Problem Description

You are given a binary string s (containing only '0's and '1's). Your task is to split this string into exactly 3 non-empty parts: s1, s2, and s3, where concatenating them gives back the original string (s1 + s2 + s3 = s).

The requirement is that each of the three parts must contain the same number of '1's. You need to count how many different ways you can perform such a split.

For example, if s = "10101", one valid split could be s1 = "1", s2 = "01", s3 = "01", where each part contains exactly one '1'.

Since the answer can be very large, return the result modulo 10^9 + 7.

Key points to understand:

• The string must be split into exactly 3 non-empty parts
• Each part must have an equal number of '1's
• The order of characters within each part must be preserved (you're just choosing where to make the cuts)
• If it's impossible to split the string such that each part has the same number of '1's (for example, if the total number of '1's is not divisible by 3), return 0

Intuition

Let's think about what determines a valid split. First, we need to count the total number of '1's in the string. If this count isn't divisible by 3, it's impossible to split the string into three parts with equal '1's, so we return 0.

If the total count is divisible by 3, each part must contain exactly count/3 ones. Now, the key insight is that we don't need to try all possible splits - we just need to find the valid ranges where we can place our two cut points.

Consider the special case where there are no '1's at all. In this case, any split works! We just need to choose 2 positions out of n-1 possible cutting positions, which gives us C(n-1, 2) = (n-1) * (n-2) / 2 ways.

For the general case with '1's present, let's say each part needs exactly k ones. We need to find:

• Where can we place the first cut? It must be after exactly k ones.
• Where can we place the second cut? It must be after exactly 2k ones.

The trick is recognizing that there might be '0's between the k-th and (k+1)-th one, giving us flexibility in where to cut. For example, if we have "...1000001..." and we need to cut after the first '1', we can cut at any of the positions where the '0's are.

So we find:

• i1: the position of the k-th '1' (minimum position for first cut)
• i2: the position of the (k+1)-th '1' (maximum position for first cut, exclusive)
• j1: the position of the 2k-th '1' (minimum position for second cut)
• j2: the position of the (2k+1)-th '1' (maximum position for second cut, exclusive)

The number of valid ways to split is then (i2 - i1) * (j2 - j1) because we can independently choose any valid position for the first cut and any valid position for the second cut.

Learn more about Math patterns.

Solution Approach

The implementation follows the intuition we developed by systematically finding the valid cut positions.

Step 1: Count the total number of '1's

We first traverse the string and count all '1's. If this count cannot be divided by 3 (i.e., count % 3 != 0), we immediately return 0 as it's impossible to split equally.

Step 2: Handle the special case of no '1's

If count = 0, all characters are '0's. We can place our two cuts at any positions among the n-1 possible positions between characters. The number of ways is C(n-1, 2) = (n-1) * (n-2) / 2.

Step 3: Find the boundary positions

For the general case, we calculate cnt = count / 3, which is the number of '1's each part must have.

We implement a helper function find(x) that returns the index of the x-th '1' in the string. This function iterates through the string, counting '1's until it reaches the target count.

Using this function, we find four critical positions:

• i1 = find(cnt): Position of the cnt-th '1' (earliest possible position for first cut)
• i2 = find(cnt + 1): Position of the (cnt+1)-th '1' (latest possible position for first cut + 1)
• j1 = find(cnt * 2): Position of the (2*cnt)-th '1' (earliest possible position for second cut)
• j2 = find(cnt * 2 + 1): Position of the (2*cnt+1)-th '1' (latest possible position for second cut + 1)

Step 4: Calculate the number of ways

The first cut can be placed at any position from i1 to i2-1 (inclusive), giving us (i2 - i1) choices. The second cut can be placed at any position from j1 to j2-1 (inclusive), giving us (j2 - j1) choices.

Since these choices are independent, the total number of ways is (i2 - i1) * (j2 - j1).

Step 5: Return the result modulo 10^9 + 7

Since the answer can be very large, we return the result modulo 10^9 + 7.

The time complexity is O(n) as we traverse the string a constant number of times, and the space complexity is O(1) as we only use a few variables to store positions and counts.

Example Walkthrough

Let's walk through the solution with the string s = "10101".

