Dada una array de N * M que contiene { 0 , 1 , # }. La tarea es encontrar el valor máximo de fuerza basado en las siguientes reglas:
- La fuerza inicial es cero .
- Si encuentra un 0 , la Fuerza disminuye en 2.
- Si encuentra un 1 , la Fuerza aumenta en 5.
- Si encuentra un # , salta al comienzo de una nueva fila sin perder fuerza.
Nota: Tienes que recorrer cada fila de la array en orden de arriba hacia abajo de izquierda a derecha.
Ejemplo:
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
Acercarse:
- Atraviese la array mat[][] desde i = [0, N], j = [0, M] y verifique:
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.
- En cada paso, almacene el valor máximo de fuerza hasta ahora e imprima la fuerza al final.
A continuación se muestra la implementación del enfoque anterior:
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>
Producción:
14
Publicación traducida automáticamente
Artículo escrito por thakurabhaysingh445 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA