Solution Approach

The implementation uses a sliding window with prefix state compression to efficiently find the optimal substring for each character pair.

Main Loop Structure: We enumerate all possible character pairs (a, b) where a ≠ b. For each pair, we process the string using a sliding window approach.

Key Variables:

• curA, curB: Count of characters a and b in the current window [l+1, r]
• preA, preB: Cumulative count of characters a and b before position l (i.e., in [0, l])
• l: Position before the left boundary of the window (initialized to -1)
• r: Right boundary of the window
• t[2][2]: 2D array storing minimum values of preA - preB for each parity combination

Sliding Window Logic:

1. We iterate through the string with r as the right pointer
2. For each position r, we update curA and curB based on whether the current character equals a or b
3. We shrink the window from the left when two conditions are met:
   • Window size is at least k: r - l >= k
   • Character b appears at least twice in the window: curB - preB >= 2 (ensuring non-zero even frequency)

Window Shrinking: When shrinking, we:

1. Record the state at the left boundary: t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB)
2. Move l forward and update preA and preB accordingly

Answer Calculation: For each position r, we calculate the potential answer using:

ans = max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1])

This formula works because:

• curA - curB is the total frequency difference from position 0 to r
• We subtract t[(curA & 1) ^ 1][curB & 1] to get the frequency difference in the substring
• (curA & 1) ^ 1 ensures we find a prefix with opposite parity for a (making the substring count odd)
• curB & 1 ensures we find a prefix with the same parity for b (making the substring count even)

Why This Works: The substring [l+1, r] has:

• Frequency of a = curA - preA
• Frequency of b = curB - preB
• Frequency difference = (curA - preA) - (curB - preB) = (curA - curB) - (preA - preB)

By storing the minimum preA - preB for each parity state, we maximize the frequency difference while ensuring the parity constraints are satisfied.

The algorithm processes each character pair in O(n) time, giving an overall time complexity of O(5 × 4 × n) = O(n).

Example Walkthrough

Let's walk through a concrete example with s = "02211" and k = 2.

We'll focus on one character pair: a = '1' (want odd frequency) and b = '2' (want even frequency).

Initial Setup:

• l = -1 (position before the window starts)
• preA = 0, preB = 0 (no characters before position -1)
• curA = 0, curB = 0 (cumulative counts)
• t[2][2] = [[INF, INF], [INF, INF]] (minimum values for each parity state)

Step-by-step processing:

Position r = 0 (char = '0'):

• Neither 'a' nor 'b', so curA = 0, curB = 0
• Window size: 0 - (-1) = 1 < k, don't shrink
• Skip answer calculation (window too small)

Position r = 1 (char = '2'):

• This is 'b', so curB = 1
• Window size: 1 - (-1) = 2 >= k, can potentially shrink
• Check if curB - preB >= 2: 1 - 0 = 1 < 2, don't shrink yet
• Skip answer calculation (need even frequency for 'b', but have odd)

Position r = 2 (char = '2'):

• This is 'b', so curB = 2
• Window size: 2 - (-1) = 3 >= k
• Check if curB - preB >= 2: 2 - 0 = 2 >= 2, can shrink!
• Shrink window:
  • Record state: t[preA & 1][preB & 1] = t[0][0] = min(INF, 0 - 0) = 0
  • Move l from -1 to 0
  • Update prefix counts: preA = 0, preB = 0 (char at position 0 is '0')
• Calculate answer:
  • Need: curA odd (currently 0, even) and curB even (currently 2, even ✓)
  • Look for prefix with opposite parity for A: t[(0 & 1) ^ 1][2 & 1] = t[1][0]
  • t[1][0] = INF, so no valid answer yet

Position r = 3 (char = '1'):

• This is 'a', so curA = 1
• Window size: 3 - 0 = 3 >= k
• Check if curB - preB >= 2: 2 - 0 = 2 >= 2, can shrink!
• Shrink window:
  • Record state: t[0][0] = min(0, 0 - 0) = 0
  • Move l from 0 to 1
  • Update prefix counts: preA = 0, preB = 1 (char at position 1 is '2')
• Calculate answer:
  • Need: curA odd (currently 1, odd ✓) and curB even (currently 2, even ✓)
  • Look for prefix with opposite parity for A and same for B: t[(1 & 1) ^ 1][2 & 1] = t[0][0] = 0
  • Answer = (curA - curB) - t[0][0] = (1 - 2) - 0 = -1

Position r = 4 (char = '1'):

• This is 'a', so curA = 2
• Window size: 4 - 1 = 3 >= k
• Check if curB - preB >= 2: 2 - 1 = 1 < 2, don't shrink
• Calculate answer:
  • Need: curA odd (currently 2, even) and curB even (currently 2, even ✓)
  • Look for prefix with opposite parity for A: t[(2 & 1) ^ 1][2 & 1] = t[1][0] = INF
  • No valid answer from this position

Result for this pair (a='1', b='2'): Maximum difference = -1

The algorithm would continue checking all other character pairs. For instance, with a = '2' and b = '1', we might find a substring "2211" where '2' appears 2 times (even, but we want odd) and '1' appears 2 times (even ✓), but this wouldn't be valid since '2' needs odd frequency.

