# Solving Leetcode Interviews in Seconds with AI: Smallest Index With Equal Value


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2057" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a 0-indexed integer array nums, return the smallest index i of nums such that i mod 10 == nums[i], or -1 if such index does not exist. x mod y denotes the remainder when x is divided by y.   Example 1:  Input: nums = [0,1,2] Output: 0 Explanation:  i=0: 0 mod 10 = 0 == nums[0]. i=1: 1 mod 10 = 1 == nums[1]. i=2: 2 mod 10 = 2 == nums[2]. All indices have i mod 10 == nums[i], so we return the smallest index 0.  Example 2:  Input: nums = [4,3,2,1] Output: 2 Explanation:  i=0: 0 mod 10 = 0 != nums[0]. i=1: 1 mod 10 = 1 != nums[1]. i=2: 2 mod 10 = 2 == nums[2]. i=3: 3 mod 10 = 3 != nums[3]. 2 is the only index which has i mod 10 == nums[i].  Example 3:  Input: nums = [1,2,3,4,5,6,7,8,9,0] Output: -1 Explanation: No index satisfies i mod 10 == nums[i].    Constraints:  1 <= nums.length <= 100 0 <= nums[i] <= 9  

	# Explanation
	Here's an efficient solution to find the smallest index `i` in the `nums` array such that `i mod 10 == nums[i]`.

*   **Iterate and Check:** Loop through the `nums` array, checking the condition `i % 10 == nums[i]` for each index `i`.
*   **Early Return:** If the condition is met, immediately return the index `i` as we're looking for the smallest such index.
*   **Handle No Match:** If the loop completes without finding a matching index, return -1.

*   **Time Complexity:** O(n), where n is the length of the nums array. **Space Complexity:** O(1).

	
	# Code
	```python
	def smallest_equal(nums):
    """
    Finds the smallest index i such that i mod 10 == nums[i].

    Args:
        nums: A list of integers.

    Returns:
        The smallest index i such that i mod 10 == nums[i], or -1 if it doesn't exist.
    """
    for i in range(len(nums)):
        if i % 10 == nums[i]:
            return i
    return -1
	```