# Solving Leetcode Interviews in Seconds with AI: Remove Boxes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "546" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 several boxes with different colors represented by different positive numbers. You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (i.e., composed of k boxes, k >= 1), remove them and get k * k points. Return the maximum points you can get.   Example 1:  Input: boxes = [1,3,2,2,2,3,4,3,1] Output: 23 Explanation: [1, 3, 2, 2, 2, 3, 4, 3, 1]  ----> [1, 3, 3, 4, 3, 1] (3*3=9 points)  ----> [1, 3, 3, 3, 1] (1*1=1 points)  ----> [1, 1] (3*3=9 points)  ----> [] (2*2=4 points)  Example 2:  Input: boxes = [1,1,1] Output: 9  Example 3:  Input: boxes = [1] Output: 1    Constraints:  1 <= boxes.length <= 100 1 <= boxes[i] <= 100  

	# Explanation
	Here's a breakdown of the approach and the Python code:

*   **Dynamic Programming with Memoization:** The core idea is to use dynamic programming to explore all possible box removal sequences. A 3D DP table `dp[l][r][k]` stores the maximum points obtainable from the sub-array `boxes[l:r+1]` when there are `k` boxes of the same color as `boxes[l]` already attached to the left of `boxes[l]`. This memoization avoids redundant calculations.

*   **State Transition:** The key is how to transition between states. We have two main choices at each state `(l, r, k)`: either remove the `k+1` boxes of color `boxes[l]` immediately, or find another position `m` (where `l < m <= r`) such that `boxes[l] == boxes[m]` and combine the `k+1` boxes at `l` with the boxes at `m`, effectively creating a larger group and deferring the removal of the `l`-colored boxes.

*   **Optimization:** Memoization avoids recomputation of overlapping subproblems.

*   **Complexity:**
    *   Runtime Complexity: O(n^4)
    *   Storage Complexity: O(n^3)

	
	# Code
	```python
	def removeBoxes(boxes):
    n = len(boxes)
    dp = {}  # Memoization table (l, r, k) -> max_points

    def solve(l, r, k):
        if l > r:
            return 0
        if (l, r, k) in dp:
            return dp[(l, r, k)]

        # Optimization: Skip consecutive same-color boxes on the left
        while l + 1 <= r and boxes[l] == boxes[l + 1]:
            l += 1
            k += 1

        # Option 1: Remove the boxes[l] group immediately
        res = (k + 1) * (k + 1) + solve(l + 1, r, 0)

        # Option 2: Find another group with the same color
        for m in range(l + 1, r + 1):
            if boxes[l] == boxes[m]:
                res = max(res, solve(l + 1, m - 1, 0) + solve(m, r, k + 1))

        dp[(l, r, k)] = res
        return res

    return solve(0, n - 1, 0)
	```