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Problem solution · Java

Count K-Subsequences of a String With Maximum Beauty Solved

Count K-Subsequences of a String With Maximum Beauty Solved: 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
55 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 Count K-Subsequences of a String With Maximum Beauty Solved, 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

  • 55 lines of Java from the credited upstream file 2842.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 5 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 codeCount K-Subsequences of a String With Maximum Beauty Solved · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int countKSubsequencesWithMaxBeauty(String s, int k) {    Map<Character, Integer> count = new HashMap<>();    for (final char c : s.toCharArray())      count.merge(c, 1, Integer::sum);    if (count.size() < k)      return 0;     long ans = 1;     for (Pair<Integer, Integer> pair : getFreqCountPairs(count)) {      final int fc = pair.getKey();      final int numOfChars = pair.getValue();      if (numOfChars >= k) {        ans *= nCk(numOfChars, k) * modPow(fc, k);        return (int) (ans % MOD);      }      ans *= modPow(fc, numOfChars);      ans %= MOD;      k -= numOfChars;    }     return (int) ans;  }   private static final int MOD = 1_000_000_007;   private List<Pair<Integer, Integer>> getFreqCountPairs(Map<Character, Integer> count) {    // freqCount := (f(c), # of chars with f(c))    Map<Integer, Integer> freqCount = new HashMap<>();    for (final int value : count.values())      freqCount.merge(value, 1, Integer::sum);    List<Pair<Integer, Integer>> freqCountPairs = new ArrayList<>();    for (Map.Entry<Integer, Integer> entry : freqCount.entrySet())      freqCountPairs.add(new Pair<>(entry.getKey(), entry.getValue()));    freqCountPairs.sort((a, b) -> b.getKey().compareTo(a.getKey()));    return freqCountPairs;  }   private long nCk(int n, int k) {    long res = 1;    for (int i = 1; i <= k; ++i)      res = res * (n - i + 1) / i;    return res;  }   private int modPow(long x, long n) {    if (n == 0)      return 1;    if (n % 2 == 1)      return (int) (x * modPow(x % MOD, (n - 1)) % MOD);    return modPow(x * x % MOD, (n / 2)) % MOD;  }}

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