Problem Description

You are given a digit string s consisting of characters. The goal is to check if you can split this string into consecutive substrings that meet specific conditions.

A value-equal string is defined as a string where all characters are identical. For example:

• "1111" is value-equal (all 1's)
• "33" is value-equal (all 3's)
• "123" is NOT value-equal (different characters)

The task is to decompose the input string s into consecutive value-equal substrings with these requirements:

• Exactly one substring must have a length of 2
• All other substrings must have a length of 3
• The substrings must be consecutive (no gaps or overlaps)
• Each substring must be value-equal

Return true if such a decomposition is possible, otherwise return false.

For example:

• If s = "000111000", this can be decomposed as "000" + "111" + "000" (all length 3, but missing the required length-2 substring), so it would return false
• If s = "00011111222", this could be decomposed as "000" + "11" + "111" + "222", which has exactly one length-2 substring and the rest are length-3, so it would return true

The key challenge is ensuring that when you group consecutive identical characters, the groups can be arranged to have exactly one group of size 2 and all others of size 3.

Intuition

When we look at this problem, we need to group consecutive identical characters and then check if these groups can be split into valid lengths (one group of length 2, rest of length 3).

The key insight is that when we have a group of consecutive identical characters, we can split them in different ways. For example, if we have "11111" (5 ones), we can split it as "11" + "111" (length 2 + length 3) or keep it as one group of length 5.

Let's think about what happens with different group lengths:

• Length 1: Cannot form either a 2 or 3, so it's impossible
• Length 2: Perfect for our required single length-2 substring
• Length 3: Perfect for a length-3 substring
• Length 4: Can be split as 2+2, but we need exactly one length-2, so this is problematic
• Length 5: Can be split as 2+3, giving us one length-2 and one length-3
• Length 6: Can be split as 3+3, giving us two length-3 substrings
• Length 7: Can be split as 2+2+3, but that gives us two length-2 substrings (not allowed)
• Length 8: Can be split as 2+3+3, giving us one length-2 and two length-3

We can see a pattern emerging. For any group of length n:

• If n % 3 == 0: We can split it into groups of 3
• If n % 3 == 1: We have a remainder of 1, which cannot form a valid group (neither 2 nor 3)
• If n % 3 == 2: We can use one length-2 substring and the rest as length-3

This leads to the solution strategy: traverse the string, identify each group of consecutive identical characters, and check if its length modulo 3 equals 1 (impossible case). Count how many groups have length modulo 3 equal to 2 (these contribute a length-2 substring). We need exactly one such group for the string to be decomposable according to the rules.

Solution Approach

The solution uses a two-pointer approach to identify and process groups of consecutive identical characters.

We initialize two variables:

• i: Starting position of the current group
• cnt2: Counter to track how many groups have contributed a length-2 substring

The algorithm works as follows:

1. Traverse the string using pointer i that marks the start of each group of identical characters.

2. Find the end of current group: For each position i, use pointer j to find where the current group ends by advancing j while s[j] == s[i]. This gives us a group of length (j - i).

3. Check the group length modulo 3:

   • If (j - i) % 3 == 1: This group has a remainder of 1 when divided by 3, which cannot be decomposed into valid substrings (neither 2 nor 3). Return false immediately.

   • If (j - i) % 3 == 2: This group can contribute one length-2 substring. Increment cnt2 to track this. If cnt2 > 1, we have more than one length-2 substring, which violates the requirement, so return false.

   • If (j - i) % 3 == 0: This group can be perfectly divided into length-3 substrings, so we continue without any special action.

4. Move to the next group: Set i = j to start processing the next group.

5. Final validation: After processing all groups, check if cnt2 == 1. We need exactly one length-2 substring for a valid decomposition.

The time complexity is O(n) where n is the length of the string, as we traverse the string once. The space complexity is O(1) as we only use a constant amount of extra space for variables.

This approach efficiently checks whether the string can be decomposed according to the rules without actually creating the substrings, making it both time and space efficient.

Example Walkthrough

Let's walk through the solution with the example s = "00011111222".

