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Problem solution · Java

Count K-Reducible Numbers Less Than N

Count K-Reducible Numbers Less Than N: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
40 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Count K-Reducible Numbers Less Than N, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 40 lines of Java from the credited upstream file 3352.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount K-Reducible Numbers Less Than N · JavaJava
Use this to learn the idea, then write your own version.
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;  }}

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