Rutas con número máximo de ‘a’ de (1, 1) a (X, Y) vertical u horizontalmente

Posted on by Rudeus Greyrat

Dada una array NXN que consta de caracteres. También se dan consultas Q, donde cada consulta contiene una coordenada (X, Y). Para cada consulta, busque todas las rutas de (1, 1) a (X, Y) moviéndose vertical u horizontalmente y tome la ruta que tenga el número máximo de  a . La tarea es imprimir el número de caracteres que no son ‘a’ a lo largo de esa ruta. 

Ejemplos : 

Input: mat[][] = {{'a', 'b', 'a'}, 
                             {'a', 'c', 'd'}, 
                             {'b', 'a', 'b'}} 
            Queries: 
            X = 1, Y = 3
            X = 3, Y = 3
Output: 
1st query: 1
2nd query: 2

Query-1: There is only one path from (1, 1) to (1, 3)
i.e., "aba" and the number of characters 
which are not 'a' is 1. 
Query-2: The path which has the maximum number of 'a'
in it is "aabab", hence non 'a' 
characters are 2.

El problema es una variante del problema Dp de ruta de costo mínimo . Necesitamos calcular previamente la array DP[][] y luego la respuesta será DP[X][Y], por lo tanto, todas las consultas se pueden responder en O(1). Si el carácter de la posición del índice (1, 1) no es ‘a’, aumente el valor de dp[1][1] en 1. Luego simplemente itere para filas y columnas, y haga un DP de costo mínimo considerando ‘a ‘ como 1 y el carácter que no es ‘a’ como 0. Dado que la array DP[][] almacena la ruta de costo mínimo desde (1, 1) a cualquier índice (i, j), por lo tanto, la consulta se puede responder en O(1 ). 

A continuación se muestra la implementación del enfoque anterior:  

C++

// C++ program to find paths with maximum number
// of 'a' from (1, 1) to (X, Y) vertically
// or horizontally
 
#include <bits/stdc++.h>
using namespace std;
 
const int n = 3;
int dp[n][n];
 
// Function to answer queries
void answerQueries(pair<int, int> queries[], int q)
{
    // Iterate till query
    for (int i = 0; i < q; i++) {
 
        // Decrease to get 0-based indexing
        int x = queries[i].first;
        x--;
        int y = queries[i].second;
        y--;
 
        // Print answer
        cout << dp[x][y] << endl;
    }
}
 
// Function that pre-computes the dp array
void pre_compute(char a[][n])
{
    // Check for the first character
    if (a[0][0] == 'a')
        dp[0][0] = 0;
    else
        dp[0][0] = 1;
 
    // Iterate in row and columns
    for (int row = 0; row < n; row++) {
        for (int col = 0; col < n; col++) {
            // If not first row or not first column
            if (row != 0 || col != 0)
                dp[row][col] = INT_MAX;
 
            // Not first row
            if (row != 0) {
                dp[row][col] = min(dp[row][col],
                                   dp[row - 1][col]);
            }
 
            // Not first column
            if (col != 0) {
                dp[row][col] = min(dp[row][col],
                                   dp[row][col - 1]);
            }
 
            // If it is not 'a' then increase by 1
            if (a[row][col] != 'a' && (row != 0 || col != 0))
                dp[row][col] += 1;
        }
    }
}
 
// Driver code
int main()
{
    // character N X N array
    char a[][3] = { { 'a', 'b', 'a' },
                    { 'a', 'c', 'd' },
                    { 'b', 'a', 'b' } };
 
    // queries
    pair<int, int> queries[] = { { 1, 3 }, { 3, 3 } };
 
    // number of queries
    int q = 2;
 
    // function call to pre-compute
    pre_compute(a);
 
    // function call to answer every query
    answerQueries(queries, q);
}

Java

// Java program to find paths with maximum number
// of 'a' from (1, 1) to (X, Y) vertically
// or horizontally
class GFG
{
static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
static int n = 3;
static int [][]dp = new int[n][n];
 
