Introduction

In this blog post, we will explore how to solve the LeetCode problem "945" 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 an integer array nums. In one move, you can pick an index i where 0 <= i < nums.length and increment nums[i] by 1. Return the minimum number of moves to make every value in nums unique. The test cases are generated so that the answer fits in a 32-bit integer. Example 1: Input: nums = [1,2,2] Output: 1 Explanation: After 1 move, the array could be [1, 2, 3]. Example 2: Input: nums = [3,2,1,2,1,7] Output: 6 Explanation: After 6 moves, the array could be [3, 4, 1, 2, 5, 7]. It can be shown that it is impossible for the array to have all unique values with 5 or less moves. Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 105

Explanation

Here's the solution to the problem, focusing on efficiency and optimality:

• Sorting: Sort the input array to easily identify adjacent duplicates.

• Greedy Increment: Iterate through the sorted array. If an element is less than or equal to the previous element, increment it until it's unique and count the moves. The target unique value will be one greater than the previous element processed.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) excluding the input array (in-place modification). Some sorting algorithms might use O(log n) space.

Code

    def minIncrementForUnique(nums):
    """
    Calculates the minimum number of moves to make all values in nums unique.

    Args:
        nums: An integer array.

    Returns:
        The minimum number of moves.
    """
    nums.sort()
    moves = 0
    taken = 0  # last taken number
    for i in range(len(nums)):
        if i == 0:
            taken = nums[i]
        else:
            if nums[i] <= taken:
                moves += (taken - nums[i] + 1)
                taken += 1
            else:
                taken = nums[i]

    return moves