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Find the N-th Value After K Seconds

Medium

Asked by:

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Topics:

ArraysDynamic Programming

You are given two integers n and k.

Initially, you start with an array a of n integers where a[i] = 1 for all 0 <= i <= n - 1. After each second, you simultaneously update each element to be the sum of all its preceding elements plus the element itself. For example, after one second, a[0] remains the same, a[1] becomes a[0] + a[1], a[2] becomes a[0] + a[1] + a[2], and so on.

Return the value of a[n - 1] after k seconds.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: n = 4, k = 5

Output: 56

Explanation:

Second	State After
0	[1,1,1,1]
1	[1,2,3,4]
2	[1,3,6,10]
3	[1,4,10,20]
4	[1,5,15,35]
5	[1,6,21,56]

Example 2:

Input: n = 5, k = 3

Output: 35

Explanation:

Second	State After
0	[1,1,1,1,1]
1	[1,2,3,4,5]
2	[1,3,6,10,15]
3	[1,4,10,20,35]

Constraints:

1 <= n, k <= 1000

Try coding it in LeetCode

Find the N-th Value After K Seconds

Medium

Asked by:

5 views

Topics:

ArraysDynamic Programming

You are given two integers n and k.

Return the value of a[n - 1] after k seconds.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: n = 4, k = 5

Output: 56

Explanation:

Second	State After
0	[1,1,1,1]
1	[1,2,3,4]
2	[1,3,6,10]
3	[1,4,10,20]
4	[1,5,15,35]
5	[1,6,21,56]

Example 2:

Input: n = 5, k = 3

Output: 35

Explanation:

Second	State After
0	[1,1,1,1,1]
1	[1,2,3,4,5]
2	[1,3,6,10,15]
3	[1,4,10,20,35]

Constraints:

1 <= n, k <= 1000

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 constraints on the size of the input array 'values', and the values of 'k' and 'n'?
Can the elements in the 'values' array be negative, zero, or floating-point numbers, or are they strictly positive integers?
If 'n' is out of bounds (i.e., less than 0 or greater than or equal to the length of 'values') at any point during the calculation, what should I return?
For edge cases like an empty input array or an array with only one element, how should the function behave?
Is 'k' guaranteed to be non-negative, or could it be zero or negative? How should I handle those cases?

Brute Force Solution

Approach

The brute force method simulates the process step by step. We'll track the value after each second, updating it according to the given rules, until we reach the desired number of seconds. This is like manually following instructions at each time interval.

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

Begin with the initial value.
For each second that passes, calculate the new value based on the specific instructions given for how the value changes each second.
Repeat this process of calculating the new value, one second at a time, until you have done it for the required number of seconds.
The value you have after the final second is the answer.

Code Implementation

def find_nth_value_after_k_seconds(initial_value, k_seconds, change_rule):    current_value = initial_value

    # Simulate each second that passes
    for second in range(k_seconds):

        # Apply the change rule to update the value.
        current_value = change_rule(current_value)

    # Return the final value after K seconds.
    return current_value

Big(O) Analysis

Time Complexity

O(K) – The brute force method iterates for K seconds, updating the value in each iteration. Each update takes constant time, as it involves applying a fixed set of rules to the current value. Since the loop runs K times, the time complexity is directly proportional to K, resulting in O(K) complexity. N, representing the potential size or range of values, is not a factor in determining the number of iterations.

Space Complexity

O(1) – The brute force method described only uses a single variable to store the current value. It updates this value in place for each second up to K. No additional data structures like lists or hashmaps are used, and the number of seconds (K) does not affect the amount of auxiliary space used. Therefore, the space complexity is constant.

Optimal Solution

Approach

The challenge involves repeatedly changing a value based on a set of rules. Instead of simulating the changes second by second, we look for repeating patterns to quickly figure out the final value. This is done by figuring out how the value changes in a cycle, and then using the cycles to jump ahead in time.

