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Maximise matrix sum by following the given Path
  • Difficulty Level : Expert
  • Last Updated : 11 May, 2021

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:
 

 

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++




// 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;
}

Java




// 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 */

Python3




# 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

C#




// 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..... */

Javascript




<script>
    // Javascript implementation of the approach
     
    let n = 3;
   
    // To store the states of the DP
    let dp = new Array(n);
    let v = new Array(n);
     
    for(let i = 0; i < n; i++)
    {
        dp[i] = new Array(n);
        v[i] = new Array(n);
        for(let j = 0; j < n; j++)
        {
            dp[i][j] = new Array(2);
            v[i][j] = new Array(2);
            for(let k = 0; k < 2; k++)
            {
                dp[i][j][k] = 0;
                v[i][j][k] = 0;
            }
        }
    }
   
    // Function to return the maximum
    // of the three integers
    function max(a, b, c)
       
    {
        let m = a;
        if (m < b)
        {
            m = b;
        }
        if (m < c)
        {
            m = c;
        }
        return m;
    }
   
    // Function to return the maximum score
    function maxScore(arr, i, j, 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];
    }
     
    let arr = [[1, 1, 1],
               [1, 5, 1],
               [1, 1, 1]];
 
    document.write(maxScore(arr, 0, 0, 0));
 
</script>
Output: 
15

 




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