Problem Description

You need to find how many strings of length n can be rearranged to contain "leet" as a substring.

A string is considered "good" if:

1. It contains only lowercase English characters (a-z)
2. Its characters can be rearranged to form a string that has "leet" as a substring

For a string to be rearrangeable into containing "leet", it must have:

• At least 1 letter 'l'
• At least 2 letters 'e'
• At least 1 letter 't'

The problem asks you to count all possible strings of length n that meet these criteria. Since we're dealing with arrangements, the order of characters in the original string doesn't matter - what matters is having the minimum required frequency of each character.

For example:

• "lteer" is good because it has 1 'l', 2 'e's, 1 't', and 1 'r'. These can be rearranged to "leetr" which contains "leet"
• "letl" is not good because it only has 1 'e', but we need at least 2 'e's to form "leet"

The solution uses dynamic programming with memoization to track:

• i: remaining positions to fill in the string
• l: whether we have at least 1 'l' (capped at 1)
• e: how many 'e's we have (capped at 2, since we need exactly 2)
• t: whether we have at least 1 't' (capped at 1)

At each position, we can either:

• Add one of 23 other lowercase letters (not l, e, or t)
• Add an 'l' and update our 'l' count
• Add an 'e' and update our 'e' count
• Add a 't' and update our 't' count

The answer should be returned modulo 10^9 + 7 to handle large numbers.

Intuition

The key insight is that we don't care about the exact arrangement of characters in our string - we only care about having enough of each required character to form "leet" somewhere in a rearrangement.

Think about it this way: if we have a string with at least 1 'l', 2 'e's, and 1 't', we can always rearrange it to place "leet" together, then put any remaining characters before or after. So the problem reduces to counting strings with the right character frequencies.

We can build the string character by character. At each position, we have 26 choices (any lowercase letter). As we build, we need to track whether we've satisfied our requirements:

• Have we added at least 1 'l'?
• Have we added at least 2 'e's?
• Have we added at least 1 't'?

This naturally leads to a recursive solution where our state is (remaining_positions, l_count, e_count, t_count). But tracking exact counts would be inefficient - we don't care if we have 3 'e's or 10 'e's, as long as we have at least 2. So we cap our tracking:

• l is either 0 (need more) or 1 (have enough)
• e can be 0, 1, or 2 (need exactly 2)
• t is either 0 (need more) or 1 (have enough)

At each step with i positions remaining, we can:

1. Add any of the 23 "other" letters (not l, e, or t) - this doesn't change our requirement tracking
2. Add an 'l' - this might change l from 0 to 1
3. Add an 'e' - this might increment e up to 2
4. Add a 't' - this might change t from 0 to 1

The base case is when i = 0 (string is complete). We return 1 if we've met all requirements (l = 1, e = 2, t = 1), otherwise 0.

Since we'll revisit the same states multiple times (e.g., reaching state (5, 1, 2, 0) through different paths), we use memoization to avoid redundant calculations.

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

The solution implements a top-down dynamic programming approach with memoization using Python's @cache decorator.

State Definition:

• dfs(i, l, e, t) represents the number of valid strings when:
  • i: remaining positions to fill
  • l: count of 'l' characters (capped at 1)
  • e: count of 'e' characters (capped at 2)
  • t: count of 't' characters (capped at 1)

Base Case: When i = 0, the string is complete. We check if we have exactly the required characters:

if i == 0:
    return int(l == 1 and e == 2 and t == 1)

This returns 1 if all requirements are met, 0 otherwise.

Recursive Transitions: For each remaining position, we have 4 choices:

1. Add one of 23 other characters (not 'l', 'e', or 't'):

   a = dfs(i - 1, l, e, t) * 23 % mod

   Since there are 26 lowercase letters total and we exclude 'l', 'e', 't', we have 23 choices. The counts remain unchanged.

