# Teaching Kids Programming – Largest Sum of Non-Adjacent Numbers (2D Version – Disappearing Coins in a Matrix)

Teaching Kids Programming – Largest Sum of Non-Adjacent Numbers (2D Version – Disappearing Coins in a Matrix) | ninjasquad

Teaching Kids Programming: Videos on Data Structures and Algorithms

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.

Constraints
n, m ≤ 250 where n and m are the number of rows and columns in matrix
Example 1
Input

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

Output
22
Explanation
We can pick cells with the coins 7, 5, and 8 and 2.

Hints:
You cannot take two adjacent rows/cols. Does this sound familiar?
Solve the problem for each single row and get maximum points for each row, then create a new array with these values and perform the same operation on the new array

### Largest Sum of Non-Adjacent Numbers in a Matrix via Dynamic Programming Algorithm

The 1D version of this problem: maximize the sum of non-adjacent numbers (when we pick i-th number, two numbers next to it (i-1, i+1) disappear). We can solve this via Dynamic Programming Algorithm:

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We can implement this in a few ways: Top Down Dynamic Programming (Recursion via Memoization aka the @cache keyword to use hash table to store):

 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 ``` ```def f(nums):     n = len(nums)     if n == 0:         return 0         @cache     def g(i):         if i == 0:             return nums[0]         if i == 1:             return max(nums[0], nums[1])         return max(g(i - 2) + nums[i], g(i - 1))           return g(len(nums) - 1)```
```def f(nums):
n = len(nums)
if n == 0:
return 0

@cache
def g(i):
if i == 0:
return nums[0]
if i == 1:
return max(nums[0], nums[1])
return max(g(i - 2) + nums[i], g(i - 1))

return g(len(nums) - 1)```

We can use the Bottom Up to implement the Dynamic Programming Equation, specifically using array to store the DP values:

 ```1 2 3 4 5 6 7 8 9 10 11 12 ``` ```def f(nums):     n = len(nums)     if not n:         return 0     if n == 1:         return nums[0]     if n == 2:         return max(nums)     dp = [nums[0], max(nums[0], nums[1])]     for i in range(2, n):         dp.append(max(dp[-2] + nums[i], dp[-1]))     return dp[-1]```
```def f(nums):
n = len(nums)
if not n:
return 0
if n == 1:
return nums[0]
if n == 2:
return max(nums)
dp = [nums[0], max(nums[0], nums[1])]
for i in range(2, n):
dp.append(max(dp[-2] + nums[i], dp[-1]))
return dp[-1]```

And, similar to Fibonacci Numbers, we only need to keep the previous 2 items, and thus we can optimize the space to O(1):

 ```1 2 3 4 5 6 7 8 9 10 ``` ```def f(nums):     n = len(nums)     if n == 0:         return 0     if n == 1:         return nums[0]     a, b = 0, 0     for i in nums:         a, b = b, max(a + i, b)     return b```
```def f(nums):
n = len(nums)
if n == 0:
return 0
if n == 1:
return nums[0]
a, b = 0, 0
for i in nums:
a, b = b, max(a + i, b)
return b```

With this f function() to solve the 1-D version, we then obtain the list of numbers aka largest sums for each row, and since all the numbers at neighbouring rows disappear, row-1 and row+1, it means that we can apply again the f function to return the largest sum.

 ```1 ``` `return f([f(x) for x in M])`
`return f([f(x) for x in M])`

The overall time complexity is O(RC) where R is the number of rows, and C is the number of columns respectively, and the space complexity (if we have the optimal implementation of the last f function which has O(1) space), is O(R) as we need to obtain the 1-D largest sum for each row, and store them in an array before we apply again the f function to it.

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