// Function to answer queries
static void answerQueries(pair queries[], int q)
{
    // Iterate till query
    for (int i = 0; i < q; i++)
    {
 
        // Decrease to get 0-based indexing
        int x = queries[i].first;
        x--;
        int y = queries[i].second;
        y--;
 
        // Print answer
        System.out.println(dp[x][y]);
    }
}
 
// Function that pre-computes the dp array
static void pre_compute(char a[][])
{
    // Check for the first character
    if (a[0][0] == 'a')
        dp[0][0] = 0;
    else
        dp[0][0] = 1;
 
    // Iterate in row and columns
    for (int row = 0; row < n; row++)
    {
        for (int col = 0; col < n; col++)
        {
            // If not first row or not first column
            if (row != 0 || col != 0)
                dp[row][col] = Integer.MAX_VALUE;
 
            // Not first row
            if (row != 0)
            {
                dp[row][col] = Math.min(dp[row][col],
                                        dp[row - 1][col]);
            }
 
            // Not first column
            if (col != 0)
            {
                dp[row][col] = Math.min(dp[row][col],
                                        dp[row][col - 1]);
            }
 
            // If it is not 'a' then increase by 1
            if (a[row][col] != 'a' && (row != 0 || col != 0))
                dp[row][col] += 1;
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    // character N X N array
    char a[][] = {{ 'a', 'b', 'a' },
                  { 'a', 'c', 'd' },
                  { 'b', 'a', 'b' }};
 
    // queries
    pair queries[] = { new pair( 1, 3 ),
                       new pair(3, 3 ) };
 
    // number of queries
    int q = 2;
 
    // function call to pre-compute
    pre_compute(a);
 
    // function call to answer every query
    answerQueries(queries, q);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find paths with maximum number
# of 'a' from (1, 1) to (X, Y) vertically
# or horizontally
n = 3
dp = [[0 for i in range(n)]
         for j in range(n)]
 
# Function to answer queries
def answerQueries(queries, q):
 
    # Iterate till query
    for i in range(q):
         
        # Decrease to get 0-based indexing
        x = queries[i][0]
        x -= 1
        y = queries[i][1]
        y -= 1
 
        # Print answer
        print(dp[x][y])
 
# Function that pre-computes the dp array
def pre_compute(a):
 
    # Check for the first character
    if a[0][0] == 'a':
        dp[0][0] = 0
    else:
        dp[0][0] = 1
 
    # Iterate in row and columns
    for row in range(n):
        for col in range(n):
             
            # If not first row or not first column
            if (row != 0 or col != 0):
                dp[row][col] = 9999
 
            # Not first row
            if (row != 0):
                dp[row][col] = min(dp[row][col],
                                   dp[row - 1][col])
 
            # Not first column
            if (col != 0):
                dp[row][col] = min(dp[row][col],
                                   dp[row][col - 1])
 
            # If it is not 'a' then increase by 1
            if (a[row][col] != 'a' and (row != 0 or
                                        col != 0)):
                dp[row][col] += 1
 
# Driver code
if __name__ == '__main__':
     
    # Character N X N array
    a = [ ('a', 'b', 'a'),
          ('a', 'c', 'd'),
          ('b', 'a', 'b') ]
 
    # Queries
    queries = [ (1, 3), (3, 3) ]
 
    # Number of queries
    q = 2
 
    # Function call to pre-compute
    pre_compute(a)
 
    # function call to answer every query
    answerQueries(queries, q)
 
# This code is contributed by kirtishsurangalikar

C#

// C# program to find paths with maximum number
// of 'a' from (1, 1) to (X, Y) vertically
// or horizontally
using System;
 
class GFG
{
class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
static int n = 3;
static int [,]dp = new int[n, n];
 
// Function to answer queries
static void answerQueries(pair []queries, int q)
{
    // Iterate till query
    for (int i = 0; i < q; i++)
    {
 
        // Decrease to get 0-based indexing
        int x = queries[i].first;
        x--;
        int y = queries[i].second;
        y--;
 
