3179. Find the N-th Value After K Seconds

MediumArrayMathCombinatoricsPrefix SumSimulation

Problem Description

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 10^9 + 7.

Intuition

The problem can be understood as simulating the repeated updates of an integer array over k seconds. Initially, each element of the array is 1, and every second, each element is updated to include itself and all preceding elements.

Considering the constraints, the approach chosen is to directly simulate the process. For each second, iterate through the array starting from the second element (index 1) and update each element to the sum of itself and the element before it. This effectively accumulates values toward the end of the array over each second. Continue this process until k seconds have elapsed.

The final result is the value of the last element in the array after these updates, which is returned after taking modulo 10^9 + 7 to manage large numbers and fit within typical integer range constraints.

Learn more about Math, Combinatorics and Prefix Sum patterns.

Solution Approach

The solution utilizes a simulation approach to solve the problem. This approach is feasible due to the constraints, as n can be a maximum of 1000, allowing for direct manipulation and updating of arrays.

Initialize an Array: Start with an array a of size n, where each element is initialized to 1. This represents the initial state before any updates.
Simulate for k seconds:
- Repeat the updating process k times.
- For each second, iterate through the array from the second element to the last:
  - Update each element a[i] to be the sum of itself and the previous element a[i-1]. This can be expressed as a[i] = (a[i] + a[i - 1]) % mod, where mod is 10^9 + 7. The use of the modulo operation ensures that the numbers do not exceed the specified limits due to potentially large values.
Return Result: After completing the updates for k seconds, return the last element of the array a[n - 1].

This approach makes use of a simple array and iterative logic to ensure that each element is updated according to the rules given in the problem statement. The key operation here is summing up preceding elements including the current one in each step, and applying the modulo operation to handle large numbers.

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Example Walkthrough

Let's illustrate the solution approach with a small example where n = 3 and k = 2.

Initialization:
- Start with an array a of size 3, initialized as a = [1, 1, 1].
Simulating the First Second:
- Update elements starting from the second position:
  - a[1] = (a[1] + a[0]) % (10^9 + 7) = (1 + 1) % (10^9 + 7) = 2
  - a[2] = (a[2] + a[1]) % (10^9 + 7) = (1 + 2) % (10^9 + 7) = 3
- After the first second, the array becomes a = [1, 2, 3].
Simulating the Second Second:
- Update elements starting from the second position:
  - a[1] = (a[1] + a[0]) % (10^9 + 7) = (2 + 1) % (10^9 + 7) = 3
  - a[2] = (a[2] + a[1]) % (10^9 + 7) = (3 + 3) % (10^9 + 7) = 6
- After the second second, the array becomes a = [1, 3, 6].
Result:
- After k = 2 seconds, the value of a[n-1] = a[2] is 6.

Thus, the result returned is 6, which is the final value at the last position of the array after applying the updates for 2 seconds.

Solution Implementation

1class Solution:
2    def valueAfterKSeconds(self, n: int, k: int) -> int:
3        # Initialize an array 'a' of size 'n' with all elements as 1
4        a = [1] * n
5        # Define the modulus to prevent overflow
6        mod = 10**9 + 7
7      
8        # Repeat the process 'k' times
9        for _ in range(k):
10            # Update each element starting from the second one
11            for i in range(1, n):
12                # Compute the cumulative sum for 'a[i]'
13                a[i] = (a[i] + a[i - 1]) % mod
14      
15        # Return the value of the last element, which represents the result after 'k' seconds
16        return a[n - 1]
17

1class Solution {
2    public int valueAfterKSeconds(int n, int k) {
3        final int MOD = (int) 1e9 + 7; // Define a constant for the modulo operation to prevent integer overflow.
4        int[] array = new int[n]; // Create an array of size n to store current values.
5        Arrays.fill(array, 1); // Initialize all elements of the array to 1.
6
7        // Repeat the operation for k seconds.
8        while (k-- > 0) {
9            // Iterate over the array starting from the second element.
10            for (int i = 1; i < n; ++i) {
11                // Update the current element with the sum of its current value and the value of the previous element modulo MOD.
12                array[i] = (array[i] + array[i - 1]) % MOD;
13            }
14        }
15
16        // Return the last element of the array as the result.
17        return array[n - 1];
18    }
19}
20

1#include <vector>
2
3class Solution {
4public:
5    int valueAfterKSeconds(int n, int k) {
6        // The modulo value to keep numbers manageable
7        const int MOD = 1e9 + 7;
8      
9        // Initialize a vector of size n with all elements set to 1
10        std::vector<int> sequence(n, 1);
11
12        // Loop for k seconds
13        while (k-- > 0) {
14            // Update the sequence
15            for (int i = 1; i < n; ++i) {
16                // Update each element to be the sum of itself and the previous element, modulo MOD
17                sequence[i] = (sequence[i] + sequence[i - 1]) % MOD;
18            }
19        }
20      
21        // Return the last element of the sequence
22        return sequence[n - 1];
23    }
24};
25

1// This function calculates the value of the nth element after k iterations.
2function valueAfterKSeconds(n: number, k: number): number {
3    // Initialize an array of size n filled with 1s
4    const a: number[] = Array(n).fill(1);
5    // Set the modulus value
6    const mod: number = 10 ** 9 + 7;
7  
8    // Iterate for k seconds
9    while (k--) {
10        // Update the array based on the sum of previous elements modulo mod
11        for (let i = 1; i < n; ++i) {
12            a[i] = (a[i] + a[i - 1]) % mod;
13        }
14    }
15  
16    // Return the final value of the nth element
17    return a[n - 1];
18}
19

Time and Space Complexity

The time complexity of the given code is O(n * k). This is because there are two nested loops: the outer loop iterates k times, and for each iteration, the inner loop iterates n - 1 times. Therefore, the total number of operations is proportional to n * k.

The space complexity of the code is O(n). This is because the code maintains a single list a of size n, which dictates the space used.

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

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