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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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def count_valid_strings(N):
    MOD = 10**9 + 7
   
    T, A, N, R, U, S = range(6)
   
    dp = [[0] * 6 for _ in range(N + 1)]
   
    dp[1][T] = 1
    dp[1][N] = 1
    dp[1][S] = 1
   
    for i in range(1, N):
        dp[i + 1][A] = (dp[i][T]) % MOD
        dp[i + 1][N] = (dp[i][A]) % MOD
        dp[i + 1][R] = (dp[i][A] + dp[i][N]) % MOD
        dp[i + 1][U] = (dp[i][R]) % MOD
        dp[i + 1][S] = (dp[i][U]) % MOD
        dp[i + 1][T] = (dp[i][S]) % MOD
        dp[i + 1][A] = (dp[i + 1][A] + dp[i][S]) % MOD
        dp[i + 1][N] = (dp[i + 1][N] + dp[i][S]) % MOD
        dp[i + 1][R] = (dp[i + 1][R] + dp[i][S]) % MOD
        dp[i + 1][U] = (dp[i + 1][U] + dp[i][S]) % MOD
   
    result = sum(dp[N][c] for c in [T, A, N, R, U, S]) % MOD
    return result


Thoughtwork โœ…
import math
def sieve(n):
    is_prime = [True] * (n + 1)
    is_prime[0] = is_prime[1] = False
    for i in range(2, int(n**0.5) + 1):
        if is_prime[i]:
            for j in range(i * i, n + 1, i):
                is_prime[j] = False
    return [i for i in range(2, n + 1) if is_prime[i]]

def count_nearly_primes(L, R):
    limit = int(math.sqrt(R))
    primes = sieve(limit)
   
    nearly_primes_count = 0
    for prime in primes:
        nearly_prime = prime * prime
        if L < nearly_prime < R:
            nearly_primes_count += 1
   
    return nearly_primes_count

Count of Nearly P โœ…
Thoughtwork
def find(N, K, X):
    MOD = 1000000007

    dp = [[0] * (K + 1) for _ in range(N + 1)]
    dp[1][1] = 1

    for i in range(2, N + 1):
        for j in range(1, K + 1):
            dp[i][j] = sum(dp[i-1][p] for p in range(1, K + 1) if p != j) % MOD
   
    return dp[N][X]


Number of Arrays โœ…
Thoughtwork
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