Find maximum sum from top to bottom row with no adjacent diagonal elements

Given a matrix A[][] of N * M, the task is to find the maximum sum from the top row to the bottom row after selecting one element from each row with no adjacent diagonal element.

Examples:

Input: A = { {1, 2, 3, 4}, {8, 7, 6, 5}, {10, 11, 12, 13} }
Output: 25
Explanation:
Selected elements to give maximum sum –
Row 0 = 4
Row 1 = 8
Row 2 = 13
Sum = 25



Input: A = { {1, 6}, {5, 3}, {11, 7} }
Output: 17
Explanation:
Selected elements to give maximum sum –
Row 0 = 1
Row 1 = 5
Row 2 = 11

Explanation: For Selecting any element if we have selected A[i][j], then elements A[i+1][j+1] and A[i+1][j-1] cannot be selected.
In the Given example Select the maximum element from the top-row where 4 is maximum in this case, then element A[1][2] cannot be selected which is 6, select the element 8 which maximum from the availiable options. Similiary element 11 cannot be selected from the 3rd row. Select the element 13 to get the maximum sum which is 25.

Naive Approach: Generate all the combinations of N elements after choosing 1 element from every row and select the combination which produces maximum sum.
Efficient Approach: The idea is to use the concept of Dynamic programming in bottom up manner. Begin with the bottom most row of the given matrix and repeat the below process untill we reach the top most row.

Below is the implementation of the above approach.

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to find
// maximum sum from top to bottom
// row with no adjacent diagonal elements
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the maximum
// path sum from top to bottom row
int maxSum(vector<vector<int> >& V,
                       int n, int m){
    int ans = 0;
    for (int i = n - 2; i >= 0; --i) {
        // Create an auxilary array of next row
        // with the element and it's position
        vector<pair<int, int> > aux;
  
        for (int j = 0; j < m; ++j) {
            aux.push_back({ V[i + 1][j],
                                    j });
        }
  
        // Sort the auxilary array
        sort(aux.begin(), aux.end());
        reverse(aux.begin(), aux.end());
  
        // Find maximum from row above to
        // be added to the current element
        for (int j = 0; j < m; ++j) {
              
            // Find the maximum element from
            // the next row that can be added
            // to current row element
            for (int k = 0; k < m; ++k) {
                if (aux[k].second - j == 0 ||
                   abs(aux[k].second - j) > 1) {
                    V[i][j] += aux[k].first;
                    break;
                }
            }
        }
    }
  
    // Find the maximum sum
    for (int i = 0; i < m; ++i) {
        ans = max(ans, V[0][i]);
    }
    return ans;
}
  
// Driver Code
int main()
{
  
    vector<vector<int> > V{{ 1, 2, 3, 4 },
                           { 8, 7, 6, 5 },
                           { 10, 11, 12, 13 }};
    int n = V.size();
    int m = V[0].size();
  
    // Function to find maximum path
    cout << maxSum(V, n, m);
  
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation to find
# maximum sum from top to bottom
# row with no adjacent diagonal elements
  
# Function to find the maximum
# path sum from top to bottom row
def maxSum(V, n, m):
    ans = 0
    for i in range(n - 2, -1, -1):
          
        # Create an auxilary array of next row
        # with the element and it's position
        aux = []
  
        for j in range(m):
            aux.append([V[i + 1][j], j])
  
        # Sort the auxilary array
        aux = sorted(aux)
        aux = aux[::-1]
  
        # Find maximum from row above to
        # be added to the current element
        for j in range(m):
  
            # Find the maximum element from
            # the next row that can be added
            # to current row element
            for k in range(m):
                if (aux[k][1] - j == 0 or
                abs(aux[k][1] - j) > 1):
                    V[i][j] += aux[k][0]
                    break
  
    # Find the maximum sum
    for i in range(m):
        ans = max(ans, V[0][i])
  
    return ans
  
# Driver Code
if __name__ == '__main__':
  
    V=[[ 1, 2, 3, 4 ],
    [ 8, 7, 6, 5 ],
    [ 10, 11, 12, 13]]
    n = len(V)
    m = len(V[0])
  
    # Function to find maximum path
    print(maxSum(V, n, m))
  
# This code is contributed by mohit kumar 29
chevron_right

Output:
25

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details





Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29

Article Tags :