences: number[], lower: number, upper: number): number {
17    // Reset the global variables for a new function call
18    runningSum = 0;
19    minSum = 0;
20    maxSum = 0;
21
22    // Iterate over the differences array
23    for (let difference of differences) {
24        // Accumulate the sum of differences
25        runningSum += difference;
26      
27        // Update the minimum and maximum sums if necessary
28        minSum = Math.min(minSum, runningSum);
29        maxSum = Math.max(maxSum, runningSum);
30    }
31  
32    // Calculate the range of the final array values
33    let validRange: number = upper - lower - (maxSum - minSum) + 1;
34
35    // Return the number of valid arrays, ensuring the number is not negative
36    return Math.max(0, validRange);
37}
38
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Time and Space Complexity

The given Python function numberOfArrays calculates how many valid arrays can be generated from a list of differences within the given upper and lower bounds.

Time Complexity: The time complexity of the numberOfArrays function is O(n), where n is the length of the differences list. This is because we iterate through each element of differences exactly once, performing constant-time operations (addition, minimum, maximum) at each step.

Space Complexity: The space complexity of the function is O(1) as the function uses a fixed number of integer variables (num, mi, mx) and does not allocate any additional space that grows with the input size. The space used for the input differences list does not count towards the space complexity of the function itself as it is provided as input.

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