2478. Number of Beautiful Partitions LeetCode Solution

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Problem Statement
Complexity Analysis
Number of Beautiful Partitions solution in C++
Number of Beautiful Partitions solution in Java
Number of Beautiful Partitions solution in Python
Additional Resources

2478. Number of Beautiful Partitions LeetCode Solution image

Problem Statement of Number of Beautiful Partitions

You are given a string s that consists of the digits ‘1’ to ‘9’ and two integers k and minLength.
A partition of s is called beautiful if:

s is partitioned into k non-intersecting substrings.
Each substring has a length of at least minLength.
Each substring starts with a prime digit and ends with a non-prime digit. Prime digits are ‘2’, ‘3’, ‘5’, and ‘7’, and the rest of the digits are non-prime.

Return the number of beautiful partitions of s. Since the answer may be very large, return it modulo 109 + 7.
A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = “23542185131”, k = 3, minLength = 2
Output: 3
Explanation: There exists three ways to create a beautiful partition:
“2354 | 218 | 5131”
“2354 | 21851 | 31”
“2354218 | 51 | 31”

Example 2:

Input: s = “23542185131”, k = 3, minLength = 3
Output: 1
Explanation: There exists one way to create a beautiful partition: “2354 | 218 | 5131”.

Example 3:

Input: s = “3312958”, k = 3, minLength = 1
Output: 1
Explanation: There exists one way to create a beautiful partition: “331 | 29 | 58”.

Constraints:

1 <= k, minLength <= s.length <= 1000
s consists of the digits '1' to '9'.

Complexity Analysis

Time Complexity: O(nk)
Space Complexity: O(nk)

2478. Number of Beautiful Partitions LeetCode Solution in C++

class Solution {
 public:
  int beautifulPartitions(string s, int k, int minLength) {
    if (!isPrime(s.front()) || isPrime(s.back()))
      return 0;
    vector<vector<int>> mem(s.length(), vector<int>(k, -1));
    return beautifulPartitions(s, minLength, k - 1, minLength, mem);
  }

 private:
  static constexpr int kMod = 1'000'000'007;

  // Returns the number of beautiful partitions of s[i..n) with k bars (|) left.
  int beautifulPartitions(const string& s, int i, int k, int minLength,
                          vector<vector<int>>& mem) {
    if (i <= s.length() && k == 0)
      return 1;
    if (i >= s.length())
      return 0;
    if (mem[i][k] != -1)
      return mem[i][k];

    // Don't split between s[i - 1] and s[i].
    int res = beautifulPartitions(s, i + 1, k, minLength, mem) % kMod;

    // Split between s[i - 1] and s[i].
    if (isPrime(s[i]) && !isPrime(s[i - 1]))
      res += beautifulPartitions(s, i + minLength, k - 1, minLength, mem);

    return mem[i][k] = res % kMod;
  }

  bool isPrime(char c) {
    return c == '2' || c == '3' || c == '5' || c == '7';
  }
};
/* code provided by PROGIEZ */

2478. Number of Beautiful Partitions LeetCode Solution in Java

class Solution {
  public int beautifulPartitions(String s, int k, int minLength) {
    if (!isPrime(s.charAt(0)) || isPrime(s.charAt(s.length() - 1)))
      return 0;
    Integer[][] mem = new Integer[s.length()][k];
    return beautifulPartitions(s, minLength, k - 1, minLength, mem);
  }

  private static final int kMod = 1_000_000_007;

  // Returns the number of beautiful partitions of s[i..n) with k bars (|) left.
  private int beautifulPartitions(final String s, int i, int k, int minLength, Integer[][] mem) {
    if (i <= s.length() && k == 0)
      return 1;
    if (i >= s.length())
      return 0;
    if (mem[i][k] != null)
      return mem[i][k];

    // Don't split between s[i - 1] and s[i].
    int ans = beautifulPartitions(s, i + 1, k, minLength, mem) % kMod;

    // Split between s[i - 1] and s[i].
    if (isPrime(s.charAt(i)) && !isPrime(s.charAt(i - 1)))
      ans += beautifulPartitions(s, i + minLength, k - 1, minLength, mem);

    return mem[i][k] = ans % kMod;
  }

  private boolean isPrime(char c) {
    return c == '2' || c == '3' || c == '5' || c == '7';
  }
}
// code provided by PROGIEZ

2478. Number of Beautiful Partitions LeetCode Solution in Python

class Solution:
  def beautifulPartitions(self, s: str, k: int, minLength: int) -> int:
    def isPrime(c: str) -> bool:
      return c in '2357'

    if not isPrime(s[0]) or isPrime(s[-1]):
      return 0

    kMod = 1_000_000_007

    @lru_cache(None)
    def dp(i: int, k: int) -> int:
      """
      Returns the number of beautiful partitions of s[i..n) with k bars (|)
      left.
      """
      if i <= len(s) and k == 0:
        return 1
      if i >= len(s):
        return 0

      # Don't split between s[i - 1] and s[i].
      ans = dp(i + 1, k) % kMod

      # Split between s[i - 1] and s[i].
      if isPrime(s[i]) and not isPrime(s[i - 1]):
        ans += dp(i + minLength, k - 1)

      return ans % kMod

    return dp(minLength, k - 1)
 # code by PROGIEZ