LeetCode #3352 — HARD

Count K-Reducible Numbers Less Than N

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given a binary string s representing a number n in its binary form.

You are also given an integer k.

An integer x is called k-reducible if performing the following operation at most k times reduces it to 1:

  • Replace x with the count of set bits in its binary representation.

For example, the binary representation of 6 is "110". Applying the operation once reduces it to 2 (since "110" has two set bits). Applying the operation again to 2 (binary "10") reduces it to 1 (since "10" has one set bit).

Return an integer denoting the number of positive integers less than n that are k-reducible.

Since the answer may be too large, return it modulo 109 + 7.

Example 1:

Input: s = "111", k = 1

Output: 3

Explanation:

n = 7. The 1-reducible integers less than 7 are 1, 2, and 4.

Example 2:

Input: s = "1000", k = 2

Output: 6

Explanation:

n = 8. The 2-reducible integers less than 8 are 1, 2, 3, 4, 5, and 6.

Example 3:

Input: s = "1", k = 3

Output: 0

Explanation:

There are no positive integers less than n = 1, so the answer is 0.

Constraints:

  • 1 <= s.length <= 800
  • s has no leading zeros.
  • s consists only of the characters '0' and '1'.
  • 1 <= k <= 5
Patterns Used
📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a binary string s representing a number n in its binary form. You are also given an integer k. An integer x is called k-reducible if performing the following operation at most k times reduces it to 1: Replace x with the count of set bits in its binary representation. For example, the binary representation of 6 is "110". Applying the operation once reduces it to 2 (since "110" has two set bits). Applying the operation again to 2 (binary "10") reduces it to 1 (since "10" has one set bit). Return an integer denoting the number of positive integers less than n that are k-reducible. Since the answer may be too large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming

Example 1

"111"
1

Example 2

"1000"
2

Example 3

"1"
3
Step 02

Core Insight

What unlocks the optimal approach

  • You can precompute number of operations required to convert a number with <code>x</code> bits to 1.
  • Use digit dp.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3352: Count K-Reducible Numbers Less Than N
class Solution {
  public int countKReducibleNumbers(String s, int k) {
    Integer[][][] mem = new Integer[s.length()][s.length() + 1][2];
    return count(s, 0, 0, true, k, getOps(s), mem) - 1; // - 0
  }

  private static final int MOD = 1_000_000_007;

  // Returns the number of positive integers less than n that are k-reducible,
  // considering the i-th digit, where `setBits` is the number of set bits in
  // the current number, and `tight` indicates if the current digit is
  // tightly bound.
  private int count(String s, int i, int setBits, boolean tight, int k, int[] ops,
                    Integer[][][] mem) {
    if (i == s.length())
      return (ops[setBits] < k && !tight) ? 1 : 0;
    if (mem[i][setBits][tight ? 1 : 0] != null)
      return mem[i][setBits][tight ? 1 : 0];

    int res = 0;
    final int maxDigit = tight ? s.charAt(i) - '0' : 1;

    for (int d = 0; d <= maxDigit; ++d) {
      final boolean nextTight = tight && (d == maxDigit);
      res += count(s, i + 1, setBits + d, nextTight, k, ops, mem);
      res %= MOD;
    }

    return mem[i][setBits][tight ? 1 : 0] = res;
  }

  // Returns the number of operations to reduce a number to 0.
  private int[] getOps(String s) {
    int[] ops = new int[s.length() + 1];
    for (int num = 2; num <= s.length(); ++num)
      ops[num] = 1 + ops[Integer.bitCount(num)];
    return ops;
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Related Problems
#62Unique Pathsmedium→#70Climbing Stairseasy→#96Unique Binary Search Treesmedium→#233Number of Digit Onehard→#241Different Ways to Add Parenthesesmedium→