Problem Description

You start with an array a of length n where every element is initialized to 1. That is, a[i] = 1 for all positions from 0 to n-1.

Every second, you perform a transformation on the array. In this transformation, each element is updated to become the sum of itself plus all elements that come before it in the array. The updates happen simultaneously for all elements.

For example, if you have an array [1, 1, 1, 1]:

• After 1 second: [1, 2, 3, 4]

  • a[0] stays 1 (no elements before it)
  • a[1] becomes 1 + 1 = 2 (sum of a[0] and itself)
  • a[2] becomes 1 + 1 + 1 = 3 (sum of a[0], a[1], and itself)
  • a[3] becomes 1 + 1 + 1 + 1 = 4 (sum of all elements)

• After 2 seconds: [1, 3, 6, 10]

  • a[0] stays 1
  • a[1] becomes 1 + 2 = 3
  • a[2] becomes 1 + 2 + 3 = 6
  • a[3] becomes 1 + 2 + 3 + 4 = 10

Your task is to find the value of the last element a[n-1] after performing this transformation for k seconds.

Since the final answer can be very large, return it modulo 10^9 + 7.

Intuition

Looking at the transformation pattern, we can observe that in each second, every element becomes a cumulative sum up to its position. This is essentially computing prefix sums repeatedly.

When we update the array, element a[i] becomes the sum of all elements from a[0] to a[i]. This operation is equivalent to computing the prefix sum of the current array. If we repeat this process k times, we're essentially computing prefix sums of prefix sums.

The key insight is that since n is limited to at most 1000, and we need to perform this operation k times, we can directly simulate the process without worrying about time complexity. Each transformation takes O(n) time to compute all prefix sums, and we do this k times, giving us O(n * k) total time complexity.

The pattern that emerges is interesting from a mathematical perspective - this process is related to Pascal's triangle and combinatorial coefficients. After k transformations, the value at position i is related to the binomial coefficient C(k+i, i). However, given the constraints, direct simulation is the most straightforward approach.

To implement the simulation, we maintain an array and in each iteration, we update each element (except the first one which always stays 1) by adding the previous element to it. This gives us the cumulative sum effect we need. We use modulo 10^9 + 7 at each step to prevent integer overflow and get the final answer in the required format.

Learn more about Math, Combinatorics and Prefix Sum patterns.

Solution Approach

The solution uses direct simulation to solve the problem. Here's how the implementation works:

Initialization: We create an array a of size n and initialize all elements to 1. This represents the initial state where a[i] = 1 for all positions.

a = [1] * n
mod = 10**9 + 7

Simulation Loop: We run a loop for k iterations, where each iteration represents one second of transformation.

for _ in range(k):
    for i in range(1, n):
        a[i] = (a[i] + a[i - 1]) % mod

In each iteration:

• We iterate through the array from index 1 to n-1 (skipping index 0 since it always remains 1)
• For each position i, we update a[i] to be the sum of its current value and the value at a[i-1]
• This update pattern automatically gives us the cumulative sum effect because:
  • When we update a[1], it becomes a[1] + a[0]
  • When we update a[2], it becomes a[2] + a[1], but a[1] already contains the sum of original a[0] and a[1]
  • This continues, creating the prefix sum for each position

Modulo Operation: We apply modulo 10^9 + 7 at each step to prevent integer overflow and ensure the result stays within bounds.

Return Value: After k seconds of simulation, we return a[n-1], which is the last element of the array.

The beauty of this approach is its simplicity - by updating elements from left to right and adding the previous element to the current one, we naturally compute the prefix sums in-place without needing additional space for storing intermediate results.

Example Walkthrough

Let's walk through a small example with n = 4 and k = 3 to illustrate the solution approach.