2. Add character 'l':

   b = dfs(i - 1, min(1, l + 1), e, t)

   We use min(1, l + 1) to cap the count at 1 since we only need to track whether we have at least one 'l'.

3. Add character 'e':

   c = dfs(i - 1, l, min(2, e + 1), t)

   We use min(2, e + 1) to cap at 2 since we need exactly 2 'e's for "leet".

4. Add character 't':

   d = dfs(i - 1, l, e, min(1, t + 1))

   Similar to 'l', we cap at 1.

Final Calculation: The total number of ways is the sum of all four choices:

return (a + b + c + d) % mod

Memoization: The @cache decorator automatically memoizes the function results, preventing redundant calculations for the same state. This is crucial since many different paths can lead to the same state.

Time Complexity: O(n × 2 × 3 × 2) = O(n) since we have at most n values for i, 2 for l, 3 for e, and 2 for t.

Space Complexity: O(n) for the memoization cache and recursion stack.

The answer is obtained by calling dfs(n, 0, 0, 0), starting with an empty string (0 of each required character) and n positions to fill.

Example Walkthrough

Let's walk through a small example with n = 5 to illustrate how the solution works.

We start with dfs(5, 0, 0, 0) - we have 5 positions to fill and haven't added any required characters yet.

Step 1: From state (5, 0, 0, 0) We have 4 choices for the first character:

• Add one of 23 other letters → leads to dfs(4, 0, 0, 0) × 23
• Add 'l' → leads to dfs(4, 1, 0, 0)
• Add 'e' → leads to dfs(4, 0, 1, 0)
• Add 't' → leads to dfs(4, 0, 0, 1)

Let's follow one path: adding 'l' first.

Step 2: From state (4, 1, 0, 0) Now we have 'l' covered (l=1). Our choices:

• Add one of 23 other letters → dfs(3, 1, 0, 0) × 23
• Add another 'l' → dfs(3, 1, 0, 0) (l stays at 1, capped)
• Add 'e' → dfs(3, 1, 1, 0)
• Add 't' → dfs(3, 1, 0, 1)

Let's add 'e' next.

Step 3: From state (3, 1, 1, 0) We have 1 'l' and 1 'e'. Our choices:

• Add one of 23 other letters → dfs(2, 1, 1, 0) × 23
• Add 'l' → dfs(2, 1, 1, 0) (already have enough 'l')
• Add 'e' → dfs(2, 1, 2, 0) (e becomes 2)
• Add 't' → dfs(2, 1, 1, 1)

Let's add another 'e'.

Step 4: From state (2, 1, 2, 0) We have 1 'l' and 2 'e's. We still need a 't'. Our choices:

• Add one of 23 other letters → dfs(1, 1, 2, 0) × 23
• Add 'l' → dfs(1, 1, 2, 0)
• Add 'e' → dfs(1, 1, 2, 0) (e stays at 2, capped)
• Add 't' → dfs(1, 1, 2, 1) ✓ This completes our requirements!

Let's add 't'.

Step 5: From state (1, 1, 2, 1) We have all required characters! With 1 position left:

• Add any of 23 other letters → dfs(0, 1, 2, 1) × 23 = 1 × 23 = 23
• Add 'l' → dfs(0, 1, 2, 1) = 1
• Add 'e' → dfs(0, 1, 2, 1) = 1
• Add 't' → dfs(0, 1, 2, 1) = 1

Total for this state: 23 + 1 + 1 + 1 = 26

Base Case Example: When we reach dfs(0, 1, 2, 1):

• i = 0 (string is complete)
• l = 1 ✓ (have at least 1 'l')
• e = 2 ✓ (have exactly 2 'e's)
• t = 1 ✓ (have at least 1 't')
• Returns 1 (this is a valid string)

If we had reached dfs(0, 1, 1, 1) instead:

• e = 1 ✗ (need 2 'e's but only have 1)
• Returns 0 (invalid string)

The path we traced ('l', 'e', 'e', 't', any character) represents strings like "leetx", "leeta", etc. The algorithm counts all possible paths that lead to valid strings, using memoization to avoid recalculating states that can be reached through different character orderings.

