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Problem solution · Java

Find the Count of Good Integers

Find the Count of Good Integers: a Java solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
45 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

Sorting and greedy selection

For Find the Count of Good Integers, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 45 lines of Java from the credited upstream file 3272.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 4 loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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 codeFind the Count of Good Integers · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long countGoodIntegers(int n, int k) {    final int halfLength = (n + 1) / 2;    final int minHalf = (int) Math.pow(10, halfLength - 1);    final int maxHalf = (int) Math.pow(10, halfLength);    long ans = 0;    Set<String> seen = new HashSet<>();     for (int num = minHalf; num < maxHalf; ++num) {      final String firstHalf = String.valueOf(num);      final String secondHalf = new StringBuilder(firstHalf).reverse().toString();      final String palindrome = firstHalf + secondHalf.substring(n % 2);      if (Long.parseLong(palindrome) % k != 0)        continue;      char[] sortedDigits = palindrome.toCharArray();      Arrays.sort(sortedDigits);      String sortedDigitsStr = new String(sortedDigits);      if (seen.contains(sortedDigitsStr))        continue;      seen.add(sortedDigitsStr);      int[] digitCount = new int[10];      for (char c : palindrome.toCharArray())        ++digitCount[c - '0'];      // Leading zeros are not allowed, so the first digit is special.      final int firstDigitChoices = n - digitCount[0];      long permutations = firstDigitChoices * factorial(n - 1);      // For each repeated digit, divide by the factorial of the frequency since      // permutations that swap identical digits don't create a new number.      for (final int freq : digitCount)        if (freq > 1)          permutations /= factorial(freq);      ans += permutations;    }     return ans;  }   private long factorial(int n) {    long res = 1;    for (int i = 2; i <= n; ++i)      res *= i;    return res;  }}

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