Problem Description

You are given two strings word1 and word2 of equal length. Your goal is to transform word1 into word2 using the minimum number of operations.

To achieve this transformation, you need to divide word1 into one or more contiguous substrings. For each substring, you can perform the following operations:

1. Replace: Change any single character in the substring to another lowercase English letter
2. Swap: Exchange any two characters within the substring
3. Reverse: Reverse the entire substring

Important constraints:

• Each operation counts as exactly one operation
• Each character in a substring can be involved in at most one operation (a character cannot be used in multiple replace, swap, or reverse operations)
• The substrings must be contiguous and non-overlapping
• The division of word1 into substrings is part of your strategy

The problem asks you to find the minimum total number of operations needed across all substrings to transform word1 into word2.

For example, if you have word1 = "abc" and word2 = "cba", you could:

• Treat the entire string as one substring and reverse it (1 operation)
• Or divide it into smaller substrings and apply different operations

The challenge is to find the optimal way to divide the string and apply operations to minimize the total count.

Intuition

The key insight is that we need to find the optimal way to partition word1 into substrings and determine the best operations for each substring. This naturally leads to a dynamic programming approach where we build up the solution for longer prefixes using solutions for shorter ones.

Let's think about what happens when we process a substring from index j to i-1. For this substring, we have two main choices:

1. Don't reverse it and apply other operations as needed
2. Reverse it first (costs 1 operation), then apply other operations

Why does the order matter? If we reverse first, the positions of characters change, affecting which characters need to be modified. However, swaps and replacements can be done in any order after the reversal since they're independent operations on different characters.

For counting the minimum operations needed for a substring (with or without reversal), we can use a greedy matching strategy. When we compare corresponding positions between word1 and word2:

• If characters match, no operation is needed
• If they don't match, we have two options:
  • Replace this character (always costs 1 operation)
  • Try to find another mismatched position where we can swap to fix both positions (costs 1 operation for 2 fixes)

The clever part is tracking potential swap pairs. When we see a mismatch (a, b) at position i (where word1[i] = a and word2[i] = b), we check if we've previously seen the reverse mismatch (b, a). If yes, we can pair them up as a single swap operation. If not, we record this mismatch for potential future pairing.

This greedy pairing strategy works because:

• A swap between two positions fixes both mismatches with one operation
• A replacement only fixes one mismatch with one operation
• Therefore, whenever we can pair mismatches, we should do so immediately

The dynamic programming formula becomes: f[i] = min over all j < i of (f[j] + cost of transforming substring[j:i-1]), where the cost is the minimum between processing the substring normally or reversing it first.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The solution uses dynamic programming combined with a greedy character matching strategy. Let's break down the implementation:

Dynamic Programming Setup

We define f[i] as the minimum number of operations to transform the first i characters of word1 to the first i characters of word2. We initialize:

• f[0] = 0 (empty strings require no operations)
• f[i] = inf for all other positions initially

For each position i from 1 to n, we try all possible split points j where 0 ≤ j < i. This means we're considering transforming the substring word1[j:i] to word2[j:i] as a single unit.

The calc Function

The helper function calc(l, r, rev) computes the minimum operations needed to transform word1[l:r+1] to word2[l:r+1]:

• l, r define the substring boundaries (inclusive)
• rev indicates whether we reverse the substring first

The function implements a greedy matching strategy:

def calc(l: int, r: int, rev: bool) -> int:
    cnt = Counter()  # Tracks unpaired mismatches
    res = 0          # Operation count
  
    for i in range(l, r + 1):
        # If reversed, map position i to its reversed position
        j = r - (i - l) if rev else i
        a, b = word1[j], word2[i]
      
        if a != b:  # Characters don't match
            if cnt[(b, a)] > 0:
                # Found a matching pair for swap
                cnt[(b, a)] -= 1
            else:
                # Record this mismatch for potential future pairing
                cnt[(a, b)] += 1
                res += 1
    return res

The key insight: when we encounter a mismatch (a, b) at position i:

• We check if we've seen the reverse mismatch (b, a) before
• If yes, we can pair them as a single swap operation (decrement the counter)
• If no, we count this as a new operation and store it for potential future pairing

Main DP Transition

For each position i and split point j:

t = min(calc(j, i - 1, False), 1 + calc(j, i - 1, True))
f[i] = min(f[i], f[j] + t)

This considers:

1. Not reversing the substring: cost is calc(j, i-1, False)
2. Reversing the substring: cost is 1 + calc(j, i-1, True) (1 for the reverse operation itself)

We take the minimum of these two options and add it to f[j] (the cost of transforming the first j characters).

