Maximise matrix sum by following the given Path

Given a 2d-matrix mat[][] consisting of positive integers, the task is to find the maximum score we can reach if we have to go to cell mat[0][N – 1] starting from mat[0][0]. We have to cover the matrix in two phases:

  1. Phase 1: If we are at cell mat[i][j] then we can only go to cells mat[i][j + 1] or mat[i + 1][j] without changing the phase else we can go to cell mat[i – 1][j] and switch to phase 2.
  2. Phase 2: If we are at cell mat[i][j] then we can only go to cells mat[i][j + 1] or mat[i – 1][j].
    We can not go out of bounds and the switching between phases will occur at most once.

Note: We may be able to visit the cells of column in which we switch the phase twice.

Examples:

Input: mat[][] = {
{1, 1, 1},
{1, 5, 1},
{1, 1, 1}}
Output: 15
Path: (0, 0) -> (0, 1) -> (1, 1) -> (2, 1) -> (1, 1) -> (0, 1) -> (0, 2)
Phase 1: (0, 0) -> (0, 1) -> (1, 1) -> (2, 1)
Phase 2: (2, 1) -> (1, 1) -> (0, 1) -> (0, 2)
Total score = 1 + 1 + 5 + 1 + 5 + 1 + 1 = 15

Input: mat[][] = {
{1, 1, 1},
{1, 1, 1},
{1, 1, 1}}
Output: 7

Prerequisite: Maximum sum path in a matrix from top to bottom.

Approach: This problem can be solved using dynamic programming Let’s suppose we are at the cell mat[i][j] and S is the shrink factor. If its 0, then we are in phase-1 (growing phase) else we are in phase-2 (shrinking phase). Thus, S can take only two values. Set S = 1 as soon as we take a step down.
dp[i][j][S] will be defined as maximum score we can get if we get from cell mat[i][j] to mat[0][N – 1]. Now, let’s discuss the paths we can take.
Let us assume we are at cell mat[i][j].

  • Case 1: When S = 0 then we have three possible paths,
    1. Go to cell mat[i + 1][j].
    2. Go to cell mat[i][j + 1].
    3. Go to cell mat[i – 1][j] and update S = 1.
  • Case 2: When S = 1 then we have two possible paths,
    1. Go to cell mat[i – 1][j].
    2. Go to cell mat[i][j + 1].

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define n 3
  
// To store the states of the DP
int dp[n][n][2];
bool v[n][n][2];
  
// Function to return the maximum
// of the three integers
int max(int a, int b, int c)
{
    int m = a;
    if (m < b)
        m = b;
    if (m < c)
        m = c;
    return m;
}
  
// Function to return the maximum score
int maxScore(int arr[][n], int i, int j, int s)
{
    // Base cases
    if (i > n - 1 || i < 0 || j > n - 1)
        return 0;
    if (i == 0 and j == n - 1)
        return arr[i][j];
  
    // If the state has already
    // been solved then return it
    if (v[i][j][s])
        return dp[i][j][s];
  
    // Marking the state as solved
    v[i][j][s] = 1;
  
    // Growing phase
    if (!s)
        dp[i][j][s] = arr[i][j] + max(maxScore(arr, i + 1, j, s),
                                      maxScore(arr, i, j + 1, s),
                                      maxScore(arr, i - 1, j, !s));
  
    // Shrinking phase
    else
        dp[i][j][s] = arr[i][j] + max(maxScore(arr, i - 1, j, s),
                                      maxScore(arr, i, j + 1, s));
  
    // Returning the solved state
    return dp[i][j][s];
}
  
// Driver code
int main()
{
    int arr[n][n] = { { 1, 1, 1 },
                      { 1, 5, 1 },
                      { 1, 1, 1 } };
  
    cout << maxScore(arr, 0, 0, 0);
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
  
    static int n = 3;
  
    // To store the states of the DP
    static int[][][] dp = new int[n][n][2];
    static boolean[][][] v = new boolean[n][n][2];
  
    // Function to return the maximum
    // of the three integers
    static int max(int a, int b, int c)
      
    {
        int m = a;
        if (m < b)
        {
            m = b;
        }
        if (m < c) 
        {
            m = c;
        }
        return m;
    }
  
// Function to return the maximum score
    static int maxScore(int arr[][], int i, int j, int s) 
    {
        // Base cases
        if (i > n - 1 || i < 0 || j > n - 1
        {
            return 0;
        }
        if (i == 0 && j == n - 1
        {
            return arr[i][j];
        }
  
        // If the state has already
        // been solved then return it
        if (v[i][j][s])
          