Step 1: Count total '1's

• Traverse the string: '1' (count=1), '0', '1' (count=2), '0', '1' (count=3)
• Total count = 3, which is divisible by 3 ✓
• Each part needs: cnt = 3/3 = 1 one

Step 2: Check for special case

• Since count ≠ 0, we proceed to the general case

Step 3: Find boundary positions

• Find the 1st '1' (cnt=1): at index 0 → i1 = 0
• Find the 2nd '1' (cnt+1=2): at index 2 → i2 = 2
• Find the 2nd '1' (cnt*2=2): at index 2 → j1 = 2
• Find the 3rd '1' (cnt*2+1=3): at index 4 → j2 = 4

Step 4: Calculate valid splits

• First cut can be placed at positions: [0, 1] (from i1=0 to i2-1=1)
  • Position 0: splits to "1" | "0101"
  • Position 1: splits to "10" | "101"
• Second cut can be placed at positions: [2, 3] (from j1=2 to j2-1=3)
  • Position 2: splits the remainder to "1" | "01" or "10" | "1"
  • Position 3: splits the remainder to "101" | "" or "1010" | "1"

Let's verify all combinations:

1. First cut at 0, second cut at 2: "1" | "01" | "01" ✓ (each has 1 one)
2. First cut at 0, second cut at 3: "1" | "010" | "1" ✓ (each has 1 one)
3. First cut at 1, second cut at 2: "10" | "1" | "01" ✓ (each has 1 one)
4. First cut at 1, second cut at 3: "10" | "10" | "1" ✓ (each has 1 one)

Number of ways = (i2 - i1) × (j2 - j1) = (2 - 0) × (4 - 2) = 2 × 2 = 4

Step 5: Return result

• Return 4 % (10^9 + 7) = 4

Solution Implementation

Time and Space Complexity

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

• The initial calculation sum(c == '1' for c in s) iterates through the entire string once: O(n)
• The find() function is called at most 4 times (for finding i1, i2, j1, j2), and each call iterates through the string in the worst case: 4 * O(n) = O(n)
• When cnt == 0, the calculation ((n - 1) * (n - 2) // 2) % mod is done in constant time: O(1)
• All other operations (modulo, multiplication, subtraction) are constant time: O(1)

Overall, the algorithm makes a constant number of passes through the string, resulting in O(n) time complexity.

Space Complexity: O(1)

• The algorithm uses only a fixed number of variables: cnt, m, n, mod, i1, i2, j1, j2, and the local variables in the find() function (t, i, c)
• The find() function doesn't create any additional data structures
• The generator expression sum(c == '1' for c in s) doesn't create an intermediate list, it processes elements one at a time
• No recursive calls are made, so there's no additional stack space usage

The space usage doesn't scale with the input size, making it O(1) constant space complexity.

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

Common Pitfalls

1. Off-by-One Error in Split Position Calculation

A common mistake is misunderstanding what the split positions represent. When we find positions like first_part_end and second_part_start, developers often confuse whether these are inclusive or exclusive boundaries.

The Pitfall:

# Incorrect interpretation might lead to:
first_split_options = second_part_start - first_part_end - 1  # Wrong!
# or
first_split_options = second_part_start - first_part_end + 1  # Wrong!

Why it happens: The confusion arises because first_part_end is the index of the last '1' in the first part, and second_part_start is the index of the first '1' in the second part. The valid cut positions are all indices from first_part_end (inclusive) to second_part_start - 1 (inclusive).

The Correct Approach:

# The cut can be placed right after first_part_end up to right before second_part_start
# This gives us (second_part_start - first_part_end) valid positions
first_split_options = second_part_start - first_part_end

2. Integer Overflow in the Special Case

When all characters are '0's, we calculate C(n-1, 2) = (n-1) * (n-2) / 2. For large strings, this multiplication can cause integer overflow before applying the modulo operation.

The Pitfall:

# For very large n, this might overflow in some languages
return (string_length - 1) * (string_length - 2) // 2 % MOD

The Solution:

# Apply modulo during the calculation to prevent overflow
result = ((string_length - 1) * (string_length - 2) // 2) % MOD
# Or even safer:
result = (((string_length - 1) % MOD) * ((string_length - 2) % MOD) // 2) % MOD

3. Incorrect Handling of Edge Cases

Another pitfall is not properly handling the case where find_position_of_nth_one might not find enough '1's, especially when calculating third_part_start.

The Pitfall:

# If total_ones equals 2*ones_per_part exactly, there's no (2*ones_per_part + 1)th '1'
third_part_start = find_position_of_nth_one(ones_per_part * 2 + 1)  # Returns -1 or crashes

The Solution:

# Handle the boundary case explicitly
if ones_per_part * 2 + 1 > total_ones:
    third_part_start = string_length  # Use string length as the boundary
else:
    third_part_start = find_position_of_nth_one(ones_per_part * 2 + 1)

4. Misunderstanding the Problem Requirements

Some developers might think they need to find all possible combinations of three substrings with equal '1's, rather than understanding that the three parts must be contiguous segments of the original string.

The Pitfall: Trying to use combinations or permutations to select characters from different positions, which would break the string's original order.

The Correct Understanding: The problem asks for contiguous splits only. We're choosing two cut positions in the string, not rearranging characters. The string "10101" can only be split by choosing where to place two cuts, maintaining the original character sequence in each part.