Problem Description

You are given a string croakOfFrogs that represents sounds made by multiple frogs croaking simultaneously. Each frog must make the complete sound "croak" by producing the letters 'c', 'r', 'o', 'a', and 'k' in that exact sequential order.

The key points of this problem are:

1. Multiple frogs can croak at the same time - This means the string might contain interleaved croaks from different frogs. For example, "ccroak" could represent two frogs where one starts croaking ('c') before the other completes.

2. Each frog must complete the entire sequence - A valid croak requires all five letters 'c', 'r', 'o', 'a', 'k' to appear in order for that specific frog. A frog cannot skip letters or produce them out of order.

3. Find the minimum number of frogs needed - You need to determine the smallest number of different frogs that could have produced the given string of sounds.

4. Return -1 for invalid strings - If the given string cannot be formed by any combination of valid croaks, return -1.

For example:

• "croakcroak" requires only 1 frog (one frog croaks twice sequentially)
• "crcoakroak" requires 2 frogs (the second frog starts with 'c' before the first frog finishes)
• "croakcrook" is invalid and should return -1 (the second sequence has 'o' before 'a')

The challenge is to track how many frogs are croaking concurrently at any point and determine if the given string represents a valid combination of complete croaks.

Intuition

The key insight is to think of this problem as tracking the state of multiple frogs as they progress through their croaks. Each frog must go through the sequence 'c' -> 'r' -> 'o' -> 'a' -> 'k' in order.

We can visualize this as frogs moving through different stages:

• Stage 0: Waiting to start (hasn't made any sound yet)
• Stage 1: Said 'c', waiting to say 'r'
• Stage 2: Said 'cr', waiting to say 'o'
• Stage 3: Said 'cro', waiting to say 'a'
• Stage 4: Said 'croa', waiting to say 'k'
• Back to Stage 0: Completed 'croak', can start again

When we encounter a letter in the string, we need to find a frog that's ready to make that sound. For example:

• If we see 'c', we need a free frog (at stage 0) to start croaking
• If we see 'r', we need a frog that has already said 'c' (at stage 1)
• If we see 'o', we need a frog that has already said 'cr' (at stage 2)
• And so on...

This leads us to track how many frogs are at each stage. We can use an array cnt where cnt[i] represents the number of frogs waiting to say the letter at index i in "croak".

The number of active frogs (those currently in the middle of croaking) tells us how many frogs we need at minimum. When a frog starts croaking (says 'c'), we increase the active count. When a frog completes (says 'k'), we decrease the active count. The maximum number of active frogs at any point is our answer.

The validation happens naturally: if we encounter a letter but no frog is ready to say it (the previous stage count is 0), then the string is invalid. Also, if the string length isn't divisible by 5, or if some frogs remain incomplete at the end, the string is invalid.

Solution Approach

First, we validate that the string length is a multiple of 5. Since each complete croak has exactly 5 letters, any valid combination must have a length divisible by 5. If not, we immediately return -1.

We create a mapping from each letter in "croak" to its index position:

• 'c' → 0
• 'r' → 1
• 'o' → 2
• 'a' → 3
• 'k' → 4

This is done using: idx = {c: i for i, c in enumerate('croak')}

We maintain an array cnt of size 5 where cnt[i] tracks how many frogs are waiting to say the letter at index i+1 in the sequence. Initially, all values are 0.

We also track:

• x: The number of frogs currently in the middle of croaking (active frogs)
• ans: The maximum number of active frogs seen so far (this will be our answer)

For each character in croakOfFrogs, we:

1. Get its index i using our mapping

2. Increment cnt[i] by 1 (one more frog has reached this stage)

3. Handle based on the value of i:

   If i = 0 (letter is 'c'):

   • A new frog starts croaking
   • Increment x by 1 (one more active frog)
   • Update ans = max(ans, x) to track the peak number of frogs

   If i > 0 (letter is 'r', 'o', 'a', or 'k'):

   • Check if cnt[i-1] > 0 (is there a frog at the previous stage?)
   • If no frog is available (cnt[i-1] = 0), return -1 as it's invalid
   • Otherwise, decrement cnt[i-1] by 1 (move that frog forward)
   • If i = 4 (letter is 'k'), the frog completes its croak:
     • Decrement x by 1 (one less active frog)

After processing all characters, we check if x = 0. If all frogs have completed their croaks, x should be 0, and we return ans. Otherwise, some frogs are incomplete, so we return -1.

The algorithm runs in O(n) time where n is the length of the string, and uses O(1) space for the fixed-size array and variables.

Example Walkthrough

Let's walk through the string "ccroak" to illustrate the solution approach.

