LeetCode #3790 — MEDIUM

Smallest All-Ones Multiple

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

You are given a positive integer k.

Find the smallest integer n divisible by k that consists of only the digit 1 in its decimal representation (e.g., 1, 11, 111, ...).

Return an integer denoting the number of digits in the decimal representation of n. If no such n exists, return -1.

Example 1:

Input: k = 3

Output: 3

Explanation:

n = 111 because 111 is divisible by 3, but 1 and 11 are not. The length of n = 111 is 3.

Example 2:

Input: k = 7

Output: 6

Explanation:

n = 111111. The length of n = 111111 is 6.

Example 3:

Input: k = 2

Output: -1

Explanation:

There does not exist a valid n that is a multiple of 2.

Constraints:

2 <= k <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer k. Find the smallest integer n divisible by k that consists of only the digit 1 in its decimal representation (e.g., 1, 11, 111, ...). Return an integer denoting the number of digits in the decimal representation of n. If no such n exists, return -1.

Baseline thinking

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

Pattern signal: Hash Map · Math

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

Notice that <code>n % k</code> should be <code>0</code>.
Build the number digit-by-digit using only modulo <code>k</code>. Start with remainder <code>rem = 1 % k</code>. Repeatedly update <code>rem = (rem * 10 + 1) % k</code> while counting how many <code>1</code>s have been appended.
Continue until <code>rem == 0</code> or a remainder repeats (which indicates a cycle and that no such <code>n</code> exists).

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 #3790: Smallest All-Ones Multiple
class Solution {
    public int minAllOneMultiple(int k) {
        if ((k & 1) == 0) {
            return -1;
        }

        int x = 1 % k;
        int ans = 1;

        for (int i = 0; i < k; i++) {
            x = (x * 10 + 1) % k;
            ans++;
            if (x == 0) {
                return ans;
            }
        }

        return -1;
    }
}

// Accepted solution for LeetCode #3790: Smallest All-Ones Multiple
func minAllOneMultiple(k int) int {
	if k&1 == 0 {
		return -1
	}

	x := 1 % k
	ans := 1

	for i := 0; i < k; i++ {
		x = (x*10 + 1) % k
		ans++
		if x == 0 {
			return ans
		}
	}

	return -1
}

# Accepted solution for LeetCode #3790: Smallest All-Ones Multiple
class Solution:
    def minAllOneMultiple(self, k: int) -> int:
        if k % 2 == 0:
            return -1
        x = 1 % k
        ans = 1
        for _ in range(k):
            x = (x * 10 + 1) % k
            ans += 1
            if x == 0:
                return ans
        return -1

// Accepted solution for LeetCode #3790: Smallest All-Ones Multiple
// 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 #3790: Smallest All-Ones Multiple
// class Solution {
//     public int minAllOneMultiple(int k) {
//         if ((k & 1) == 0) {
//             return -1;
//         }
// 
//         int x = 1 % k;
//         int ans = 1;
// 
//         for (int i = 0; i < k; i++) {
//             x = (x * 10 + 1) % k;
//             ans++;
//             if (x == 0) {
//                 return ans;
//             }
//         }
// 
//         return -1;
//     }
// }

// Accepted solution for LeetCode #3790: Smallest All-Ones Multiple
function minAllOneMultiple(k: number): number {
    if ((k & 1) === 0) {
        return -1;
    }

    let x = 1 % k;
    let ans = 1;

    for (let i = 0; i < k; i++) {
        x = (x * 10 + 1) % k;
        ans++;
        if (x === 0) {
            return ans;
        }
    }

    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(k)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.