Dinner Plate Stacks - Leetcode Solution

Problem Statement:

You are given a list of integers representing the diameter of each dinner plate stack. You need to find the minimum number of moves required to make all the stacks equal, where a move is defined as removing one plate from the top of one stack and placing it on top of another stack.

Example: Input: [1,2,3] Output: 2 Explanation: We can make the stacks equal in 2 moves. First, we remove a plate from the top of stack 3 and place it on top of stack 1. The stacks now look like: [2,2,2]. Then, we remove a plate from the top of stack 3 and place it on top of stack 2. The final stacks look like: [2,3,1]. Note that there are other ways to make the stacks equal in 2 moves.

Solution:

The problem can be solved using a simple mathematical calculation. We need to find the minimum number of moves required to make all the stacks equal. Let's take a closer look at the stacks and try to understand how we can achieve this goal:

[5, 2, 1, 2, 5]

First, we need to find the largest stack. In this example, the largest stack has a diameter of 5. We can start by removing a plate from the top of this stack and place it on top of another stack.

[4, 2, 1, 2, 5]

Now we have one stack with a diameter of 4 and two other stacks with a diameter of 2. We need to repeat this process until all the stacks have the same diameter.

[3, 2, 1, 3, 4]

[2, 2, 1, 4, 3]

[1, 3, 1, 4, 2]

[1, 2, 2, 3, 2]

[1, 1, 3, 2, 3]

[1, 1, 2, 3, 3]

[1, 1, 1, 4, 2]

[1, 1, 1, 3, 3]

[1, 1, 1, 2, 4]

Finally, we have all the stacks with a diameter of 4. The minimum number of moves required to make all the stacks equal is 8.

Let's look at one more example:

[1,2,3]

First, we need to find the largest stack. In this example, the largest stack has a diameter of 3. We can start by removing a plate from the top of this stack and place it on top of another stack.

[1,2,2]

Now we have one stack with a diameter of 1, one stack with a diameter of 2, and one stack with a diameter of 2. We need to repeat this process until all the stacks have the same diameter.

[1,1,3]

[1,2,2]

[2,1,2]

[2,2,1]

[3,1,1]

Finally, we have all the stacks with a diameter of 2. The minimum number of moves required to make all the stacks equal is 2.

So, the algorithm to solve the Dinner Plate Stacks problem is:

1. Find the largest stack.
2. Remove a plate from the top of this stack and place it on top of another stack.
3. Repeat steps 1 and 2 until all the stacks have the same diameter.

The time complexity of this algorithm is O(nlogn) due to the sorting step required to find the largest stack. The space complexity is O(1) as we are not using any extra space apart from the input list.

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