Time Complexity

• The outer DP loop runs O(n) times
• For each position, we try O(n) split points
• Each calc function call processes O(n) characters
• Overall: O(n³) time complexity

The solution efficiently combines dynamic programming for optimal partitioning with a greedy strategy for character matching within each partition.

Example Walkthrough

Let's walk through a concrete example with word1 = "abcd" and word2 = "dcba".

We'll build our DP table f[i] where f[i] represents the minimum operations to transform the first i characters.

Initial state: f[0] = 0 (empty strings need no operations)

Computing f[1]: Transform "a" to "d"

• Only option: substring [0:0]
• Without reverse: "a" → "d" needs 1 replacement
• With reverse: "a" → "d" still needs 1 replacement (reversing single char doesn't help)
• f[1] = f[0] + 1 = 1

Computing f[2]: Transform "ab" to "dc"

• Option 1: Use substring [0:1] (entire string)
  • Without reverse: "ab" → "dc"
    • Position 0: 'a' ≠ 'd', record mismatch (a,d), res = 1
    • Position 1: 'b' ≠ 'c', record mismatch (b,c), res = 2
    • Total: 2 operations
  • With reverse: "ba" → "dc"
    • Position 0: 'b' ≠ 'd', record mismatch (b,d), res = 1
    • Position 1: 'a' ≠ 'c', record mismatch (a,c), res = 2
    • Total: 1 (reverse) + 2 = 3 operations
  • Best for this option: f[0] + 2 = 2
• Option 2: Split at position 1, use substring [1:1]
  • Previous work: f[1] = 1
  • Transform "b" to "c": needs 1 replacement
  • Total: f[1] + 1 = 2
• f[2] = 2

Computing f[3]: Transform "abc" to "dcb"

• Option 1: Use substring [0:2] (entire string)
  • Without reverse: "abc" → "dcb"
    • Position 0: 'a' ≠ 'd', record mismatch (a,d), res = 1
    • Position 1: 'b' ≠ 'c', record mismatch (b,c), res = 2
    • Position 2: 'c' ≠ 'b', check for (c,b) - not found, but wait! We have (b,c) from position 1
    • We can pair positions 1 and 2 as a swap! Decrement counter for (b,c)
    • Total: 2 operations (1 replacement for 'a'→'d', 1 swap for positions 1&2)
  • With reverse: "cba" → "dcb"
    • Position 0: 'c' ≠ 'd', record mismatch (c,d), res = 1
    • Position 1: 'b' = 'c'? No, 'b' ≠ 'c', record mismatch (b,c), res = 2
    • Position 2: 'a' ≠ 'b', record mismatch (a,b), res = 3
    • Total: 1 (reverse) + 3 = 4 operations
  • Best for this option: f[0] + 2 = 2
• Other split options would give higher costs
• f[3] = 2

Computing f[4]: Transform "abcd" to "dcba"

• Option 1: Use substring [0:3] (entire string)
  • Without reverse: "abcd" → "dcba"
    • Position 0: 'a' ≠ 'd', record mismatch (a,d), res = 1
    • Position 1: 'b' ≠ 'c', record mismatch (b,c), res = 2
    • Position 2: 'c' ≠ 'b', found (b,c) in counter, pair them! counter[(b,c)] -= 1
    • Position 3: 'd' ≠ 'a', found (a,d) in counter, pair them! counter[(a,d)] -= 1
    • Total: 2 operations (two swaps: positions 0↔3 and 1↔2)
  • With reverse: "dcba" → "dcba"
    • All positions match perfectly!
    • Total: 1 (reverse) + 0 = 1 operation
  • Best for this option: f[0] + 1 = 1
• This is optimal! Reversing the entire string gives us the answer in just 1 operation.
• f[4] = 1

Final answer: 1 operation (reverse the entire string)

This example demonstrates how the algorithm considers different partitioning strategies and efficiently counts operations using the greedy pairing approach for swaps.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^3)

The analysis breaks down as follows:

• The outer loop in the main function runs n times (from 1 to n)
• The inner loop runs up to i times for each iteration of the outer loop
• For each pair (j, i), the calc function is called twice (once with rev=False and once with rev=True)
• Each calc function iterates through i - j elements, which can be up to n elements in the worst case
• The Counter operations inside calc (checking, updating) take O(1) time since we're dealing with character pairs

Therefore, the total time complexity is O(n) * O(n) * O(n) = O(n^3)

Space Complexity: O(n + |Σ|^2)

The space usage consists of:

• The f array of size n + 1: O(n)
• The Counter object in calc function stores character pairs (a, b): In the worst case, this can store up to |Σ|^2 different pairs where |Σ| = 26 (the size of the lowercase English alphabet)
• Other variables use O(1) space

Therefore, the total space complexity is O(n + |Σ|^2) where |Σ| = 26 in this problem.

