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 available options. Similarly, 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 until we reach the top most row. 

  • Create an auxiliary array of the bottom most row elements with it corresponding indexes.
  • Sort the auxiliary array.
  • Iterate over the auxiliary array and for each element find maximum element from the row above to be added to the current element to produce maximum sum such that for every A[i][j] the selected element is not A[i-1][j+1] or A[i-1][j-1].
  • Repeat this until reach to the topmost row of the given matrix array.
  • Find the maximum element from the top row to get the maximum sum

Below is the implementation of the above approach. 

C++

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// 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 auxiliary 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 auxiliary 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;
}

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Java

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// Java implementation to find maximum
// sum from top to bottom row with no
// adjacent diagonal elements
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
// Function to find the maximum
// path sum from top to bottom row
static int maxSum(int[][] V,
                  int n, int m)
{
    int ans = 0;
    for(int i = n - 2; i >= 0; --i)
    {
         
        // Create an auxiliary array of next row
        // with the element and it's position
        ArrayList<int[]> aux = new ArrayList<>();
         
        for(int j = 0; j < m; ++j)
        {
            aux.add(new int[]{V[i + 1][j], j});
        }
 
        // Sort the auxiliary array
        Collections.sort(aux, (a, b) -> b[0] - a[0]);
 
        // 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.get(k)[1] - j == 0 ||
                    Math.abs(aux.get(k)[1] - j) > 1)
                {
                    V[i][j] += aux.get(k)[0];
                    break;
                }
            }
        }
    }
 
    // Find the maximum sum
    for(int i = 0; i < m; ++i)
    {
        ans = Math.max(ans, V[0][i]);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int[][] V = { { 1, 2, 3, 4 },
                  { 8, 7, 6, 5 },
                  { 10, 11, 12, 13 } };
                   
    int n = V.length;
    int m = V[0].length;
     
    // Function to find maximum path
    System.out.println(maxSum(V, n, m));
}
}
 
// This code is contributed by offbeat

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Python3

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# 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 auxiliary 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 auxiliary 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

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Output: 

25









 

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