Introduction

In this blog post, we will explore how to solve the LeetCode problem "1394" 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 an array of integers arr, a lucky integer is an integer that has a frequency in the array equal to its value. Return the largest lucky integer in the array. If there is no lucky integer return -1. Example 1: Input: arr = [2,2,3,4] Output: 2 Explanation: The only lucky number in the array is 2 because frequency[2] == 2. Example 2: Input: arr = [1,2,2,3,3,3] Output: 3 Explanation: 1, 2 and 3 are all lucky numbers, return the largest of them. Example 3: Input: arr = [2,2,2,3,3] Output: -1 Explanation: There are no lucky numbers in the array. Constraints: 1 <= arr.length <= 500 1 <= arr[i] <= 500

Explanation

Here's a solution to find the largest lucky integer in an array, along with explanations:

• High-Level Approach:

  • Use a frequency counting array to efficiently determine the count of each number in the input array.
  • Iterate backwards from the largest possible lucky number (500 based on constraints) down to 1. Check if the count of the current number equals the number itself.
  • Return the first lucky number found during the backwards iteration. If none are found, return -1.

• Complexity:

  • Runtime Complexity: O(N), where N is the length of the input array.
  • Storage Complexity: O(1) - The frequency array has a fixed size of 501 regardless of the input size.

Code

    def find_lucky(arr):
    """
    Finds the largest lucky integer in an array.

    Args:
        arr: A list of integers.

    Returns:
        The largest lucky integer in the array, or -1 if there is no lucky integer.
    """

    frequency = [0] * 501  # Initialize frequency array (values 1 to 500)

    # Count the frequency of each number
    for num in arr:
        frequency[num] += 1

    # Iterate backwards from 500 to 1
    for i in range(500, 0, -1):
        if frequency[i] == i:
            return i

    return -1  # No lucky integer found