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

Common Pitfalls

1. Misunderstanding the Greedy Pairing Logic

Pitfall: The most common mistake is misunderstanding how the character pairing works in the calculate_segment_cost function. Many developers incorrectly assume that when we find a mismatch (a, b), we should look for any occurrence of b and a anywhere in the remaining substring.

Why it's wrong: The algorithm uses a clever greedy approach where it pairs mismatches as it encounters them sequentially. When we see a mismatch (a, b) at position i, we check if we've previously seen the reverse mismatch (b, a) that hasn't been paired yet. This ensures each character participates in at most one operation.

Example of incorrect approach:

# WRONG: Trying to find all possible pairs first
def wrong_calc(left, right, is_reversed):
    mismatches = []
    for i in range(left, right + 1):
        j = right - (i - left) if is_reversed else i
        if word1[j] != word2[i]:
            mismatches.append((word1[j], word2[i]))
  
    # Wrong: trying to match pairs after collecting all mismatches
    paired = 0
    used = [False] * len(mismatches)
    for i in range(len(mismatches)):
        if used[i]: continue
        for j in range(i + 1, len(mismatches)):
            if not used[j] and mismatches[i] == (mismatches[j][1], mismatches[j][0]):
                paired += 1
                used[i] = used[j] = True
                break
    return len(mismatches) - paired

Correct approach: Process characters sequentially and maintain a counter of unmatched pairs. This ensures optimal pairing while respecting the constraint that each character can only be involved in one operation.

2. Incorrect Handling of the Reverse Operation Cost

Pitfall: Forgetting to add the cost of the reverse operation itself when considering the reversed substring option.

Wrong:

# Missing the +1 for the reverse operation
segment_cost = min(
    calculate_segment_cost(start_pos, end_pos - 1, False),
    calculate_segment_cost(start_pos, end_pos - 1, True)  # WRONG: Missing +1
)

Correct:

segment_cost = min(
    calculate_segment_cost(start_pos, end_pos - 1, False),
    1 + calculate_segment_cost(start_pos, end_pos - 1, True)  # Correct: +1 for reverse
)

3. Off-by-One Errors in Index Mapping

Pitfall: When reversing a substring, incorrectly calculating the mapped index can lead to wrong character comparisons.

Wrong mapping:

# Common mistakes in reverse index calculation
word1_idx = right - word2_idx + left  # Wrong formula
# or
word1_idx = left + right - word2_idx  # Also wrong for general case

Correct mapping:

if is_reversed:
    # Correct: distance from left in word2 maps to distance from right in word1
    word1_idx = right - (word2_idx - left)
else:
    word1_idx = word2_idx

Verification: For a segment [left=2, right=5]:

• word2_idx=2 should map to word1_idx=5 (first becomes last)
• word2_idx=3 should map to word1_idx=4
• word2_idx=5 should map to word1_idx=2 (last becomes first)

4. Counter Key Order Confusion

Pitfall: Using inconsistent key ordering in the Counter, leading to failed pair matching.

Wrong:

# Inconsistent key ordering
if char1 != char2:
    if unmatched_pairs[(char1, char2)] > 0:  # Wrong: should be (char2, char1)
        unmatched_pairs[(char1, char2)] -= 1
    else:
        unmatched_pairs[(char2, char1)] += 1  # Wrong: should be (char1, char2)
        swap_count += 1

Correct: Always maintain consistent ordering - when we have mismatch (a, b), we look for previous occurrence of (b, a) and store (a, b) if not found.

Solution to Avoid These Pitfalls

1. Test with small examples: Walk through the algorithm manually with strings like "abc" → "bac" to verify the pairing logic.

2. Add assertions: Include debug assertions to verify index calculations:

assert left <= word1_idx <= right, f"Index {word1_idx} out of bounds [{left}, {right}]"

3. Consistent naming: Use clear variable names that indicate what they represent (e.g., word1_idx vs word2_idx instead of generic i, j).

4. Unit test edge cases: Test single character strings, already matching strings, and complete reversals to catch edge case bugs early.