Problem Description

Alice is trying to type a string on her computer, but she's clumsy and sometimes holds a key down too long, causing a character to be typed multiple times in a row.

You're given:

• word: the final string that appears on Alice's screen (containing the repeated characters)
• k: a positive integer representing the minimum length of the original string Alice intended to type

Your task is to find how many different original strings Alice could have been trying to type, given that:

• The original string must have at least k characters
• Any character in word could be the result of Alice pressing that key one or more times
• Consecutive identical characters in word could have come from Alice intending to type that character any number of times (from 1 up to the total count of consecutive occurrences)

For example, if word = "aaa", Alice could have intended to type:

• "a" (pressed 'a' once but held it too long, resulting in "aaa")
• "aa" (pressed 'a' twice, with at least one held too long)
• "aaa" (pressed 'a' three times)

The answer should be returned modulo 10^9 + 7 since it can be very large.

The problem essentially asks you to count all possible ways to "compress" the given string by reducing consecutive character groups, while ensuring the resulting string has at least k characters.

Intuition

The key insight is to recognize that this is a counting problem with constraints. Let's think about it step by step.

First, observe that consecutive identical characters form groups. For example, "aaabbcc" has three groups: "aaa", "bb", and "cc". For each group of size n, Alice could have intended to type anywhere from 1 to n characters. So a group of 3 identical characters gives us 3 choices.

Without any length constraint, the total number of possibilities would simply be the product of choices for each group. If we have groups of sizes 3, 2, and 2, we'd have 3 × 2 × 2 = 12 total possibilities.

But here's the twist: we need the original string to have at least k characters. This constraint makes things tricky. We can reformulate this as:

• Total possibilities = (All possibilities) - (Possibilities with length < k)

Let's denote:

• a = total number of ways without length restriction
• b = number of ways with length less than k
• Answer = a - b

For calculating a, it's straightforward - multiply the number of choices for each group.

For calculating b, we need dynamic programming. Why? Because we need to count all combinations where the total selected characters is less than k. This is a classic DP problem where we need to track:

• How many groups we've processed
• How many total characters we've selected so far

The DP state f[i][j] represents the number of ways to select exactly j characters from the first i groups. For each group with x extra selectable characters (group size - 1, since we must select at least 1), we can choose anywhere from 0 to min(x, j) additional characters.

The prefix sum optimization comes from the observation that when transitioning from group i-1 to group i, we're essentially summing over a range of previous states. Instead of iterating through each possibility, we can use prefix sums to calculate this range sum in constant time.

Learn more about Dynamic Programming and Prefix Sum patterns.

Solution Approach

Let's walk through the implementation step by step:

Step 1: Group Consecutive Characters

First, we iterate through word and identify groups of consecutive identical characters. For each group:

• Count its size
• If the group has more than 1 character, it means we have choices (we can select 1 to size characters)
• Store size - 1 in the nums array (representing extra selectable characters beyond the mandatory 1)
• Multiply ans by the group size to accumulate total possibilities without restriction
• Decrement k by 1 for each group (since we must select at least 1 character per group)

for i, c in enumerate(word):
    cur += 1
    if i == len(word) - 1 or c != word[i + 1]:
        if cur > 1:
            if k > 0:
                nums.append(cur - 1)
            ans = ans * cur % mod
        cur = 0
        k -= 1

Step 2: Check Early Termination

If k < 1 after selecting one character from each group, it means we already satisfy the minimum length requirement. Return ans directly.

if k < 1:
    return ans

Step 3: Dynamic Programming Setup

Create a 2D DP table f[m+1][k] where:

• m is the number of groups with extra selectable characters
• f[i][j] = number of ways to select exactly j characters from the first i groups
• Initialize f[0][0] = 1 (one way to select 0 characters from 0 groups)

m = len(nums)
f = [[0] * k for _ in range(m + 1)]
f[0][0] = 1

Step 4: DP Transition with Prefix Sum Optimization

For each group i with x extra selectable characters:

