Maximum strength in a Matrix after performing specified operations

Given a matrix of N * M containing {0, 1, #}. The task is to find the maximum value of strength based on following rules: 

1. Initial strength is zero.
2. If you encounter a 0, Strength decreases by 2.
3. If you encounter a 1, Strength increases by 5.
4. If you encounter a #, Jumps to the start of a new row without losing any strength.

Note: You have to traverse every row of the matrix in top-down order from left to right.
Example:

Input:
      {{1, 0, 1, 0},
       {0, #, 0, 0},
       {1, 1, 0, 0},
       {0, #, 1, 0}}

Output: 14
Explanation:
Here you starts with strength S = 0.

For the first row {1, 0, 1, 0}:
After {1} -> S = S + 5 = 5
After {0} -> S = S - 2 = 3
After {1} -> S = S + 5 = 8
After {0} -> S = S - 2 = 6

For the Second row {0, #, 0, 0}:
After {0} -> S = S - 2 = 4
After {#} -> Jump to next row.

For the Third row {1, 1, 0, 0}:
After {1} -> S = S + 5 = 9
After {1} -> S = S + 5 = 14
After {0} -> S = S - 2 = 12
After {0} -> S = S - 2 = 10

For the Fourth row {0, #, 1, 0}:
After {0} -> S = S - 2 = 8
After {#} -> Jump to next row.

So, The maximum value of S is 14 

Approach:

1. Traverse the matrix mat[][] from i = [0, N], j = [0, M] and check: 
   

If mat[i][j] = 0 then, S = S - 2.
If mat[i][j] = 1 then, S = S + 5.
If mat[i][j] = # then, jump to the next row.

2. At every step store maximum value of strength till now and Print the strength at the end. 
   

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function return the Maximum
// value of the strength
void MaxStrength(char mat[100][100],
                 int n, int m)
{
    int S = 0;
    int ans = 0;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {

            char Curr = mat[i][j];

            // If current element
            // is 1
            if (Curr == '1') {
                S += 5;
            }
            // If current element
            // is 0
            if (Curr == '0') {
                S -= 2;
            }
            // If current element
            // is '#'
            if (Curr == '#') {
                break;
            }

            // Store the value of
            // maximum strength
            // till now
            ans = max(ans, S);
        }
    }

    cout << ans;

    return;
}

// Driver code
int main()
{
    int N = 4;
    int M = 4;
    char Mat[100][100]{ { '1', '0', '1', '0' },
                        { '0', '#', '0', '0' },
                        { '1', '1', '0', '0' },
                        { '0', '#', '1', '0' } };

    MaxStrength(Mat, N, M);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;

class GFG{

// Function return the maximum
// value of the strength
static void MaxStrength(char[][] mat,
                        int n, int m)
{
    int S = 0;
    int ans = 0;

    for(int i = 0; i < n; i++)
    {
       for(int j = 0; j < m; j++)
       {
          char Curr = mat[i][j];

          // If current element
          // is 1
          if (Curr == '1')
          {
              S += 5;
          }

          // If current element
          // is 0
          if (Curr == '0')
          {
              S -= 2;
          }

          // If current element
          // is '#'
          if (Curr == '#')
          {
              break;
          }

          // Store the value of
          // maximum strength
          // till now
          ans = Math.max(ans, S);
       }
    }
    System.out.println(ans);
    return;
}

// Driver code
public static void main (String[] args)
{
    int N = 4;
    int M = 4;
    char[][] Mat = { { '1', '0', '1', '0' },
                     { '0', '#', '0', '0' },
                     { '1', '1', '0', '0' },
                     { '0', '#', '1', '0' } };

    MaxStrength(Mat, N, M);
}
}

// This code is contributed by shubhamsingh10

Python3

# python3 program for the above approach

# Function return the Maximum
# value of the strength
def MaxStrength(mat, n, m):
    S = 0
    ans = 0

    for i in range(n):
        for j in range(m):
            Curr = mat[i][j]

            # If current element
            # is 1
            if (Curr == '1'):
                S += 5

            # If current element
            # is 0
            if (Curr == '0'):
                S -= 2

            # If current element
            # is '#'
            if (Curr == '#'):
                break

            # Store the value of
            # maximum strength
            # till now
            ans = max(ans, S)

    print(ans)
    return

# Driver code
if __name__ == '__main__':

    N = 4;
    M = 4;
    Mat = [ ['1', '0', '1', '0'],
            ['0', '#', '0', '0'],
            ['1', '1', '0', '0'],
            ['0', '#', '1', '0'] ]

    MaxStrength(Mat, N, M)

# This code is contributed by Samarth

C#

// C# program for the above approach
using System;

class GFG{

// Function return the maximum
// value of the strength
static void MaxStrength(char[,] mat,
                        int n, int m)
{
    int S = 0;
    int ans = 0;

    for(int i = 0; i < n; i++)
    {
       for(int j = 0; j < m; j++)
       {
          char Curr = mat[i, j];

          // If current element
          // is 1
          if (Curr == '1')
          {
              S += 5;
          }

          // If current element
          // is 0
          if (Curr == '0')
          {
              S -= 2;
          }

          // If current element
          // is '#'
          if (Curr == '#')
          {
              break;
          }

          // Store the value of
          // maximum strength
          // till now
          ans = Math.Max(ans, S);
       }
    }
    Console.WriteLine(ans);
    return;
}

// Driver code
public static void Main(String[] args)
{
    int N = 4;
    int M = 4;
    char[,] Mat = { { '1', '0', '1', '0' },
                    { '0', '#', '0', '0' },
                    { '1', '1', '0', '0' },
                    { '0', '#', '1', '0' } };

    MaxStrength(Mat, N, M);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript program for the above approach

// Function return the Maximum
// value of the strength
function MaxStrength(mat, n, m)
{
    var S = 0;
    var ans = 0;

    for (var i = 0; i < n; i++) {
        for (var j = 0; j < m; j++) {

            var Curr = mat[i][j];

            // If current element
            // is 1
            if (Curr == '1') {
                S += 5;
            }
            // If current element
            // is 0
            if (Curr == '0') {
                S -= 2;
            }
            // If current element
            // is '#'
            if (Curr == '#') {
                break;
            }

            // Store the value of
            // maximum strength
            // till now
            ans = Math.max(ans, S);
        }
    }

    document.write( ans);

    return;
}

// Driver code
var N = 4;
var M = 4;
var Mat = [ [ '1', '0', '1', '0' ],
                    [ '0', '#', '0', '0' ],
                    [ '1', '1', '0', '0' ],
                    [ '0', '#', '1', '0' ] ];
MaxStrength(Mat, N, M);

// This code is contributed by noob2000.

</script>

Output:

14

Time Complexity: O(n*m) where n is rows and m is columns.

Space Complexity: O(1) as no extra space has been used.