How do you solve Number of Beautiful Partitions on LeetCode?

Number of Beautiful Partitions (LeetCode #2478) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(n * k) space complexity. Key insight: Understanding prime and non-prime digits is crucial.

What is the brute force solution for Number of Beautiful Partitions?

The brute force approach for Number of Beautiful Partitions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to partition the string into k substrings. It checks each substring to see if it meets the beautiful criteria, which can be very inefficient. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Number of Beautiful Partitions?

Number of Beautiful Partitions uses the following concepts: String, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Number of Beautiful Partitions?

Explain your thought process clearly. Discuss edge cases and constraints. Optimize your solution step by step.

What are common mistakes when solving Number of Beautiful Partitions?

Forgetting to check the length of substrings. Not using modulo operation for large numbers.

#2478

Number of Beautiful Partitions

Hard

StringDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k)
Space	O(1)	O(n * k)

💡

Intuition

Time O(n * k)Space O(n * k)

The optimal approach uses dynamic programming to efficiently count valid partitions. Instead of checking all combinations, we build up solutions using previously computed results.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i] represents the number of ways to partition the string up to index i.
2Step 2: Iterate through the string and for each position, check if it can form a valid substring ending at that position.
3Step 3: Update the DP array based on previous valid partitions.

solution.py19 lines

1# Full working Python code
2MOD = 10**9 + 7
3primes = {'2', '3', '5', '7'}
4non_primes = {'1', '4', '6', '8', '9'}
5
6def beautiful_partitions(s, k, minLength):
7    n = len(s)
8    dp = [[0] * (k + 1) for _ in range(n)]
9    for i in range(n):
10        if i + 1 >= minLength and s[i] in non_primes:
11            for j in range(1, k + 1):
12                if j == 1:
13                    if s[0] in primes:
14                        dp[i][j] = 1
15                else:
16                    for l in range(minLength, i + 2):
17                        if s[l - 1] in non_primes and s[l - minLength] in primes:
18                            dp[i][j] = (dp[i][j] + dp[l - 2][j - 1]) % MOD
19    return dp[n - 1][k]

ℹ

Complexity note: The complexity is linear with respect to the length of the string and the number of partitions due to the nested loops iterating through the string and partitions.

1Understanding prime and non-prime digits is crucial.
2Dynamic programming can significantly reduce the number of checks needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.