Collecting Disappearing Coins - Amazon Top Interview Questions

Problem Statement :

You are given a two-dimensional list of integers matrix where each matrix[r][c] represents the number of coins in that cell. When you pick up coins on matrix[r][c], all the coins on row r - 1 and r + 1 disappear, as well as the coins at the two cells matrix[r][c + 1] and matrix[r][c - 1]. Return the maximum number of coins that you can collect.


n, m ≤ 250 where n and m are the number of rows and columns in matrix

Example 1


matrix = [
    [1, 7, 6, 5],
    [9, 9, 3, 1],
    [4, 8, 1, 2]




We can pick cells with the coins 7, 5, and 8 and 2.

Solution :


                        Solution in C++ :

int best(vector<int>& v) {
    // return the maximum sum of elements given that you can't choose adjacent elements
    int n = v.size();
    // dp[i] is the maximum sum you can get considering exactly the first i elements
    vector<int> dp(n + 1);
    for (int i = 0; i < n; i++) {
        // don't choose element i
        dp[i + 1] = max(dp[i + 1], dp[i]);
        // choose element i
        dp[i + 1] = max(dp[i + 1], v[i] + (i == 0 ? 0 : dp[i - 1]));
    return dp[n];

int solve(vector<vector<int>>& matrix) {
    vector<int> rows;
    for (auto& row : matrix) {
    return best(rows);

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[][] matrix) {
        if (matrix == null || matrix.length == 0)
            return 0;
        int[] arr = new int[matrix.length];

        for (int i = 0; i < matrix.length; i++) arr[i] = getMaximumNonAdjacentSum(matrix[i]);

        return getMaximumNonAdjacentSum(arr);

    private int getMaximumNonAdjacentSum(int[] arr) {
        if (arr.length == 0)
            return 0;
        if (arr.length == 1)
            return arr[0];
        if (arr.length == 2)
            return Math.max(arr[0], arr[1]);

        int[] dp = new int[arr.length];
        dp[0] = arr[0];
        dp[1] = Math.max(arr[0], arr[1]);

        for (int i = 2; i < arr.length; i++) dp[i] = Math.max(dp[i - 1], arr[i] + dp[i - 2]);
        return dp[arr.length - 1];

                        Solution in Python : 
class Solution:
    def best(self, l):
        canadd = 0
        cannotadd = 0
        for elem in l:
            if isinstance(elem, list):
                elem =
            canadd, cannotadd = cannotadd, max(cannotadd, canadd + elem)
        return cannotadd

    def solve(self, matrix):

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