CPMagical CP Library

Problem solution · Java

Numbers With Repeated Digits

Numbers With Repeated Digits: 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 Numbers With Repeated Digits, 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 1012.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 codeNumbers With Repeated Digits · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int numDupDigitsAtMostN(int n) {    return n - countSpecialNumbers(n);  }   // Same as 2376. Count Special Integers  private int countSpecialNumbers(int n) {    final int digitSize = (int) Math.log10(n) + 1;    Integer[][][] mem = new Integer[digitSize + 1][1 << 10][2];    return count(String.valueOf(n), 0, 0, true, mem) - 1; // - 0;  }   // Returns the number of special integers, considering the i-th digit, where  // `used` is the bitmask of the used digits, and `tight` indicates if the  // current digit is tightly bound.  private int count(final String s, int i, int used, boolean tight, Integer[][][] mem) {    if (i == s.length())      return 1;    if (mem[i][used][tight ? 1 : 0] != null)      return mem[i][used][tight ? 1 : 0];     int res = 0;    final int maxDigit = tight ? s.charAt(i) - '0' : 9;     for (int d = 0; d <= maxDigit; ++d) {      // `d` is used.      if ((used >> d & 1) == 1)        continue;      // Use `d` now.      final boolean nextTight = tight && (d == maxDigit);      if (used == 0 && d == 0) // Don't count leading 0s as used.        res += count(s, i + 1, used, nextTight, mem);      else        res += count(s, i + 1, used | 1 << d, nextTight, mem);    }     return mem[i][used][tight ? 1 : 0] = res;  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