Problem Description

You are given two strings, word1 and word2, of equal length. Your task is to transform word1 into word2.

To do this, you can divide word1 into one or more contiguous non-empty substrings. For each such substring substr, you are allowed to perform these operations:

1. Replace: Replace the character at any one index of substr with another lowercase English letter.
2. Swap: Swap any two characters in substr.
3. Reverse Substring: Reverse the entire substring substr.

Each operation counts as one step, and each character in a substring can be used in each operation type at most once (no single index may be involved in more than one replace, one swap, or one reverse).

Return the minimum number of operations required to transform word1 into word2.

Intuition

Since we can divide word1 into substrings and perform powerful operations on them (replace, swap, reverse), it's important to find the optimal way to split word1 and select which operations to use on each substring. If a substring can be turned into the target segment of word2 with fewer operations by using a reverse at the start, we should consider that.

The key insight is to think about dynamic programming: for every position in the string, keep track of the minimum number of operations needed to convert the first part of word1 up to that point. For each substring, try both normal and reversed forms to see which requires fewer steps. Within a substring, count mismatches between word1 and word2 and see if swaps can help reduce replace operations (since swapping can pair up mismatches). This way, we always combine substrings in the most efficient order, exploring all possible splits and using the substring operations smartly.

Solution Approach

We use dynamic programming to solve this problem efficiently. Define f[i] as the minimum number of operations needed to convert the first i characters of word1 into the first i characters of word2. Our goal is to compute f[n], where n is the length of the words.

For each position i (1 to n), we consider all possible previous split points j (0 to i-1), which represent dividing word1[0:i] into two parts: word1[0:j] (already converted) and word1[j:i] (current substring to process). For each substring word1[j:i], we check two scenarios:

• Transform it as-is (rev=False)
• Reverse it first, then transform (rev=True, costing one operation for reversing)

For either case, we use a helper function calc(l, r, rev) to determine the minimum number of operations required to turn word1[l:r] (or its reverse) into word2[l:r]. Inside calc, we use a counter to identify character mismatches. If there are mismatched pairs (like a in word1 and b in word2), a swap can sometimes fix two characters in one operation. We use a dictionary to match up such pairs and count the number of changes.

The core DP formula is:

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

• calc(j, i-1, False): Cost to make substring match without reversing.
• 1 + calc(j, i-1, True): One operation for reversing, plus the cost to match after reverse.

We initialize f[0] = 0 and iterate through each position, updating f[i] based on the best possible split.

Finally, the answer is f[n], the minimum number of operations to convert word1 to word2.

Example Walkthrough

Consider word1 = "abca" and word2 = "caba".

Let's walk through how the dynamic programming approach works step-by-step:

1. Initialization: Set f[0] = 0. We'll compute f[i] for each i from 1 to 4.

2. Step-by-step DP computation:

   i = 1

   • Only possible split: j=0, substring: "a" vs "c"
   • No reverse: need 1 replace (a → c). Cost = 1
   • Reverse is the same as original (since length 1), also cost = 1
   • f[1] = min(0 + 1, 0 + 1 + 0) = 1

   i = 2

   • Possible splits:
     • j=0, substr: "ab" vs "ca"
       • No reverse: 'a' vs 'c' (replace), 'b' vs 'a' (replace), total 2 replaces = 2
       • Reversed: "ba" vs "ca"
         • 'b' vs 'c' (replace), 'a' vs 'a' (match), total 1 replace
         • Need 1 reverse operation, so total = 1 (reverse) + 1 (replace) = 2
       • Min cost for this split: 2
     • j=1, substr: "b" vs "a"
       • No reverse: need 1 replace
       • Reversed: same as original, cost = 1
       • Min cost for this split: 1
     • Total:
       • f[2] = min(0 + 2, 1 + 1) = 2

   i = 3

   • Possible splits:
     • j=0, substr: "abc" vs "cab"
       • No reverse: positions 0: a vs c (replace), 1: b vs a (replace), 2: c vs b (replace) → 3 replaces
       • Try swaps:
         • 'a' wanted to be 'c' at pos 0, 'b' wanted to be 'a' at pos 1, 'c' wanted to be 'b' at pos 2.
         • Possible swap: swap pos 0 and pos 2, "cba" → now at pos 0, c matches c, at pos 2, a vs b (still needs replace), at pos 1, b vs a (replace), total: 1 swap + 2 replaces = 3 steps
       • Reversed: "cba" vs "cab"
         • Positions: 0: c vs c (match), 1: b vs a (replace), 2: a vs b (replace) → 2 replaces
         • Plus 1 for reverse = 3
       • Min cost: 3
     • j=1, substr: "bc" vs "ab"
       • No reverse: b vs a (replace), c vs b (replace) = 2
       • Try swaps: no better option
       • Reversed: "cb" vs "ab": c vs a (replace), b vs b (match) = 1 replace + 1 reverse = 2
       • Take better one: 2
       • Add DP: f[1] + 2 = 1 + 2 = 3
     • j=2, substr: "c" vs "b"
       • Needs 1 replace
       • DP: f[2] + 1 = 2 + 1 = 3
     • So f[3] = min(3, 3, 3) = 3

   i = 4

   • Try all splits:
     • j=0, substr: "abca" vs "caba":
       • No reverse: a/c, b/a, c/b, a/a; need 3 replaces
       • Try swaps: swap pos 0 (a) and pos 2 (c): "cb a a" vs "caba"
         • Now c/c, b/a, a/b, a/a; b/a and a/b: can swap pos 1 and 2 to form c a b a vs caba; now only need 0 replaces, 2 swaps (0,2 and 1,2)
         • However, swapping twice not allowed at same position in one op; must check constraints, so in this example, just use replace for each pair.
       • Reversed: "acba" vs "caba":
         • a/c (replace), c/a (replace), b/b (match), a/a (match) = 2 replaces + 1 reverse = 3
     • j=1, substr: "bca" vs "aba":
       • No reverse: b/a, c/b, a/a → 2 replaces
       • Swaps: swap b and c: cba vs aba → c/a (replace), b/b, a/a ⇒ 1 replace + 1 swap = 2
       • Reverse: "acb" vs "aba": a/a, c/b, b/a: needs c/b and b/a (can swap), so 1 swap + 1 reverse = 2
       • DP: f[1] + 2 = 1+2=3
     • j=2, substr: "ca" vs "ba":
       • No reverse: c/b (replace), a/a (match) = 1 replace
       • Reverse: a c vs b a: a/b (replace), c/a (replace) = 2 + 1(rev)=3
       • DP: f[2] + 1 = 2+1=3
     • j=3, substr: "a" vs "a": match
       • No op needed
       • DP: f[3]+0=3+0=3
     • So overall, f[4] = 3

Result: The minimum number of operations to convert "abca" to "caba" is 3.

This example shows how the DP approach considers all ways to split the string and whether reversing or swapping can reduce operations for each substring.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n^3 + |\Sigma|^2), where n is the length of the input strings and |\Sigma| is the size of the character set (which is 26 for lowercase English letters). This complexity arises because for each of the O(n^2) pairs (j, i), the function calc is called, which itself takes O(n) time due to the traversal over substring. The |\Sigma|^2 term comes from the size of the counter used for all possible character pairs.

The space complexity of the code is O(n + |\Sigma|^2), where O(n) accounts for the DP array f, and O(|\Sigma|^2) comes from the Counter used in calc for all character pair combinations.