Introduction

In this blog post, we will explore how to solve the LeetCode problem "2963" 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

You are given a 0-indexed array nums consisting of positive integers. A partition of an array into one or more contiguous subarrays is called good if no two subarrays contain the same number. Return the total number of good partitions of nums. Since the answer may be large, return it modulo 109 + 7. Example 1: Input: nums = [1,2,3,4] Output: 8 Explanation: The 8 possible good partitions are: ([1], [2], [3], [4]), ([1], [2], [3,4]), ([1], [2,3], [4]), ([1], [2,3,4]), ([1,2], [3], [4]), ([1,2], [3,4]), ([1,2,3], [4]), and ([1,2,3,4]). Example 2: Input: nums = [1,1,1,1] Output: 1 Explanation: The only possible good partition is: ([1,1,1,1]). Example 3: Input: nums = [1,2,1,3] Output: 2 Explanation: The 2 possible good partitions are: ([1,2,1], [3]) and ([1,2,1,3]). Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109

Explanation

Here's a breakdown of the solution and the code:

• High-Level Approach:

  • Iterate through the array, keeping track of the last occurrence of each number using a dictionary (or hash map).
  • Maintain a "furthest reach" pointer, indicating the maximum index that must be included in the current subarray to avoid repeating numbers in future subarrays.
  • The number of good partitions is 2^{(number of subarrays - 1)}, because each split point has two options: to split or not to split.

• Complexity:

  • Runtime Complexity: O(n), where n is the length of the input array.
  • Storage Complexity: O(n), in the worst case, to store the last occurrences of all unique numbers.

Code

    def numberOfGoodPartitions(nums):
    """
    Calculates the number of good partitions of an array.

    Args:
        nums: A list of positive integers.

    Returns:
        The total number of good partitions modulo 10^9 + 7.
    """

    n = len(nums)
    last_occurrence = {}
    for i in range(n):
        last_occurrence[nums[i]] = i

    furthest_reach = 0
    subarrays = 0
    for i in range(n):
        furthest_reach = max(furthest_reach, last_occurrence[nums[i]])
        if i == furthest_reach:
            subarrays += 1

    if subarrays == 0:
        return 0  # Handle edge case when input is empty

    MOD = 10**9 + 7
    if subarrays == 1 :
        return 1
    else:
        return pow(2, subarrays - 1, MOD)