chore: add LeetCode daily solution

Author: TechnoBlogger14o3

Medium/2025-11-25-1015-Smallest-Integer-Divisible-by-K/explanation.md

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+# Smallest Integer Divisible by K (Medium)
+
+**Problem ID:** 1015  
+**Date:** 2025-11-25  
+**Link:** https://leetcode.com/problems/smallest-integer-divisible-by-k/
+
+## Approach
+
+To solve the problem of finding the length of the smallest positive integer \( n \) that consists solely of the digit '1' and is divisible by a given integer \( k \), we can utilize a breadth-first search (BFS) approach to explore potential candidates for \( n \).
+
+### Approach:
+
+1. **Understanding the Problem**: 
+   - The number \( n \) can be represented as a sequence of '1's: \( 1, 11, 111, \ldots \). The goal is to find the smallest such number that is divisible by \( k \).
+
+2. **Using Modulo for Divisibility**:
+   - Instead of generating large numbers directly, we can work with their remainders when divided by \( k \). This is because if \( n \equiv 0 \mod k \), then \( n \) is divisible by \( k \).
+
+3. **BFS for Exploration**:
+   - We can treat each number formed by '1's as a node in a graph. Starting from the number '1', we can generate new numbers by appending another '1' to the current number.
+   - For each number represented by its remainder \( r \) (where \( r = n \mod k \)), we can generate two new remainders:
+     - \( (r \times 10 + 1) \mod k \) (this represents appending another '1').
+   - We will keep track of the length of the number (or the number of '1's) in the BFS queue.
+
+4. **Cycle Prevention**:
+   - To avoid processing the same remainder multiple times, we maintain a set of visited remainders.
+
+5. **Termination**:
+   - The BFS continues until we either find a remainder of \( 0 \) (indicating divisibility by \( k \)) or exhaust all possibilities without finding such a remainder.
+
+### Data Structures:
+- A queue to facilitate the BFS process.
+- A set to track visited remainders to prevent cycles.
+
+### Time Complexity:
+- The time complexity is \( O(k) \) since there are at most \( k \) different remainders (from \( 0 \) to \( k-1 \)) that we can encounter.
+
+### Space Complexity:
+- The space complexity is also \( O(k) \) due to the storage of visited remainders.
+
+### Conclusion:
+By using BFS and focusing on remainders rather than constructing large numbers, we efficiently find the length of the smallest integer composed solely of '1's that is divisible by \( k \). If no such integer exists, we return -1.

Medium/2025-11-25-1015-Smallest-Integer-Divisible-by-K/solution.java

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+class Solution {
+    public int smallestRepunitDivByK(int k) {
+        if (k % 2 == 0 || k % 5 == 0) {
+            return -1;
+        }
+        
+        int length = 1;
+        int remainder = 1 % k;
+        
+        while (remainder != 0) {
+            remainder = (remainder * 10 + 1) % k;
+            length++;
+        }
+        
+        return length;
+    }
+}

Medium/2025-11-25-1015-Smallest-Integer-Divisible-by-K/solution.js

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+var smallestRepunitDivByK = function(k) {
+    if (k % 2 === 0 || k % 5 === 0) return -1;
+    
+    let n = 1;
+    let remainder = 1 % k;
+    
+    while (remainder !== 0) {
+        remainder = (remainder * 10 + 1) % k;
+        n++;
+    }
+    
+    return n;
+};

Medium/2025-11-25-1015-Smallest-Integer-Divisible-by-K/solution.py

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+class Solution:
+    def smallestRepunitDivByK(self, k: int) -> int:
+        if k % 2 == 0 or k % 5 == 0:
+            return -1
+        
+        length = 1
+        num = 1 % k
+        
+        while num != 0:
+            num = (num * 10 + 1) % k
+            length += 1
+            
+        return length

README.md

@@ -289,3 +289,4 @@ Through completing the Blind 75 and NeetCode 150, you will have mastered:
 - 2025-11-22 — [Find Minimum Operations to Make All Elements Divisible by Three](https://leetcode.com/problems/find-minimum-operations-to-make-all-elements-divisible-by-three/) (Easy) → `Easy/2025-11-22-3190-Find-Minimum-Operations-to-Make-All-Elements-Divisible-by-Three`
 - 2025-11-23 — [Greatest Sum Divisible by Three](https://leetcode.com/problems/greatest-sum-divisible-by-three/) (Medium) → `Medium/2025-11-23-1262-Greatest-Sum-Divisible-by-Three`
 - 2025-11-24 — [Binary Prefix Divisible By 5](https://leetcode.com/problems/binary-prefix-divisible-by-5/) (Easy) → `Easy/2025-11-24-1018-Binary-Prefix-Divisible-By-5`
+- 2025-11-25 — [Smallest Integer Divisible by K](https://leetcode.com/problems/smallest-integer-divisible-by-k/) (Medium) → `Medium/2025-11-25-1015-Smallest-Integer-Divisible-by-K`