        // Print answer
        Console.WriteLine(dp[x, y]);
    }
}
 
// Function that pre-computes the dp array
static void pre_compute(char [,]a)
{
    // Check for the first character
    if (a[0, 0] == 'a')
        dp[0, 0] = 0;
    else
        dp[0, 0] = 1;
 
    // Iterate in row and columns
    for (int row = 0; row < n; row++)
    {
        for (int col = 0; col < n; col++)
        {
            // If not first row or not first column
            if (row != 0 || col != 0)
                dp[row, col] = int.MaxValue;
 
            // Not first row
            if (row != 0)
            {
                dp[row, col] = Math.Min(dp[row, col],
                                        dp[row - 1, col]);
            }
 
            // Not first column
            if (col != 0)
            {
                dp[row, col] = Math.Min(dp[row, col],
                                        dp[row, col - 1]);
            }
 
            // If it is not 'a' then increase by 1
            if (a[row, col] != 'a' &&
               (row != 0 || col != 0))
                dp[row, col] += 1;
        }
    }
}
 
// Driver code
public static void Main(String[] args)
{
    // character N X N array
    char [,]a = {{ 'a', 'b', 'a' },
                 { 'a', 'c', 'd' },
                 { 'b', 'a', 'b' }};
 
    // queries
    pair []queries = { new pair(1, 3),
                       new pair(3, 3) };
 
    // number of queries
    int q = 2;
 
    // function call to pre-compute
    pre_compute(a);
 
    // function call to answer every query
    answerQueries(queries, q);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to find paths with maximum number
// of 'a' from (1, 1) to (X, Y) vertically
// or horizontally
 
class pair
{
    constructor(first,second)
    {
        this.first = first;
        this.second = second;
    }
}
 
let n = 3;
 
let dp=new Array(n);
for(let i=0;i<n;i++)
{
    dp[i]=new Array(n);
    for(let j=0;j<n;j++)
    {
        dp[i][j]=0;
    }
}
 
// Function to answer queries
function answerQueries(queries,q)
{
    // Iterate till query
    for (let i = 0; i < q; i++)
    {
  
        // Decrease to get 0-based indexing
        let x = queries[i].first;
        x--;
        let y = queries[i].second;
        y--;
  
        // Print answer
        document.write(dp[x][y]+"<br>");
    }
}
 
// Function that pre-computes the dp array
function pre_compute(a)
{
    // Check for the first character
    if (a[0][0] == 'a')
        dp[0][0] = 0;
    else
        dp[0][0] = 1;
  
    // Iterate in row and columns
    for (let row = 0; row < n; row++)
    {
        for (let col = 0; col < n; col++)
        {
            // If not first row or not first column
            if (row != 0 || col != 0)
                dp[row][col] = Number.MAX_VALUE;
  
            // Not first row
            if (row != 0)
            {
                dp[row][col] = Math.min(dp[row][col],
                                        dp[row - 1][col]);
            }
  
            // Not first column
            if (col != 0)
            {
                dp[row][col] = Math.min(dp[row][col],
                                        dp[row][col - 1]);
            }
  
            // If it is not 'a' then increase by 1
            if (a[row][col] != 'a' && (row != 0 || col != 0))
                dp[row][col] += 1;
        }
    }
}
 
// Driver code
let a=[[ 'a', 'b', 'a' ],
                  [ 'a', 'c', 'd' ],
                  [ 'b', 'a', 'b' ]];
// queries
let queries = [ new pair( 1, 3 ),
new pair(3, 3 ) ];
 
// number of queries
let q = 2;
 
// function call to pre-compute
pre_compute(a);
 
// function call to answer every query
answerQueries(queries, q);
 
// This code is contributed by rag2127
</script>
Producción

1
2

Complejidad de tiempo : O(N 2 ) para precálculo y O(1) para cada consulta. 
Espacio Auxiliar : O(N 2 )
 

Publicación traducida automáticamente

Artículo escrito por Striver y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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