Initial Setup:

• String length: 6 (not divisible by 5, but let's continue for illustration)
• idx = {'c': 0, 'r': 1, 'o': 2, 'a': 3, 'k': 4}
• cnt = [0, 0, 0, 0, 0] (tracks frogs at each stage)
• x = 0 (active frogs)
• ans = 0 (maximum frogs needed)

Processing each character:

Character 1: 'c'

• i = 0 (index of 'c' in "croak")
• cnt[0]++ → cnt = [1, 0, 0, 0, 0] (one frog at stage 'c')
• Since i = 0, a new frog starts croaking
• x = 1 (one active frog)
• ans = max(0, 1) = 1

Character 2: 'c'

• i = 0
• cnt[0]++ → cnt = [2, 0, 0, 0, 0] (two frogs at stage 'c')
• Since i = 0, another frog starts croaking
• x = 2 (two active frogs)
• ans = max(1, 2) = 2

Character 3: 'r'

• i = 1 (index of 'r' in "croak")
• cnt[1]++ → cnt = [2, 1, 0, 0, 0]
• Since i > 0, check if cnt[0] > 0 (is there a frog ready to say 'r'?)
• Yes! cnt[0] = 2 > 0
• cnt[0]-- → cnt = [1, 1, 0, 0, 0] (move one frog from 'c' to 'r')

Character 4: 'o'

• i = 2
• cnt[2]++ → cnt = [1, 1, 1, 0, 0]
• Check if cnt[1] > 0 (is there a frog ready to say 'o'?)
• Yes! cnt[1] = 1 > 0
• cnt[1]-- → cnt = [1, 0, 1, 0, 0] (move one frog from 'r' to 'o')

Character 5: 'a'

• i = 3
• cnt[3]++ → cnt = [1, 0, 1, 1, 0]
• Check if cnt[2] > 0 (is there a frog ready to say 'a'?)
• Yes! cnt[2] = 1 > 0
• cnt[2]-- → cnt = [1, 0, 0, 1, 0] (move one frog from 'o' to 'a')

Character 6: 'k'

• i = 4
• cnt[4]++ → cnt = [1, 0, 0, 1, 1]
• Check if cnt[3] > 0 (is there a frog ready to say 'k'?)
• Yes! cnt[3] = 1 > 0
• cnt[3]-- → cnt = [1, 0, 0, 0, 1] (move one frog from 'a' to 'k')
• Since i = 4, this frog completes its croak
• x = 1 (one active frog remains)

Final Check:

• x = 1 ≠ 0 (one frog hasn't completed its croak)
• Return -1 (invalid string)

The algorithm correctly identifies that "ccroak" is invalid because one frog started with 'c' but never completed its croak. We needed 2 frogs at the peak (when both had started), but the string is invalid as one frog remains incomplete.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string croakOfFrogs. The algorithm iterates through each character in the input string exactly once using the for loop. Within each iteration, all operations (dictionary lookup with idx.get, array indexing, arithmetic operations, and comparisons) are performed in constant time O(1).

The space complexity is O(1) or O(C) where C represents the size of the character set used. In this implementation:

• The idx dictionary stores a mapping for 5 characters ('c', 'r', 'o', 'a', 'k'), requiring O(5) = O(1) space
• The cnt array has a fixed size of 5 elements, requiring O(5) = O(1) space
• Additional variables ans and x use O(1) space

Since the space usage doesn't depend on the input size n but only on the fixed character set size (which is 5 for "croak"), the space complexity is constant. The reference answer mentions C=26 considering the full lowercase alphabet set, but the actual implementation only uses 5 characters, making the practical space complexity O(1).

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

Common Pitfalls

1. Not Properly Tracking Frog Transitions Between Stages

A common mistake is incorrectly managing how frogs move through the croaking sequence. When processing a character at position i > 0, developers often forget to decrement the count at position i-1 after incrementing at position i. This leads to double-counting frogs and incorrect results.

Incorrect approach:

# Wrong: Only incrementing without decrementing the previous stage
stage_count[position] += 1
if position > 0 and stage_count[position - 1] == 0:
    return -1
# Missing: stage_count[position - 1] -= 1

Solution: Always remember that when a frog progresses from one stage to the next, it leaves the previous stage. The correct implementation should both increment the current stage AND decrement the previous stage count.

2. Misunderstanding When to Update Active Frog Count

Another pitfall is updating active_frogs at the wrong positions. Some implementations incorrectly increment active_frogs for every character or decrement it at positions other than 'k'.

Incorrect approach:

# Wrong: Incrementing active_frogs for every character
for char in croakOfFrogs:
    position = char_to_index[char]
    active_frogs += 1  # Wrong!
    stage_count[position] += 1

Solution:

• Only increment active_frogs when position == 0 (character 'c' - frog starts)
• Only decrement active_frogs when position == 4 (character 'k' - frog completes)

3. Forgetting to Track Maximum Concurrent Frogs

Some solutions correctly track active frogs but forget to maintain the maximum value throughout the iteration, returning active_frogs instead of the peak value.

Incorrect approach:

# Wrong: Returning current active_frogs instead of maximum
return active_frogs  # This will always be 0 for valid inputs!

Solution: Always update max_frogs = max(max_frogs, active_frogs) whenever a new frog starts croaking (at position 0), and return max_frogs at the end.

4. Not Handling the Stage Count Array Correctly

The stage_count array represents frogs waiting at each stage, but it's easy to misinterpret what each index means. Remember that stage_count[i] represents frogs that have just completed the character at position i and are ready for position i+1.

Key insight: When processing 'c' (position 0), we don't need to check stage_count[-1] because 'c' starts a new croak. The frog at stage_count[0] is ready to say 'r', not 'c'.