Check If Digits Are Equal in String After Operations II

Check If Digits Are Equal in String After Operations II

What is this problem about?

The "Check If Digits Are Equal in String After Operations II interview question" is a significantly harder version of the digit reduction problem. You are given a string of digits and perform the same operation: replace each adjacent pair $(s[i], s[i+1])$ with their sum modulo 10. However, the initial string can be extremely long (up to $10^5$), making direct simulation impossible. You need to calculate the final two digits using a mathematical shortcut.

Why is this asked in interviews?

This "Check If Digits Are Equal coding problem" is asked by companies like ADP to test a candidate's mastery of Combinatorics and Number Theory. It evaluates whether you can recognize that the reduction process follows the pattern of Pascal's Triangle. It tests your ability to use Lucas' Theorem or Prime Factorization to compute large binomial coefficients modulo a non-prime number (10).

Algorithmic pattern used

This problem uses the Binomial Coefficient (Combinatorics) pattern.

1. Pascal's Connection: The final two digits depend on the original digits. If $n$ is the length of the string, the final two digits are derived from the sums $\sum s[i] imes \binom{n-2}{i}$ for $i$ from 0 to $n-2$ and $i$ from 1 to $n-1$.
2. Modulo 10: Since 10 is not prime ($2 imes 5$), you must calculate the coefficients modulo 2 and modulo 5 separately and then combine them using the Chinese Remainder Theorem (CRT).
3. Lucas' Theorem: To find $\binom{n}{k} \pmod p$ where $p$ is prime, Lucas' Theorem provides an $O(\log_p n)$ way to compute the result.

Example explanation

Suppose $n=4$, string is "1234". Final two digits are $d_1, d_2$.

• $d_1$ is derived from $s[0], s[1], s[2]$ with coefficients $\binom{2}{0}, \binom{2}{1}, \binom{2}{2}$ (1, 2, 1).
• $d_1 = (1 imes 1 + 2 imes 2 + 1 imes 3) \pmod{10} = 1 + 4 + 3 = 8$.
• $d_2$ is derived from $s[1], s[2], s[3]$ with coefficients (1, 2, 1).
• $d_2 = (1 imes 2 + 2 imes 3 + 1 imes 4) \pmod{10} = 2 + 6 + 4 = 12 \pmod{10} = 2$. Check: $8 eq 2$. Result: False.

Common mistakes candidates make

• Simulation: Trying to simulate the process for $N=10^5$, which leads to $O(N^2)$ and a Time Limit Exceeded error.
• Modulo Complexity: Forgetting that $\binom{n}{k} \pmod{10}$ requires splitting into prime factors because 10 is composite.
• Factorial Pre-calculation: Not using Prime Factorization to handle the "division" in the binomial formula $\frac{n!}{k!(n-k)!}$ when working under a modulo.

Interview preparation tip

Master the relationship between "iterative summing" and binomial coefficients. This "Math interview pattern" is the key to many "Hard" level reduction problems. Learn how to calculate combinations modulo small primes using Lucas' Theorem.