# Solving Leetcode Interviews in Seconds with AI: Find the N-th Value After K Seconds


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3179" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the solution:

*   The problem can be solved using combinatorics. The value of `a[n-1]` after `k` seconds is equivalent to the binomial coefficient (n+k-1 choose k).
*   We can precompute factorials and their inverses modulo (10^9 + 7) to efficiently calculate binomial coefficients.
*   The result of each step should be returned modulo (10^9 + 7).

*   **Runtime Complexity:** O(n + k), **Storage Complexity:** O(n + k)

	
	# Code
	```python
	def solve():
    n, k = map(int, input().split())
    MOD = 10**9 + 7

    def power(a, b):
        res = 1
        a %= MOD
        while b > 0:
            if b % 2 == 1:
                res = (res * a) % MOD
            a = (a * a) % MOD
            b //= 2
        return res

    def modInverse(n):
        return power(n, MOD - 2)

    def combinations(n, k):
        if k < 0 or k > n:
            return 0
        if k == 0 or k == n:
            return 1
        if k > n // 2:
            k = n - k
        
        num = fact[n]
        den = (fact[k] * fact[n - k]) % MOD
        return (num * modInverse(den)) % MOD
    
    max_val = n + k - 1
    fact = [1] * (max_val + 1)
    for i in range(2, max_val + 1):
        fact[i] = (fact[i - 1] * i) % MOD

    print(combinations(n + k - 1, k))

solve()
	```