LeetCode #3029 — MEDIUM

Minimum Time to Revert Word to Initial State I

Move from brute-force thinking to an efficient approach using string matching strategy.

The Problem

Problem Statement

You are given a 0-indexed string word and an integer k.

At every second, you must perform the following operations:

Remove the first k characters of word.
Add any k characters to the end of word.

Note that you do not necessarily need to add the same characters that you removed. However, you must perform both operations at every second.

Return the minimum time greater than zero required for word to revert to its initial state.

Example 1:

Input: word = "abacaba", k = 3
Output: 2
Explanation: At the 1st second, we remove characters "aba" from the prefix of word, and add characters "bac" to the end of word. Thus, word becomes equal to "cababac".
At the 2nd second, we remove characters "cab" from the prefix of word, and add "aba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 2 seconds is the minimum time greater than zero required for word to revert to its initial state.

Example 2:

Input: word = "abacaba", k = 4
Output: 1
Explanation: At the 1st second, we remove characters "abac" from the prefix of word, and add characters "caba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 1 second is the minimum time greater than zero required for word to revert to its initial state.

Example 3:

Input: word = "abcbabcd", k = 2
Output: 4
Explanation: At every second, we will remove the first 2 characters of word, and add the same characters to the end of word.
After 4 seconds, word becomes equal to "abcbabcd" and reverts to its initial state.
It can be shown that 4 seconds is the minimum time greater than zero required for word to revert to its initial state.

Constraints:

1 <= word.length <= 50
1 <= k <= word.length
word consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string word and an integer k. At every second, you must perform the following operations: Remove the first k characters of word. Add any k characters to the end of word. Note that you do not necessarily need to add the same characters that you removed. However, you must perform both operations at every second. Return the minimum time greater than zero required for word to revert to its initial state.

Baseline thinking

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

Pattern signal: String Matching

Example 1

"abacaba"
3

Example 2

"abacaba"
4

Example 3

"abcbabcd"
2

Core Insight

What unlocks the optimal approach

Find the longest suffix which is also a prefix and the length is multiple of <code>k</code>.

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

Upper-end input sizes

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 #3029: Minimum Time to Revert Word to Initial State I
class Solution {
    public int minimumTimeToInitialState(String word, int k) {
        int n = word.length();
        for (int i = k; i < n; i += k) {
            if (word.substring(i).equals(word.substring(0, n - i))) {
                return i / k;
            }
        }
        return (n + k - 1) / k;
    }
}

// Accepted solution for LeetCode #3029: Minimum Time to Revert Word to Initial State I
func minimumTimeToInitialState(word string, k int) int {
	n := len(word)
	for i := k; i < n; i += k {
		if word[i:] == word[:n-i] {
			return i / k
		}
	}
	return (n + k - 1) / k
}

# Accepted solution for LeetCode #3029: Minimum Time to Revert Word to Initial State I
class Solution:
    def minimumTimeToInitialState(self, word: str, k: int) -> int:
        n = len(word)
        for i in range(k, n, k):
            if word[i:] == word[:-i]:
                return i // k
        return (n + k - 1) // k

// Accepted solution for LeetCode #3029: Minimum Time to Revert Word to Initial State I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3029: Minimum Time to Revert Word to Initial State I
// class Solution {
//     public int minimumTimeToInitialState(String word, int k) {
//         int n = word.length();
//         for (int i = k; i < n; i += k) {
//             if (word.substring(i).equals(word.substring(0, n - i))) {
//                 return i / k;
//             }
//         }
//         return (n + k - 1) / k;
//     }
// }

// Accepted solution for LeetCode #3029: Minimum Time to Revert Word to Initial State I
function minimumTimeToInitialState(word: string, k: number): number {
    const n = word.length;
    for (let i = k; i < n; i += k) {
        if (word.slice(i) === word.slice(0, -i)) {
            return Math.floor(i / k);
        }
    }
    return Math.floor((n + k - 1) / k);
}

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^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × m) time

O(1) space

At each of the n starting positions in the text, compare up to m characters with the pattern. If a mismatch occurs, shift by one and restart. Worst case (e.g., searching "aab" in "aaaa...a") checks m characters at nearly every position: O(n × m).

KMP / Z-ALGO

O(n + m) time

O(m) space

KMP and Z-algorithm preprocess the pattern in O(m) to build a failure/Z-array, then scan the text in O(n) — never backtracking. Total: O(n + m). Rabin-Karp uses rolling hashes for O(n + m) expected time. All beat the O(n × m) brute force of checking every position.

Shortcut: Preprocessing avoids backtracking → O(n + m). The failure function is the key insight.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.