Step 1: Initialize variables

• i = 0 (starting position)
• cnt2 = 0 (count of groups contributing a length-2 substring)

Step 2: Process first group (000)

• Start at i = 0, character is '0'
• Find end: j advances while s[j] == '0', stops at j = 3
• Group length: 3 - 0 = 3
• Check: 3 % 3 = 0 (perfectly divisible by 3, no action needed)
• Move: i = 3

Step 3: Process second group (11111)

• Start at i = 3, character is '1'
• Find end: j advances while s[j] == '1', stops at j = 8
• Group length: 8 - 3 = 5
• Check: 5 % 3 = 2 (remainder 2, can contribute one length-2)
• Action: Increment cnt2 = 1 (first length-2 substring found)
• Note: This group of 5 can be split as "11" + "111"
• Move: i = 8

Step 4: Process third group (222)

• Start at i = 8, character is '2'
• Find end: j advances while s[j] == '2', stops at j = 11 (end of string)
• Group length: 11 - 8 = 3
• Check: 3 % 3 = 0 (perfectly divisible by 3, no action needed)
• Move: i = 11 (end of string)

Step 5: Final validation

• Check: cnt2 == 1? Yes! We have exactly one length-2 substring
• Return: true

The decomposition would be: "000" + "11" + "111" + "222" (one length-2 and three length-3 substrings).

Counter-example: If we had s = "0001111" (3 zeros, 4 ones):

• First group: length 3, 3 % 3 = 0 ✓
• Second group: length 4, 4 % 3 = 1 ✗
• Return false immediately (length 4 cannot be split into valid combinations)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the string s.

The algorithm uses a single pass through the string with two pointers i and j. The outer while loop iterates through the string, and for each position i, the inner while loop moves j forward to find consecutive identical characters. Although there are nested loops, each character in the string is visited exactly once by the pointer j. The pointer i jumps to the position of j after each group of identical characters is processed, ensuring no character is revisited. Therefore, the total number of operations is proportional to n.

Space Complexity: O(1).

The algorithm only uses a constant amount of extra space regardless of the input size. The variables used are:

• i and j: two integer pointers
• n: storing the length of the string
• cnt2: a counter for groups with remainder 2 when divided by 3

All these variables occupy constant space that doesn't scale with the input size n.

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

Common Pitfalls

Pitfall 1: Misunderstanding Group Length Requirements

The Mistake: A common misconception is thinking that each group of consecutive identical characters must be either length 2 OR length 3. This leads to incorrect logic like:

# INCORRECT approach
while i < n:
    j = i
    while j < n and s[j] == s[i]:
        j += 1
  
    group_length = j - i
  
    # Wrong: assuming each group must be exactly 2 or 3
    if group_length == 2:
        pair_count += 1
    elif group_length != 3:
        return False  # This is wrong!

Why It's Wrong: Groups can be longer than 3! For example, a group of 5 identical characters ("11111") can be split into one length-3 substring ("111") and one length-2 substring ("11"). The key insight is that groups need to be divisible into combinations of 2s and 3s.

The Correct Understanding:

• Length % 3 == 0: Can be perfectly divided into 3s (e.g., 6 → 3+3)
• Length % 3 == 1: Cannot be divided into 2s and 3s (e.g., 4 → impossible)
• Length % 3 == 2: Can be divided into 3s with one 2 left over (e.g., 5 → 3+2)

Pitfall 2: Forgetting Edge Cases with Very Short Strings

The Mistake: Not handling strings that are too short to form the required pattern:

# Missing edge case check
def isDecomposable(self, s: str) -> bool:
    # Forgot to check if string is even long enough
    # Minimum valid length is 5 (one pair + one triplet)
    i = 0
    n = len(s)
    pair_count = 0
    # ... rest of code

The Solution: While the modulo logic naturally handles this (a string of length < 5 will either have remainder 1 or won't have exactly one pair), explicitly checking can make the code clearer:

def isDecomposable(self, s: str) -> bool:
    n = len(s)
  
    # Early return for strings too short to be valid
    if n < 5:
        return False
  
    # Continue with main logic...

Pitfall 3: Incorrect Pair Counting Logic

The Mistake: Counting pairs incorrectly when a group can contribute multiple pairs:

# INCORRECT: Thinking each group can only contribute one pair maximum
if group_length % 3 == 2:
    pair_count += 1  # What if group_length is 8? (8 = 3 + 3 + 2)

Why The Given Solution Works: The solution correctly handles this because:

• A group with length % 3 == 2 contributes exactly ONE pair regardless of total length
• For example: length 5 → 3+2 (one pair), length 8 → 3+3+2 (one pair), length 11 → 3+3+3+2 (one pair)
• The algorithm correctly counts this as one pair contribution per group

Pitfall 4: Off-by-One Errors in Group Detection

The Mistake: Incorrectly identifying group boundaries:

# INCORRECT: Off-by-one error
while i < n:
    j = i + 1  # Starting j at i+1 instead of i
    while j < n and s[j] == s[i]:
        j += 1
  
    group_length = j - i  # This would be wrong if j started at i+1

The Correct Approach: Always start j at the same position as i to correctly count the group length:

j = i  # Start at the same position
while j < n and s[j] == s[i]:
    j += 1
group_length = j - i  # Correct length calculation