Maximum path sum in a triangle.

We have given numbers in form of a triangle, by starting at the top of the triangle and moving to adjacent numbers on the row below, find the maximum total from top to bottom. 
Examples : 

Input : 
 3
 7 4
 2 4 6
8 5 9 3
Output : 23
Explanation : 3 + 7 + 4 + 9 = 23 

Input :
 8
 -4 4
 2 2 6
1 1 1 1
Output : 19
Explanation : 8 + 4 + 6 + 1 = 19

We can go through the brute force by checking every possible path but that is much time taking so we should try to solve this problem with the help of dynamic programming which reduces the time complexity. 

Implementation of Recursive Approach:

14

Complexity Analysis:

• Time Complexity: O(2^N*N)
• Auxiliary Space:  O(N)

If we should left shift every element and put 0 at each empty position to make it a regular matrix, then our problem looks like minimum cost path. 
So, after converting our input triangle elements into a regular matrix we should apply the dynamic programming concept to find the maximum path sum. 

DP Top-Down
Since there are overlapping subproblems, we can avoid the repeated work done in method 1 by storing the min-cost path calculated so far using top-down approach

14

Complexity Analysis:

• Time Complexity:  O(m*n) where m = no of rows and n = no of columns
• Auxiliary Space:  O(n²)

DP(Bottom - UP)
Since there are overlapping subproblems, we can avoid the repeated work done in method 1 by storing the min-cost path calculated so far using the bottom-up approach thus reducing stack space

14

Complexity Analysis:

• Time Complexity: O(m*n) where m = no of rows and n = no of columns
• Auxiliary Space: O(n²)

Space Optimization (Without Changning input matrix)
We do not need a 2d matrix we only need a 1d array that stores the minimum of the immediate next column  and thus we can reduce space

14

Complexity Analysis:

• Time Complexity: O(m*n) where m = no of rows and n = no of columns
• Auxiliary Space:  O(n)

Space Optimization (Changing input matrix)
Applying, DP in bottom-up manner we should solve our problem as: 
Example: 

 3
 7 4
 2 4 6
8 5 9 3

Step 1 :
3 0 0 0
7 4 0 0
2 4 6 0
8 5 9 3

Step 2 :
3 0 0 0
7 4 0 0
10 13 15 0

Step 3 :
3 0 0 0
20 19 0 0

Step 4:
23 0 0 0

output : 23

14

Complexity Analysis:

• Time Complexity: O(m*n) where m = no of rows and n = no of columns
• Auxiliary Space:  O(1)