LeetCode #3579 — HARD

Minimum Steps to Convert String with Operations

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

For this, divide word1 into one or more contiguous substrings. For each substring substr you can perform the following operations:

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

Each of these counts as one operation and each character of each substring can be used in each type of operation at most once (i.e. 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.

Example 1:

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

Output: 4

Explanation:

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

For the substring "ab",
- Perform operation of type 3 on "ab" -> "ba".
- Perform operation of type 1 on "ba" -> "da".
For the substring "c" do no operations.
For the substring "df",
- Perform operation of type 1 on "df" -> "bf".
- Perform operation of type 1 on "bf" -> "be".

Example 2:

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

Output: 4

Explanation:

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

For the substring "ab",
- Perform operation of type 2 on "ab" -> "ba".
For the substring "ce",
- Perform operation of type 2 on "ce" -> "ec".
For the substring "ded",
- Perform operation of type 1 on "ded" -> "fed".
- Perform operation of type 1 on "fed" -> "fef".

Example 3:

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

Output: 2

Explanation:

Divide word1 into "abcdef". The operations are:

For the substring "abcdef",
- Perform operation of type 3 on "abcdef" -> "fedcba".
- Perform operation of type 2 on "fedcba" -> "fedabc".

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given two strings, word1 and word2, of equal length. You need to transform word1 into word2. For this, divide word1 into one or more contiguous substrings. For each substring substr you can perform the following operations: Replace: Replace the character at any one index of substr with another lowercase English letter. Swap: Swap any two characters in substr. Reverse Substring: Reverse substr. Each of these counts as one operation and each character of each substring can be used in each type of operation at most once (i.e. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Greedy

Example 1

"abcdf"
"dacbe"

Example 2

"abceded"
"baecfef"

Example 3

"abcdef"
"fedabc"

Core Insight

What unlocks the optimal approach

Use dynamic programming
Do DP on disjoint substrings of <code>word1</code>. For the DP, we try both the substring and its reversed version (just add one extra operation)
First we swap pairs like (<code>word1[i]</code>, <code>word2[i]</code>) and (<code>word1[j]</code>, <code>word2[j]</code>) where <code>word1[i] == word2[j]</code> and <code>word2[i] == word1[j]</code>
For the remaining characters, we use replace operations

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3579: Minimum Steps to Convert String with Operations
class Solution {
    public int minOperations(String word1, String word2) {
        int n = word1.length();
        int[] f = new int[n + 1];
        Arrays.fill(f, Integer.MAX_VALUE);
        f[0] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                int a = calc(word1, word2, j, i - 1, false);
                int b = 1 + calc(word1, word2, j, i - 1, true);
                int t = Math.min(a, b);
                f[i] = Math.min(f[i], f[j] + t);
            }
        }
        return f[n];
    }

    private int calc(String word1, String word2, int l, int r, boolean rev) {
        int[][] cnt = new int[26][26];
        int res = 0;
        for (int i = l; i <= r; i++) {
            int j = rev ? r - (i - l) : i;
            int a = word1.charAt(j) - 'a';
            int b = word2.charAt(i) - 'a';
            if (a != b) {
                if (cnt[b][a] > 0) {
                    cnt[b][a]--;
                } else {
                    cnt[a][b]++;
                    res++;
                }
            }
        }
        return res;
    }
}

// Accepted solution for LeetCode #3579: Minimum Steps to Convert String with Operations
func minOperations(word1 string, word2 string) int {
	n := len(word1)
	f := make([]int, n+1)
	for i := range f {
		f[i] = math.MaxInt32
	}
	f[0] = 0

	calc := func(l, r int, rev bool) int {
		var cnt [26][26]int
		res := 0

		for i := l; i <= r; i++ {
			j := i
			if rev {
				j = r - (i - l)
			}
			a := word1[j] - 'a'
			b := word2[i] - 'a'

			if a != b {
				if cnt[b][a] > 0 {
					cnt[b][a]--
				} else {
					cnt[a][b]++
					res++
				}
			}
		}

		return res
	}

	for i := 1; i <= n; i++ {
		for j := 0; j < i; j++ {
			a := calc(j, i-1, false)
			b := 1 + calc(j, i-1, true)
			t := min(a, b)
			f[i] = min(f[i], f[j]+t)
		}
	}

	return f[n]
}

# Accepted solution for LeetCode #3579: Minimum Steps to Convert String with Operations
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]

// Accepted solution for LeetCode #3579: Minimum Steps to Convert String with Operations
impl Solution {
    pub fn min_operations(word1: String, word2: String) -> i32 {
        let n = word1.len();
        let word1 = word1.as_bytes();
        let word2 = word2.as_bytes();
        let mut f = vec![i32::MAX; n + 1];
        f[0] = 0;

        for i in 1..=n {
            for j in 0..i {
                let a = Self::calc(word1, word2, j, i - 1, false);
                let b = 1 + Self::calc(word1, word2, j, i - 1, true);
                let t = a.min(b);
                f[i] = f[i].min(f[j] + t);
            }
        }

        f[n]
    }

    fn calc(word1: &[u8], word2: &[u8], l: usize, r: usize, rev: bool) -> i32 {
        let mut cnt = [[0i32; 26]; 26];
        let mut res = 0;

        for i in l..=r {
            let j = if rev { r - (i - l) } else { i };
            let a = (word1[j] - b'a') as usize;
            let b = (word2[i] - b'a') as usize;

            if a != b {
                if cnt[b][a] > 0 {
                    cnt[b][a] -= 1;
                } else {
                    cnt[a][b] += 1;
                    res += 1;
                }
            }
        }

        res
    }
}

// Accepted solution for LeetCode #3579: Minimum Steps to Convert String with Operations
function minOperations(word1: string, word2: string): number {
    const n = word1.length;
    const f = Array(n + 1).fill(Number.MAX_SAFE_INTEGER);
    f[0] = 0;

    function calc(l: number, r: number, rev: boolean): number {
        const cnt: number[][] = Array.from({ length: 26 }, () => Array(26).fill(0));
        let res = 0;

        for (let i = l; i <= r; i++) {
            const j = rev ? r - (i - l) : i;
            const a = word1.charCodeAt(j) - 97;
            const b = word2.charCodeAt(i) - 97;

            if (a !== b) {
                if (cnt[b][a] > 0) {
                    cnt[b][a]--;
                } else {
                    cnt[a][b]++;
                    res++;
                }
            }
        }

        return res;
    }

    for (let i = 1; i <= n; i++) {
        for (let j = 0; j < i; j++) {
            const a = calc(j, i - 1, false);
            const b = 1 + calc(j, i - 1, true);
            const t = Math.min(a, b);
            f[i] = Math.min(f[i], f[j] + t);
        }
    }

    return f[n];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^3 + |\Sigma|^2)

Space

O(n + |\Sigma|^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

Minimum Steps to Convert String with Operations

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

State misses one required dimension

Using greedy without proof