• Build a prefix sum array s from the previous row f[i-1]
• For each target sum j, we can select l ∈ [0, min(x, j)] characters from this group
• This means f[i][j] = sum of f[i-1][j-l] for all valid l
• Using prefix sums: f[i][j] = s[j+1] - s[j-min(x,j)]

for i, x in enumerate(nums, 1):
    s = list(accumulate(f[i - 1], initial=0))
    for j in range(k):
        f[i][j] = (s[j + 1] - s[j - min(x, j)] + mod) % mod

Step 5: Calculate Final Answer

The number of ways with length less than k is the sum of f[m][j] for all j < k. Subtract this from the total possibilities:

return (ans - sum(f[m][j] for j in range(k))) % mod

The time complexity is O(n + m × k) where n is the length of word and m is the number of groups. The space complexity is O(m × k) for the DP table.

Example Walkthrough

Let's walk through a concrete example with word = "aaabbc" and k = 3.

Step 1: Identify Groups

• Group 1: "aaa" (size 3)
• Group 2: "bb" (size 2)
• Group 3: "c" (size 1)

Step 2: Calculate Total Possibilities Without Restriction

• Group "aaa": can type 1, 2, or 3 'a's → 3 choices
• Group "bb": can type 1 or 2 'b's → 2 choices
• Group "c": can only type 1 'c' → 1 choice
• Total: 3 × 2 × 1 = 6 possibilities

These 6 possibilities are:

1. "abc" (1a + 1b + 1c)
2. "abbc" (1a + 2b + 1c)
3. "aabc" (2a + 1b + 1c)
4. "aabbc" (2a + 2b + 1c)
5. "aaabc" (3a + 1b + 1c)
6. "aaabbc" (3a + 2b + 1c)

Step 3: Adjust k and Build nums Array

• We must select at least 1 character from each group
• After mandatory selections: k = 3 - 3 = 0
• Since k = 0, we need exactly the minimum (1 from each group)
• nums = [2, 1] (extra characters we can select: 3-1=2 for "aaa", 2-1=1 for "bb")

Step 4: DP to Count Strings with Length < 3 We need to count combinations with total length less than 3 (i.e., length 0, 1, or 2 after mandatory selections).

DP table f[i][j] = ways to select exactly j extra characters from first i groups:

Initial: f[0][0] = 1

Processing Group 1 (can select 0-2 extra):

• f[1][0] = 1 (select 0 extra from "aaa")
• f[1][1] = 1 (select 1 extra from "aaa")
• f[1][2] = 1 (select 2 extra from "aaa")

Processing Group 2 (can select 0-1 extra):

• f[2][0] = f[1][0] = 1 (select 0 from both)
• f[2][1] = f[1][1] + f[1][0] = 2 (select 1 total: either 1 from first or 1 from second)
• f[2][2] = f[1][2] + f[1][1] = 2 (select 2 total: either 2 from first, or 1 from each)

Since k was adjusted to 0, we're looking for combinations with negative total length, which is 0.

Step 5: Final Answer

• Total possibilities: 6
• Possibilities with length < 3: 0
• Answer: 6 - 0 = 6

All 6 strings have at least 3 characters, so the answer is 6.

Let's try another example with word = "aaaa" and k = 2:

Groups: Just one group "aaaa" (size 4) Total possibilities: 4 (can type 1, 2, 3, or 4 'a's) After mandatory selection: k = 2 - 1 = 1 nums = [3] (can select 0-3 extra characters)

DP to count strings with length < 2:

• f[0][0] = 1
• f[1][0] = 1 (select 0 extra → string "a" has length 1)
• f[1][1] = 1 (select 1 extra → string "aa" has length 2)

Strings with length < 2: f[1][0] = 1 (just "a")

Answer: 4 - 1 = 3

The 3 valid strings are: "aa", "aaa", "aaaa" (all have length ≥ 2)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + m*k^2) where n is the length of the string word, m is the number of consecutive character groups (size of nums array), and k is the input parameter.

