Moving Stones Until Consecutive Ii - Leetcode Solution

Problem Statement:

You have an array of n stones, and a moving window (size k) scanning across the array. The window starts at index 0 and slides right until its right boundary reaches the end of the array.

Each time the window moves right by one position, you must decide to either add or subtract the stone at this position to/from your total weight.

The goal is to minimize the difference between your total weight and the target weight, by changing the weight of some stones as many times as necessary.

Return the smallest possible difference between your total weight and the target weight.

Note that some of the stones may have zero weight.

Solution:

This problem is similar to Moving Stones Until Consecutive I but with a twist that we have to maximize the score. To find the maximum score, we need to first find the total score and then iterate through the array with a sliding window of size k and find the maximum possible score by adding/subtracting stones from their current positions. This can be done by keeping track of the cumulative sum of weights of stones in the window and then using it to find the score.

Let's break down the steps required to solve this problem:

Step 1: Calculate the total score

To calculate the total score, we can iterate through the array and add up the weights of all the stones. This will give us the total weight of all the stones. Since we want to maximize the score, we will use this total weight as the target weight.

total_weight = sum(stones)

Step 2: Find the maximum possible score for each sliding window

Next, we will iterate through the array with a sliding window of size k and find the maximum possible score for each window.

To do this, we will initialize two variables - current_weight and max_score. current_weight will be used to keep track of the cumulative weight of the stones in the current window. We will initialize it to the sum of the weights of the first k stones.

max_score will be used to keep track of the maximum score we can get for the current window. We will initialize it to the absolute difference between the current_weight and the target_weight.

for i in range(k): current_weight += stones[i]

max_score = abs(current_weight - target_weight)

Next, we will slide the window to the right by one position and update the current_weight and max_score variables accordingly. We will subtract the weight of the stone that just left the window and add the weight of the stone that just entered the window to the current_weight. We will then calculate the new score as the absolute difference between the current_weight and the target_weight. If this score is greater than the current max_score, we will update max_score.

for i in range(k, len(stones)): current_weight += (stones[i] - stones[i-k]) score = abs(current_weight - target_weight) if score > max_score: max_score = score

Step 3: Return the maximum possible score

After iterating through all the windows, we will have the maximum possible score for each window. We can then return the maximum of these scores as the answer to the problem.

return max_score

Time Complexity:

The time complexity of this solution is O(n), where n is the length of the input array. This is because we iterate through the array once to calculate the total weight of all stones, and then iterate through all the windows of size k. Since each window is processed in constant time, the total time complexity is linear.

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