Count K-Reducible Numbers Less Than N

Count K-Reducible Numbers Less Than N

What is this problem about?

A number is "K-reducible" if it takes at most $k$ operations to reduce it to 1. An operation consists of replacing a number with the count of set bits (1s) in its binary representation. You are given a binary string $n$ and an integer $k$. You need to find how many positive integers less than $n$ are $k$-reducible.

Why is this asked in interviews?

This "Hard" problem is asked by companies like Schneider Electric to test proficiency in Combinatorics and Digit Dynamic Programming (Digit DP). It requires a deep understanding of how bit counts change and how to count numbers with a specific bit count using combinations (nCr). It evaluates your ability to break a complex recursive reduction process into a simple counting task.

Algorithmic pattern used

This problem follows the Digit DP and Bit Manipulation pattern.

1. Precompute how many operations it takes for any bit count $x$ to reach 1. This range is small (up to 1000).
2. For each bit count $x$ that is $k$-reducible, use Digit DP or Combinations to count how many numbers less than $n$ have exactly $x$ set bits.
3. Sum these counts together (modulo $10^9+7$).

Example explanation

Let $n = "101"$ (which is 5), $k = 1$.

• Numbers less than 5: 1, 2, 3, 4.
• 1 (binary "001"): 1 set bit. 1 op: 1 -> done. 1 is 1-reducible.
• 2 (binary "010"): 1 set bit. 1 op: 1 -> done. 2 is 1-reducible.
• 3 (binary "011"): 2 set bits. 1 op: 2. 2 is not 1. Not 1-reducible.
• 4 (binary "100"): 1 set bit. 1 op: 1 -> done. 4 is 1-reducible. Total: 3.

Common mistakes candidates make

A major challenge is correctly implementing the Digit DP or Combinatorial counting for a binary string. Another error is forgetting that the reduction process counts set bits from the previous step—for example, if a number has 3 bits, the first operation results in the number 3, not the number of bits in 3. Finally, failing to handle the "less than $n$" constraint strictly (and excluding $n$ itself) is a frequent logic bug.

Interview preparation tip

When counting numbers based on their binary properties, Combinations ($nCr$) are often faster than full DP. If you need to find how many $m$-bit numbers have $r$ bits set, it’s simply $\binom{m}{r}$. Practice building a Pascal's triangle for $nCr$ values to speed up these types of counting problems.