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, 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

    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