#3579

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Steps to Convert String with Operations

Transform word1 into word2 using minimal operations on substrings with a dynamic programming state transition approach.

Problem Statement

Given two strings, word1 and word2, of equal length, transform word1 into word2 by splitting it into contiguous substrings and applying operations to each substring. Operations allowed are replace, swap, or reverse on each substring, with each character usable only once per operation type.

Determine the minimum number of operations required to fully convert word1 into word2. Each substring can be independently manipulated, but no index may participate in more than one operation of the same type. Output the minimal steps necessary to complete the transformation.

Examples

Example 1

Input: word1 = "abcdf", word2 = "dacbe"

Output: 4

Divide word1 into "ab" , "c" , and "df" . The operations are:

Example 2

Input: word1 = "abceded", word2 = "baecfef"

Output: 4

Divide word1 into "ab" , "ce" , and "ded" . The operations are:

Example 3

Input: word1 = "abcdef", word2 = "fedabc"

Output: 2

Divide word1 into "abcdef" . The operations are:

Constraints

1 <= word1.length == word2.length <= 100
word1 and word2 consist only of lowercase English letters.

Solution Approach

Dynamic Programming Table

Define dp[i] as the minimal operations to transform the first i characters. Iterate over all possible substrings ending at i and compute dp[i] as the minimum of dp[j] plus one operation for substring word1[j:i].

Consider All Substring Operations

For each substring, evaluate replace, swap, and reverse while ensuring no character is reused for the same operation type. Update the DP table only if the operation yields fewer total steps.

Greedy State Transitions

When multiple operations result in the same substring match, prefer operations that maximize future matches to reduce total steps. This greedy tie-breaking combined with DP ensures an optimal solution.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time and space complexity depend on substring enumeration and DP table size. Naively checking all substrings yields O(n^3) time, but careful pruning and state caching reduce redundant operations.

What Interviewers Usually Probe

Check if you are correctly handling overlapping operations on characters.
Verify DP state transitions properly account for minimal operations for all substrings.
Look for potential greedy choices that can reduce total operation count.

Common Pitfalls or Variants

Common pitfalls

Reusing characters in multiple operations of the same type within a substring.
Incorrectly updating DP states when substring operations overlap.
Failing to consider reverse or swap operations that reduce overall steps.

Follow-up variants

Allow unlimited operations per character but still minimize total steps.
Restrict operations to only replace and reverse, disallow swaps.
Change problem to count maximal substring matches instead of minimal steps.

FAQ

What is the optimal approach for Minimum Steps to Convert String with Operations?

Use state transition dynamic programming with careful handling of replace, swap, and reverse operations on substrings.

Can characters be used in multiple operations on the same substring?

No, each character may only participate once per operation type within each substring.

What should I watch out for in DP updates?

Avoid double-counting operations and ensure overlapping substrings are correctly accounted for in minimal steps.

Is there a greedy shortcut for this problem?

Greedy choices help break ties when multiple operations produce equal substrings, but DP ensures overall minimal steps.

Does this problem fit the state transition DP pattern?

Yes, each substring transformation represents a DP state, and transitions model the minimal operations to reach the next state.

terminal

Solution

Solution 1: Greedy + Dynamic Programming

We define $f[i]$ as the minimum number of operations required to convert the first $i$ characters of $\textit{word1}$ to the first $i$ characters of $\textit{word2}$. The answer is $f[n]$, where $n$ is the length of both $\textit{word1}$ and $\textit{word2}$.

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class Solution:
    def minOperations(self, word1: str, word2: str) -> int:
        def calc(l: int, r: int, rev: bool) -> int:
            cnt = Counter()
            res = 0
            for i in range(l, r + 1):
                j = r - (i - l) if rev else i
                a, b = word1[j], word2[i]
                if a != b:
                    if cnt[(b, a)] > 0:
                        cnt[(b, a)] -= 1
                    else:
                        cnt[(a, b)] += 1
                        res += 1
            return res

        n = len(word1)
        f = [inf] * (n + 1)
        f[0] = 0
        for i in range(1, n + 1):
            for j in range(i):
                t = min(calc(j, i - 1, False), 1 + calc(j, i - 1, True))
                f[i] = min(f[i], f[j] + t)
        return f[n]

Continue Practicing

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