          
        {
            return dp[i][j][s];
        }
  
        // Marking the state as solved
        v[i][j][s] = true;
  
        // Growing phase
        if (s != 1)
        {
            dp[i][j][s] = arr[i][j] + Math.max(maxScore(arr, i + 1, j, s),
                                    Math.max(maxScore(arr, i, j + 1, s),
                                    maxScore(arr, i - 1, j, (s==1)?0:1)));
        } // Shrinking phase
        else 
        {
            dp[i][j][s] = arr[i][j] + Math.max(maxScore(arr, i - 1, j, s),
                    maxScore(arr, i, j + 1, s));
        }
  
        // Returning the solved state
        return dp[i][j][s];
    }
  
    // Driver code
    public static void main(String args[]) 
    {
        int arr[][] = {{1, 1, 1},
        {1, 5, 1},
        {1, 1, 1}};
  
        System.out.println(maxScore(arr, 0, 0, 0));
  
    }
}
  
/* This code contributed by PrinciRaj1992 */

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Python3

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# Python3 implementation of the approach 
import numpy as np 
  
n = 3
  
# To store the states of the DP 
dp = np.zeros((n,n,2)); 
v = np.zeros((n,n,2)); 
  
# Function to return the maximum 
# of the three integers 
def max_three(a, b, c) :
  
    m = a; 
    if (m < b) :
        m = b;
          
    if (m < c) :
        m = c; 
          
    return m; 
  
  
# Function to return the maximum score 
def maxScore(arr, i, j, s) : 
  
    # Base cases 
    if (i > n - 1 or i < 0 or j > n - 1) :
        return 0
          
    if (i == 0 and j == n - 1) :
        return arr[i][j]; 
  
    # If the state has already 
    # been solved then return it 
    if (v[i][j][s]) :
        return dp[i][j][s]; 
  
    # Marking the state as solved 
    v[i][j][s] = 1
  
    # Growing phase 
    if (not bool(s)) : 
        dp[i][j][s] = arr[i][j] + max_three(maxScore(arr, i + 1, j, s), 
                                    maxScore(arr, i, j + 1, s), 
                                    maxScore(arr, i - 1, j, not bool(s))); 
  
    # Shrinking phase 
    else :
        dp[i][j][s] = arr[i][j] + max(maxScore(arr, i - 1, j, s), 
                                    maxScore(arr, i, j + 1, s)); 
  
    # Returning the solved state 
    return dp[i][j][s]; 
  
  
# Driver code 
if __name__ == "__main__" :
  
    arr = [ [ 1, 1, 1 ], 
                    [ 1, 5, 1 ], 
                    [ 1, 1, 1 ] ,
                    ]; 
  
    print(maxScore(arr, 0, 0, 0)); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
    static int n = 3;
      
    // To store the states of the DP
    static int[,,] dp = new int[n,n,2];
    static bool[,,] v = new bool[n,n,2];
  
    // Function to return the maximum
    // of the three integers
    static int max(int a, int b, int c)
      
    {
        int m = a;
        if (m < b)
        {
            m = b;
        }
        if (m < c) 
        {
            m = c;
        }
        return m;
    }
  
    // Function to return the maximum score
    static int maxScore(int [,]arr, int i, int j, int s) 
    {
        // Base cases
        if (i > n - 1 || i < 0 || j > n - 1) 
        {
            return 0;
        }
        if ((i == 0) && (j == (n - 1))) 
        {
            return arr[i, j];
        }
  
        // If the state has already
        // been solved then return it
        if (v[i, j, s])
          
          
        {
            return dp[i, j, s];
        }
  
        // Marking the state as solved
        v[i, j, s] = true;
  
        // Growing phase
        if (s != 1)
        {
            dp[i,j,s] = arr[i,j] + Math.Max(maxScore(arr, i + 1, j, s),
                                    Math.Max(maxScore(arr, i, j + 1, s),
                                    maxScore(arr, i - 1, j, (s==1)?0:1)));
        } // Shrinking phase
        else
        {
            dp[i,j,s] = arr[i,j] + Math.Max(maxScore(arr, i - 1, j, s),
                    maxScore(arr, i, j + 1, s));
        }
  
        // Returning the solved state
        return dp[i, j, s];
    }
  
    // Driver code
    static public void Main ()
    {
        int [,]arr = {{1, 1, 1},
        {1, 5, 1},
        {1, 1, 1}};
  
        Console.WriteLine(maxScore(arr, 0, 0, 0));
    }
}
  
/* This code contributed by @Tushil..... */

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Output:

15


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