# Remove Boxes – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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:23Explanation:[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`

# C++ Remove Boxes LeetCode Solution

class Solution {

public:

int memo[200][200][200] = {};

int removeBoxes(vector<int>& boxes) {

return dp(boxes, 0, boxes.size() – 1, 0);

}

int dp(vector<int>& boxes, int l, int r, int k) {

if (l > r) return 0;

if (memo[l][r][k] > 0) return memo[l][r][k];

int lOrg = l, kOrg = k;while (l+1 <= r && boxes[l] == boxes[l+1]) { // Increase both `l` and `k` if they have consecutive colors with `boxes[l]`

l += 1;

k += 1;

}

int ans = (k+1) * (k+1) + dp(boxes, l+1, r, 0); // Remove all boxes which has the same with `boxes[l]`

for (int m = l + 1; m <= r; ++m) // Try to merge non-contiguous boxes of the same color together

if (boxes[m] == boxes[l])

ans = max(ans, dp(boxes, m, r, k+1) + dp(boxes, l+1, m-1, 0));

return memo[lOrg][r][kOrg] = ans;

}

};

# Java Remove Boxes LeetCode Solution

class Solution {

int[][][] memo;

public int removeBoxes(int[] boxes) {

int n = boxes.length;

memo = new int[n][n][n];

return dp(boxes, 0, n – 1, 0);

}

int dp(int[] boxes, int l, int r, int k) {

if (l > r) return 0;

if (memo[l][r][k] > 0) return memo[l][r][k];

int lOrg = l, kOrg = k;while (l+1 <= r && boxes[l] == boxes[l+1]) { // Increase both `l` and `k` if they have consecutive colors with `boxes[l]`

l += 1;

k += 1;

}

int ans = (k+1) * (k+1) + dp(boxes, l+1, r, 0); // Remove all boxes which has the same with `boxes[l]`

for (int m = l+1; m <= r; ++m) // Try to merge non-contiguous boxes of the same color together

if (boxes[m] == boxes[l])

ans = Math.max(ans, dp(boxes, m, r, k+1) + dp(boxes, l+1, m-1, 0));

return memo[lOrg][r][kOrg] = ans;

}

}

# Python 3 Remove Boxes LeetCode Solution

class Solution {

int[][][] memo;

public int removeBoxes(int[] boxes) {

int n = boxes.length;

memo = new int[n][n][n];

return dp(boxes, 0, n – 1, 0);

}

int dp(int[] boxes, int l, int r, int k) {

if (l > r) return 0;

if (memo[l][r][k] > 0) return memo[l][r][k];

int lOrg = l, kOrg = k;while (l+1 <= r && boxes[l] == boxes[l+1]) { // Increase both `l` and `k` if they have consecutive colors with `boxes[l]`

l += 1;

k += 1;

}

int ans = (k+1) * (k+1) + dp(boxes, l+1, r, 0); // Remove all boxes which has the same with `boxes[l]`

for (int m = l+1; m <= r; ++m) // Try to merge non-contiguous boxes of the same color together

if (boxes[m] == boxes[l])

ans = Math.max(ans, dp(boxes, m, r, k+1) + dp(boxes, l+1, m-1, 0));

return memo[lOrg][r][kOrg] = ans;

}

}

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