• The first loop iterates through the string once: O(n)
• The nested loops for dynamic programming iterate m times (for each element in nums), and for each iteration:
  • Computing prefix sum using accumulate: O(k)
  • Inner loop runs k times, and each iteration involves constant time operations with prefix sum lookup: O(k)
  • Total for this part: O(m*k)
• Final sum computation: O(k)
• Overall: O(n + m*k), and since m ≤ n, this simplifies to O(n + m*k)

However, the reference answer states O(n + k^2). This is correct when considering that m ≤ k in the worst case that matters (when we need to compute the DP table). If m > k, the function returns early with just ans, making the complexity O(n). When m ≤ k, the DP computation dominates with O(m*k) ≤ O(k^2).

Space Complexity: O(m*k) which can be simplified to O(k^2) under the same reasoning.

• The nums array stores at most m elements: O(m)
• The DP table f has dimensions (m+1) × k: O(m*k)
• The prefix sum array s has size k+1: O(k)
• Total: O(m*k), which becomes O(k^2) when m ≤ k

The reference answer's O(k^2) space complexity is correct considering that when the DP table is actually computed (i.e., when k is not reduced to below 1), we have m ≤ k, making the space complexity O(k^2).

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

Common Pitfalls

1. Incorrect Modulo Operation During Subtraction

Pitfall: When calculating the final answer by subtracting invalid combinations from total combinations, a common mistake is:

# WRONG - can produce negative results
return (total_combinations - invalid_combinations) % MOD

If invalid_combinations > total_combinations (which shouldn't happen logically but can occur due to modulo operations during intermediate calculations), this produces a negative number, leading to incorrect results in languages like Python where % preserves sign.

Solution: Always add MOD before taking modulo when subtracting:

# CORRECT - ensures positive result
return (total_combinations - invalid_combinations % MOD + MOD) % MOD

2. Off-by-One Error in k Adjustment

Pitfall: Forgetting to decrement k for each group or decrementing it incorrectly:

# WRONG - only decrements k when group size > 1
if current_group_size > 1:
    k -= 1

Every group (even single characters) contributes at least 1 character to the minimum length, so k should be decremented for each group regardless of size.

Solution: Decrement k outside the conditional check:

# CORRECT - decrements for every group
if i == len(word) - 1 or char != word[i + 1]:
    if current_group_size > 1:
        # handle extra characters
    k -= 1  # This should be outside the if statement

3. Overflow in Prefix Sum Range Calculation

Pitfall: When calculating the range sum using prefix sums, accessing negative indices:

# WRONG - can access negative index when j < min(max_extra, j)
dp[i][j] = prefix_sum[j + 1] - prefix_sum[j - min(max_extra, j)]

Solution: Ensure the index calculation is correct:

# CORRECT - max(0, ...) ensures non-negative index
start_idx = max(0, j - min(max_extra, j))
dp[i][j] = (prefix_sum[j + 1] - prefix_sum[start_idx] + MOD) % MOD

4. Not Handling Edge Case When No Groups Have Extra Characters

Pitfall: If all groups have exactly 1 character and k is still positive after processing, the DP won't execute (empty extra_chars_per_group), but we still need to return 0 since we can't meet the minimum length requirement.

Solution: Check if it's possible to meet the requirement:

if k > 0 and len(extra_chars_per_group) == 0:
    return 0  # Impossible to reach k characters

5. Integer Overflow in Total Combinations Calculation

Pitfall: Forgetting to apply modulo during multiplication:

# WRONG - can overflow in languages with fixed integer sizes
total_combinations = total_combinations * current_group_size

Solution: Apply modulo after each multiplication:

# CORRECT - prevents overflow
total_combinations = (total_combinations * current_group_size) % MOD