Introduction

In this blog post, we will explore how to solve the LeetCode problem "2537" 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 integer array nums and an integer k, return the number of good subarrays of nums. A subarray arr is good if there are at least k pairs of indices (i, j) such that i < j and arr[i] == arr[j]. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,1,1,1,1], k = 10 Output: 1 Explanation: The only good subarray is the array nums itself. Example 2: Input: nums = [3,1,4,3,2,2,4], k = 2 Output: 4 Explanation: There are 4 different good subarrays: - [3,1,4,3,2,2] that has 2 pairs. - [3,1,4,3,2,2,4] that has 3 pairs. - [1,4,3,2,2,4] that has 2 pairs. - [4,3,2,2,4] that has 2 pairs. Constraints: 1 <= nums.length <= 105 1 <= nums[i], k <= 109

Explanation

Here's the breakdown of the solution:

• Sliding Window: We utilize a sliding window approach to efficiently explore all possible subarrays. The window expands to the right, and we maintain a count of pairs within the current window.
• Pair Counting: A hash map (dictionary in Python) tracks the frequency of each number within the current window. The number of pairs contributed by each number is calculated based on its frequency.

• Window Adjustment: If the number of pairs within the window is less than k, the window expands to the right. If the number of pairs is greater than or equal to k, the window shrinks from the left until the number of pairs falls below k. Each time we shrink the window from the left, we know the current window range was a valid subarray, so we increment the solution count.

• Runtime Complexity: O(n), Storage Complexity: O(m), where n is the length of the input array and m is the number of distinct elements in the input array.

Code

    def goodSubarrays(nums, k):
    """
    Calculates the number of good subarrays in the given array.

    Args:
        nums: A list of integers.
        k: The minimum number of pairs required for a subarray to be good.

    Returns:
        The number of good subarrays.
    """

    n = len(nums)
    count = 0
    left = 0
    pairs = 0
    freq = {}

    for right in range(n):
        # Add the new element to the window
        if nums[right] in freq:
            pairs -= freq[nums[right]] * (freq[nums[right]] - 1) // 2
            freq[nums[right]] += 1
            pairs += freq[nums[right]] * (freq[nums[right]] - 1) // 2

        else:
            freq[nums[right]] = 1
            pairs += freq[nums[right]] * (freq[nums[right]] - 1) // 2

        # Shrink the window if the number of pairs is greater than or equal to k
        while pairs >= k:
            count += n - right

            if nums[left] in freq:
                pairs -= freq[nums[left]] * (freq[nums[left]] - 1) // 2
                freq[nums[left]] -= 1
                if freq[nums[left]] > 0:
                    pairs += freq[nums[left]] * (freq[nums[left]] - 1) // 2
                else:
                    del freq[nums[left]]

            left += 1

    return count