Initial Setup:

• Start with array a = [1, 1, 1, 1]
• We'll perform the transformation 3 times

First Transformation (k = 1):

• Start from index 1 and update each element by adding the previous element
• i = 1: a[1] = a[1] + a[0] = 1 + 1 = 2
• i = 2: a[2] = a[2] + a[1] = 1 + 2 = 3 (note: a[1] is already updated to 2)
• i = 3: a[3] = a[3] + a[2] = 1 + 3 = 4 (note: a[2] is already updated to 3)
• Result: a = [1, 2, 3, 4]

Second Transformation (k = 2):

• Continue with the same process on the updated array
• i = 1: a[1] = a[1] + a[0] = 2 + 1 = 3
• i = 2: a[2] = a[2] + a[1] = 3 + 3 = 6
• i = 3: a[3] = a[3] + a[2] = 4 + 6 = 10
• Result: a = [1, 3, 6, 10]

Third Transformation (k = 3):

• Apply the transformation one more time
• i = 1: a[1] = a[1] + a[0] = 3 + 1 = 4
• i = 2: a[2] = a[2] + a[1] = 6 + 4 = 10
• i = 3: a[3] = a[3] + a[2] = 10 + 10 = 20
• Result: a = [1, 4, 10, 20]

Final Answer: The last element a[3] = 20

The key insight here is that by updating elements from left to right and always adding the previously updated element, we achieve the cumulative sum effect in a single pass through the array. Each transformation builds upon the previous state, creating increasingly larger values that represent the repeated application of prefix sums.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × k), where n is the length of the array and k is the number of iterations. This is because we have two nested loops - the outer loop runs k times and the inner loop runs n-1 times (from index 1 to n-1), resulting in approximately k × n operations.

The space complexity is O(n), where n is the length of the array a. The algorithm uses a single array of size n to store the values, and only modifies this array in-place without using any additional data structures that scale with the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Apply Modulo Operation

One of the most common mistakes is forgetting to apply the modulo operation during the accumulation, leading to integer overflow for large values of n and k.

Incorrect Implementation:

def valueAfterKSeconds(self, n: int, k: int) -> int:
    array = [1] * n
    MOD = 10**9 + 7
  
    for second in range(k):
        for index in range(1, n):
            array[index] = array[index] + array[index - 1]  # No modulo here!
  
    return array[n - 1] % MOD  # Only applying modulo at the end

Why it's wrong: Even though we apply modulo at the end, intermediate values can grow exponentially large, causing integer overflow or memory issues in languages with fixed integer sizes.

Correct Approach:

array[index] = (array[index] + array[index - 1]) % MOD  # Apply modulo at each step

2. Creating a New Array Instead of In-Place Updates

Another pitfall is creating a new array for each transformation, which can lead to incorrect results.

Incorrect Implementation:

def valueAfterKSeconds(self, n: int, k: int) -> int:
    array = [1] * n
    MOD = 10**9 + 7
  
    for second in range(k):
        new_array = [0] * n
        for index in range(n):
            # Calculate sum from 0 to index
            for j in range(index + 1):
                new_array[index] += array[j]
            new_array[index] %= MOD
        array = new_array
  
    return array[n - 1]

Why it's inefficient: While this might give correct results, it's unnecessarily complex and has O(n²) time complexity per iteration instead of O(n).

Better Approach: Update the array in-place from left to right, which naturally maintains the cumulative sum property.

3. Starting the Update Loop from Index 0

A subtle mistake is starting the inner loop from index 0 instead of index 1.

Incorrect Implementation:

for second in range(k):
    for index in range(n):  # Starting from 0
        if index > 0:
            array[index] = (array[index] + array[index - 1]) % MOD

Why it's inefficient: While this works (with the additional if-check), it's cleaner to start from index 1 since array[0] always remains 1.

Cleaner Approach:

for index in range(1, n):  # Start directly from 1
    array[index] = (array[index] + array[index - 1]) % MOD

4. Not Recognizing the Mathematical Pattern

For those seeking optimization, a common pitfall is not recognizing that this problem follows a combinatorial pattern. The value at position i after k seconds equals the binomial coefficient C(k+i, i).

Mathematical Solution:

def valueAfterKSeconds(self, n: int, k: int) -> int:
    MOD = 10**9 + 7
  
    # Calculate C(k + n - 1, n - 1) using Pascal's triangle approach
    result = 1
    for i in range(1, n):
        result = result * (k + i) // i
        result %= MOD
  
    return result

However, the simulation approach is often more intuitive and sufficient for the given constraints.