Solution Implementation

Time and Space Complexity

The time complexity is O(n). The recursive function dfs has 4 parameters: i ranges from 0 to n, l can be 0 or 1, e can be 0, 1, or 2, and t can be 0 or 1. This gives us at most n × 2 × 3 × 2 = 12n unique states. Due to memoization with @cache, each state is computed only once. Each state computation involves constant time operations (4 recursive calls and modular arithmetic), so the total time complexity is O(n).

The space complexity is O(n). The memoization cache stores up to 12n states, which is O(n) space. Additionally, the recursion depth is at most n (when i goes from n to 0), contributing O(n) to the call stack space. Therefore, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect State Capping Logic

A critical pitfall is failing to properly cap the character counts or capping them at wrong values. This leads to either counting invalid strings or missing valid ones.

Wrong Implementation:

# Incorrect: Not capping the counts
def dfs(i, l, e, t):
    if i == 0:
        return int(l >= 1 and e >= 2 and t >= 1)  # Wrong check!
    # ...
    add_l = dfs(i - 1, l + 1, e, t)  # No capping!
    add_e = dfs(i - 1, l, e + 1, t)  # No capping!

Why it's wrong: Without capping, we create exponentially more states than necessary. For example, having 5 'e's vs 2 'e's doesn't matter for forming "leet", but without capping, these become different states, destroying memoization efficiency and causing TLE.

Correct Implementation:

def dfs(i, l, e, t):
    if i == 0:
        return int(l == 1 and e == 2 and t == 1)  # Exact check
    # ...
    add_l = dfs(i - 1, min(1, l + 1), e, t)  # Cap at 1
    add_e = dfs(i - 1, l, min(2, e + 1), t)  # Cap at 2
    add_t = dfs(i - 1, l, e, min(1, t + 1))  # Cap at 1

2. Forgetting to Apply Modulo at Each Step

Another common mistake is only applying modulo at the final return, which can cause integer overflow in languages with fixed integer sizes.

Wrong Implementation:

def dfs(i, l, e, t):
    # ...
    other_letters = dfs(i - 1, l, e, t) * 23  # Missing modulo!
    # ...
    return (other_letters + add_l + add_e + add_t) % MOD

Correct Implementation:

def dfs(i, l, e, t):
    # ...
    other_letters = dfs(i - 1, l, e, t) * 23 % MOD  # Apply modulo immediately
    # ...
    return (other_letters + add_l + add_e + add_t) % MOD

3. Misunderstanding the Problem Requirements

A conceptual pitfall is thinking we need to track the exact positions of 'l', 'e', 'e', 't' or that we need the substring "leet" to appear consecutively in our constructed string.

Wrong Understanding:

• Thinking we need to place "leet" as a consecutive substring during construction
• Tracking positions where each character appears
• Checking if the final string contains "leet" as-is

Correct Understanding:

• We only need to ensure the string has enough of each required character
• The characters can be rearranged later to form "leet" as a substring
• We're counting strings that CAN BE rearranged, not strings that already contain "leet"

4. Incorrect Base Case Logic

Using >= instead of == in the base case check is a subtle but critical error.

Wrong Implementation:

if i == 0:
    return int(l >= 1 and e >= 2 and t >= 1)  # Wrong!

Why it's wrong: Due to our capping logic, l can only be 0 or 1, e can only be 0, 1, or 2, and t can only be 0 or 1. Using >= would incorrectly count states where we have fewer than required characters (since 0 >= 1 is False, but so is 0 == 1). The == check ensures we have exactly the capped values that represent "at least" the required amounts.

Correct Implementation:

if i == 0:
    return int(l == 1 and e == 2 and t == 1)  # Correct!