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

First, start simulating the process, tracking how the values change over time.
Keep simulating until you notice a pattern where the sequence of values starts to repeat itself.
Once you find the repeating pattern, determine the length of the repeating cycle.
Calculate how many full cycles occur within the given time period.
Figure out how many seconds are 'left over' after all the full cycles.
Use the remainder to determine which value within the repeating cycle corresponds to the N-th value after K seconds from the beginning of a single cycle.
The value found in the cycle is the final answer.

Code Implementation

def find_nth_value(initial_values, update_rule, total_time, pattern_detection_time = 100):
    current_values = initial_values[:]
    history = []

    # Simulate to detect repeating pattern
    for time_step in range(min(total_time, pattern_detection_time)):
        history.append(current_values[:])
        new_values = update_rule(current_values)
        current_values = new_values

        # Check for pattern
        for pattern_length in range(1, len(history) // 2 + 1):
            if history[-pattern_length:] == history[-2 * pattern_length:-pattern_length]:

                # Pattern found, calculate remaining time
                remaining_time = total_time % pattern_length
                current_values = initial_values[:]

                # Simulate only the remaining time
                for _ in range(remaining_time):
                    new_values = update_rule(current_values)
                    current_values = new_values

                return current_values

    # No pattern found, simulate full time
    current_values = initial_values[:]

    # No recognizable pattern, must iterate to full time
    for _ in range(total_time):
        new_values = update_rule(current_values)
        current_values = new_values

    return current_values

Big(O) Analysis

Time Complexity

O(K) – The dominant operation is simulating the value changes until a repeating pattern is found. In the worst case, we might have to iterate through all K seconds before a pattern emerges. The pattern detection itself involves comparing current values with past values, which can be considered constant time within each iteration. Thus, the time complexity is directly proportional to K, the number of seconds, resulting in O(K).

Space Complexity

O(C) – The algorithm simulates the process and stores the values encountered until a repeating pattern is found. This is done to determine the cycle. The space complexity is determined by the number of unique values encountered before a repeating cycle is detected. Therefore, the auxiliary space required is proportional to the length of the cycle (denoted as C), where C is independent of N. Thus, the space complexity is O(C).

Edge Cases

Case	How to Handle
values is null or empty	Return 0 since there are no values to update.
k is 0	Return values[n] immediately since no updates are needed.
n is out of bounds (n < 0 or n >= values.length)	Return 0 since the index is invalid.
values.length is 1	The value at index 0 remains constant so return values[0].
Integer overflow during summation of neighboring values	Use long data type to store intermediate sums to prevent overflow.
Large k value potentially leading to performance issues	Optimize by detecting repeating patterns in the array updates or using matrix exponentiation.
All values in 'values' are 0	After one second, the array will remain all zeros; return 0.
Values contains very large numbers that can cause overflow	Check for overflow during addition, and potentially return a specific error code or throw an exception if overflow is detected and cannot be handled safely.

values is null or empty

How to Handle:

Return 0 since there are no values to update.

k is 0

How to Handle:

Return values[n] immediately since no updates are needed.

n is out of bounds (n < 0 or n >= values.length)

How to Handle:

Return 0 since the index is invalid.

values.length is 1

How to Handle:

The value at index 0 remains constant so return values[0].

Integer overflow during summation of neighboring values

How to Handle:

Use long data type to store intermediate sums to prevent overflow.

Large k value potentially leading to performance issues

How to Handle:

Optimize by detecting repeating patterns in the array updates or using matrix exponentiation.

All values in 'values' are 0

How to Handle:

After one second, the array will remain all zeros; return 0.

Values contains very large numbers that can cause overflow

How to Handle:

Check for overflow during addition, and potentially return a specific error code or throw an exception if overflow is detected and cannot be handled safely.

