LeetCode #686 — MEDIUM

Repeated String Match

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

The Problem

Problem Statement

Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b to be a substring of a after repeating it, return -1.

Notice: string "abc" repeated 0 times is "", repeated 1 time is "abc" and repeated 2 times is "abcabc".

Example 1:

Input: a = "abcd", b = "cdabcdab"
Output: 3
Explanation: We return 3 because by repeating a three times "abcdabcdabcd", b is a substring of it.

Example 2:

Input: a = "a", b = "aa"
Output: 2

Constraints:

1 <= a.length, b.length <= 10⁴
a and b consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b to be a substring of a after repeating it, return -1. Notice: string "abc" repeated 0 times is "", repeated 1 time is "abc" and repeated 2 times is "abcabc".

Baseline thinking

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

Pattern signal: String Matching

Example 1

"abcd"
"cdabcdab"

Example 2

"a"
"aa"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #686: Repeated String Match
class Solution {
    public int repeatedStringMatch(String a, String b) {
        int m = a.length(), n = b.length();
        int ans = (n + m - 1) / m;
        StringBuilder t = new StringBuilder(a.repeat(ans));
        for (int i = 0; i < 3; ++i) {
            if (t.toString().contains(b)) {
                return ans;
            }
            ++ans;
            t.append(a);
        }
        return -1;
    }
}

// Accepted solution for LeetCode #686: Repeated String Match
func repeatedStringMatch(a string, b string) int {
	m, n := len(a), len(b)
	ans := (n + m - 1) / m
	t := strings.Repeat(a, ans)
	for i := 0; i < 3; i++ {
		if strings.Contains(t, b) {
			return ans
		}
		ans++
		t += a
	}
	return -1
}

# Accepted solution for LeetCode #686: Repeated String Match
class Solution:
    def repeatedStringMatch(self, a: str, b: str) -> int:
        m, n = len(a), len(b)
        ans = ceil(n / m)
        t = [a] * ans
        for _ in range(3):
            if b in ''.join(t):
                return ans
            ans += 1
            t.append(a)
        return -1

// Accepted solution for LeetCode #686: Repeated String Match
struct Solution;

impl Solution {
    fn repeated_string_match(a: String, b: String) -> i32 {
        let mut s = String::new();
        let n = a.len();
        let m = b.len();
        if n == 0 || m == 0 {
            return -1;
        }
        let mut k = m / n;
        if k * n < m {
            k += 1;
        }
        for _ in 0..k {
            s += &a;
        }
        if s.contains(&b) {
            return k as i32;
        }
        s += &a;
        if s.contains(&b) {
            return (k + 1) as i32;
        }
        -1
    }
}

#[test]
fn test() {
    let a = "abcd".to_string();
    let b = "cdabcdab".to_string();
    assert_eq!(Solution::repeated_string_match(a, b), 3);
}

// Accepted solution for LeetCode #686: Repeated String Match
function repeatedStringMatch(a: string, b: string): number {
    const m: number = a.length,
        n: number = b.length;
    let ans: number = Math.ceil(n / m);
    let t: string = a.repeat(ans);

    for (let i = 0; i < 3; i++) {
        if (t.includes(b)) {
            return ans;
        }
        ans++;
        t += a;
    }

    return -1;
}

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 + m)

Space

O(m)

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.