The key insight is how the parity tracking ensures we only consider valid substrings: when we calculate the answer, we specifically look for a prefix state that, when "subtracted" from the current state, gives us the desired parities (odd for 'a', even for 'b').

Time and Space Complexity

Time Complexity: O(n × |Σ|²), where n is the length of string S and |Σ| is the alphabet size (5 in this problem).

The analysis breaks down as follows:

• The outer two nested loops iterate through all pairs of digits (a, b) where a ≠ b, giving us 5 × 5 - 5 = 20 iterations, which is O(|Σ|²)
• For each pair (a, b), the inner loop iterates through the string once with index r from 0 to n-1
• The while loop inside moves the left pointer l forward, but across all iterations of r, the pointer l moves at most n times total (amortized O(1) per iteration)
• All operations inside the loops (comparisons, arithmetic, array access) are O(1)
• Therefore, the total time complexity is O(|Σ|² × n)

Space Complexity: O(n + 1) = O(n)

The analysis breaks down as follows:

• The list s stores the converted integer values of the input string, requiring O(n) space
• The 2D array t is of fixed size 2 × 2, which is O(1)
• All other variables (ans, curA, curB, preA, preB, l, r, x, a, b) use constant space O(1)
• Therefore, the total space complexity is O(n)

Note: The reference answer states O(1) space complexity, which would be accurate if the input string could be processed character by character without creating the list s. However, the given code creates s = list(map(int, S)), which uses O(n) additional space.

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

Common Pitfalls

1. Incorrect Parity Matching for Valid Substrings

The Pitfall: The most common mistake is misunderstanding how the parity matching works when calculating the answer. Developers often incorrectly assume they need to check if the substring itself has odd count for a and even count for b, but the code actually checks the difference between cumulative counts and stored prefix counts.

Why it happens: The formula max_diff[(current_count_a & 1) ^ 1][current_count_b & 1] is counterintuitive. It looks for a prefix with:

• Opposite parity for a (using XOR with 1)
• Same parity for b

This ensures that the substring [left+1, right] has:

• Odd count for a: (current_count_a - prev_count_a) is odd when parities differ
• Even count for b: (current_count_b - prev_count_b) is even when parities are the same

Solution: Add explicit validation to verify the substring meets requirements:

# After calculating the potential answer, validate the substring if min_diff[(current_count_a & 1) ^ 1][current_count_b & 1] != float('inf'):  # Calculate actual substring counts  substring_count_a = current_count_a - stored_prev_count_a  substring_count_b = current_count_b - stored_prev_count_b   # Verify constraints  if substring_count_a % 2 == 1 and substring_count_b % 2 == 0 and substring_count_b > 0:  max_diff = max(max_diff, substring_count_a - substring_count_b)

2. Forgetting to Initialize min_diff Array Properly

The Pitfall: Not handling the case where no valid prefix exists yet. The min_diff array starts with float('inf') values, but we might try to use these infinite values in calculations before any valid prefix is stored.

Why it happens: The algorithm assumes that once we shrink the window enough times, we'll have valid prefixes stored. However, for small inputs or specific character distributions, we might attempt to calculate an answer before any valid prefix exists.

Solution: Check if a valid prefix exists before using it:

# Before calculating max_diff if min_diff[(current_count_a & 1) ^ 1][current_count_b & 1] != float('inf'):  max_diff = max(  max_diff,  current_count_a - current_count_b - min_diff[(current_count_a & 1) ^ 1][current_count_b & 1]  )

3. Misunderstanding the Window Shrinking Condition

The Pitfall: The condition current_count_b - prev_count_b >= 2 ensures that character b appears at least twice in the window, guaranteeing a non-zero even count. However, developers might mistakenly think this guarantees an even count, forgetting that 3, 5, 7, etc., are also possible.

Why it happens: The condition only ensures a minimum count of 2, not that the count is even. The even count is enforced through the parity matching mechanism, not this condition.

Solution: The window shrinking should continue as long as we can maintain at least 2 occurrences of b:

# More robust shrinking that ensures we keep valid windows while right - left >= k:  window_b_count = current_count_b - prev_count_b   # Only shrink if we can maintain at least 2 occurrences of b  if window_b_count >= 2:  # Store the state  min_diff[prev_count_a & 1][prev_count_b & 1] = min(  min_diff[prev_count_a & 1][prev_count_b & 1],  prev_count_a - prev_count_b  )   # Check if we can shrink further without violating constraints  next_digit = digits[left + 1] if left + 1 < len(digits) else -1  if next_digit == digit_b and window_b_count <= 2:  break # Can't shrink without losing valid b count   left += 1  prev_count_a += (digits[left] == digit_a)  prev_count_b += (digits[left] == digit_b)

4. Not Handling Edge Cases with Return Value

The Pitfall: Returning float('-inf') when no valid substring exists, which might not be the expected behavior.

Solution: Define clear behavior for when no valid substring exists:

# At the end of the function return max_diff if max_diff != float('-inf') else -1 # or 0, depending on requirements