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 constraints on the size of the input array 'values', and the values of 'k' and 'n'?
Can the elements in the 'values' array be negative, zero, or floating-point numbers, or are they strictly positive integers?
If 'n' is out of bounds (i.e., less than 0 or greater than or equal to the length of 'values') at any point during the calculation, what should I return?
For edge cases like an empty input array or an array with only one element, how should the function behave?
Is 'k' guaranteed to be non-negative, or could it be zero or negative? How should I handle those cases?

Brute Force Solution

Approach

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

Begin with the initial value.
For each second that passes, calculate the new value based on the specific instructions given for how the value changes each second.
Repeat this process of calculating the new value, one second at a time, until you have done it for the required number of seconds.
The value you have after the final second is the answer.

Code Implementation

def find_nth_value_after_k_seconds(initial_value, k_seconds, change_rule):    current_value = initial_value

    # Simulate each second that passes
    for second in range(k_seconds):

        # Apply the change rule to update the value.
        current_value = change_rule(current_value)

    # Return the final value after K seconds.
    return current_value

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, start simulating the process, tracking how the values change over time.
Keep simulating until you notice a pattern where the sequence of values starts to repeat itself.
Once you find the repeating pattern, determine the length of the repeating cycle.
Calculate how many full cycles occur within the given time period.
Figure out how many seconds are 'left over' after all the full cycles.
Use the remainder to determine which value within the repeating cycle corresponds to the N-th value after K seconds from the beginning of a single cycle.
The value found in the cycle is the final answer.

Code Implementation

def find_nth_value(initial_values, update_rule, total_time, pattern_detection_time = 100):
    current_values = initial_values[:]
    history = []

    # Simulate to detect repeating pattern
    for time_step in range(min(total_time, pattern_detection_time)):
        history.append(current_values[:])
        new_values = update_rule(current_values)
        current_values = new_values

        # Check for pattern
        for pattern_length in range(1, len(history) // 2 + 1):
            if history[-pattern_length:] == history[-2 * pattern_length:-pattern_length]:

                # Pattern found, calculate remaining time
                remaining_time = total_time % pattern_length
                current_values = initial_values[:]

                # Simulate only the remaining time
                for _ in range(remaining_time):
                    new_values = update_rule(current_values)
                    current_values = new_values

                return current_values

    # No pattern found, simulate full time
    current_values = initial_values[:]

    # No recognizable pattern, must iterate to full time
    for _ in range(total_time):
        new_values = update_rule(current_values)
        current_values = new_values

    return current_values

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
values is null or empty	Return 0 since there are no values to update.
k is 0	Return values[n] immediately since no updates are needed.
n is out of bounds (n < 0 or n >= values.length)	Return 0 since the index is invalid.
values.length is 1	The value at index 0 remains constant so return values[0].
Integer overflow during summation of neighboring values	Use long data type to store intermediate sums to prevent overflow.
Large k value potentially leading to performance issues	Optimize by detecting repeating patterns in the array updates or using matrix exponentiation.
All values in 'values' are 0	After one second, the array will remain all zeros; return 0.
Values contains very large numbers that can cause overflow	Check for overflow during addition, and potentially return a specific error code or throw an exception if overflow is detected and cannot be handled safely.

values is null or empty

How to Handle:

Return 0 since there are no values to update.

k is 0

How to Handle:

Return values[n] immediately since no updates are needed.

n is out of bounds (n < 0 or n >= values.length)

How to Handle:

Return 0 since the index is invalid.

values.length is 1

How to Handle:

The value at index 0 remains constant so return values[0].

Integer overflow during summation of neighboring values

How to Handle:

Use long data type to store intermediate sums to prevent overflow.

Large k value potentially leading to performance issues

How to Handle:

Optimize by detecting repeating patterns in the array updates or using matrix exponentiation.

All values in 'values' are 0

How to Handle:

After one second, the array will remain all zeros; return 0.

Values contains very large numbers that can cause overflow

How to Handle:

Check for overflow during addition, and potentially return a specific error code or throw an exception if overflow is detected and cannot be handled safely.