Dados dos números A y B donde ( A > B ), la tarea es verificar si B es un prefijo de A o no. Escriba “Sí” si es un prefijo De lo contrario escriba “No” .
Ejemplos:
Entrada: A = 12345, B = 12
Salida: SíEntrada: A = 12345, B = 345
Salida: No
Acercarse:
- Convierta los números dados A y B en strings str1 y str2 respectivamente.
- Recorre ambas cuerdas desde el comienzo de las cuerdas.
- Mientras recorre las strings, si en algún índice los caracteres de str1 y str2 no son iguales, imprima «No» .
- De lo contrario, escriba «Sí».
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
// Function to check if B is a
// prefix of A or not
bool checkprefix(int A, int B)
{
// Convert numbers into strings
string s1 = to_string(A);
string s2 = to_string(B);
// Find the lengths of strings
// s1 and s2
int n1 = s1.length();
int n2 = s2.length();
// Base Case
if (n1 < n2) {
return false;
}
// Traverse the strings s1 & s2
for (int i = 0; i < n2; i++) {
// If at any index characters
// are unequals then return false
if (s1[i]
!= s2[i]) {
return false;
}
}
// Return true
return true;
}
// Driver Code
int main()
{
// Given numbers
int A = 12345, B = 12;
// Function Call
bool result = checkprefix(A, B);
// If B is a prefix of A, then
// print "Yes"
if (result) {
cout << "Yes";
}
else {
cout << "No";
}
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to check if B is a
// prefix of A or not
static boolean checkprefix(int A, int B)
{
// Convert numbers into Strings
String s1 = String.valueOf(A);
String s2 = String.valueOf(B);
// Find the lengths of Strings
// s1 and s2
int n1 = s1.length();
int n2 = s2.length();
// Base Case
if (n1 < n2)
{
return false;
}
// Traverse the Strings s1 & s2
for(int i = 0; i < n2; i++)
{
// If at any index characters
// are unequals then return false
if (s1.charAt(i) != s2.charAt(1))
{
return false;
}
}
// Return true
return true;
}
// Driver Code
public static void main(String[] args)
{
// Given numbers
int A = 12345, B = 12;
// Function call
boolean result = checkprefix(A, B);
// If B is a prefix of A, then
// print "Yes"
if (!result)
{
System.out.print("Yes");
}
else
{
System.out.print("No");
}
}
}
// This code is contributed by amal kumar choubey
Python3
# Python3 program for the
# above approach
# Function to check if B is
# a prefix of A or not
def checkprefix(A, B):
# Convert numbers into strings
s1 = str(A)
s2 = str(B)
# Find the length of s1 and s2
n1 = len(s1)
n2 = len(s2)
# Base case
if n1 < n2:
return False
# Traverse the string s1 and s2
for i in range(0, n2):
# If at any index characters
# are unequal then return False
if s1[i] != s2[i]:
return False
return True
# Driver code
if __name__=='__main__':
# Given numbers
A = 12345
B = 12
# Function call
result = checkprefix(A, B)
# If B is a prefix of A ,
# then print Yes
if result:
print("Yes")
else:
print("No")
# This code is contributed by virusbuddah_
C#
// C# program for the above approach
using System;
class GFG{
// Function to check if B is a
// prefix of A or not
static bool checkprefix(int A, int B)
{
// Convert numbers into Strings
String s1 = A.ToString();
String s2 = B.ToString();
// Find the lengths of Strings
// s1 and s2
int n1 = s1.Length;
int n2 = s2.Length;
// Base Case
if (n1 < n2)
{
return false;
}
// Traverse the Strings s1 & s2
for(int i = 0; i < n2; i++)
{
// If at any index characters
// are unequals then return false
if (s1[i] != s2[i])
{
return false;
}
}
// Return true
return true;
}
// Driver Code
static public void Main ()
{
// Given numbers
int A = 12345, B = 12;
// Function call
bool result = checkprefix(A, B);
// If B is a prefix of A, then
// print "Yes"
if (result)
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
// This code is contributed by shubhamsingh10
Javascript
<script>
// javascript program for the above approach
// Function to check if B is a
// prefix of A or not
function checkprefix( A, B)
{
// Convert numbers into Strings
var s1 = A.toString();
var s2 = B.toString();
// Find the lengths of Strings
// s1 and s2
var n1 = s1.length;
var n2 = s2.length;
// Base Case
if (n1 < n2)
{
return false;
}
// Traverse the Strings s1 & s2
for(var i = 0; i < n2; i++)
{
// If at any index characters
// are unequals then return false
if (s1[i] != s2[i])
{
return false;
}
}
// Return true
return true;
}
// Driver Code
// Given numbers
var A = 12345, B = 12;
// Function call
var result = checkprefix(A, B);
// If B is a prefix of A, then
// print "Yes"
if (result)
{
document.write("Yes");
}
else
{
document.write("No");
}
</script>
Producción:
Yes
Complejidad de tiempo: O(n2), donde n2 es el tamaño de la string s2
Espacio auxiliar: O(1), ya que no se requiere espacio adicional
Usando la función incorporada: Usando la función incorporada std::boost::algorithm::starts_with() , se puede verificar si alguna string contiene el prefijo de otra string o no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
#include <boost/algorithm/string.hpp>
using namespace std;
// Function to check if B is a
// prefix of A or not
void checkprefix(int A, int B)
{
// Convert numbers into strings
string s1 = to_string(A);
string s2 = to_string(B);
bool result;
// Check if s2 is a prefix of s1
// or not using starts_with() function
result = boost::algorithm::starts_with(s1, s2);
// If result is true, print "Yes"
if (result) {
cout << "Yes";
}
else {
cout << "No";
}
}
// Driver Code
int main()
{
// Given numbers
int A = 12345, B = 12;
// Function Call
checkprefix(A, B);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to check if B is a
// prefix of A or not
static void checkprefix(int A, int B)
{
// Convert numbers into Strings
String s1 = String.valueOf(A);
String s2 = String.valueOf(B);
boolean result;
// Check if s2 is a prefix of s1
// or not using starts_with() function
result = s1.startsWith(s2);
// If result is true, print "Yes"
if (result)
{
System.out.print("Yes");
}
else
{
System.out.print("No");
}
}
// Driver Code
public static void main(String[] args)
{
// Given numbers
int A = 12345, B = 12;
// Function call
checkprefix(A, B);
}
}
// This code is contributed by amal kumar choubey
Python3
# Python3 program for the
# above approach
# Function to check if B is
# a prefix of A or not
def checkprefix(A, B):
# Convert numbers into strings
s1 = str(A)
s2 = str(B)
# Check if s2 is a prefix of s1
# or not using startswith() function
result = s1.startswith(s2)
# If result is true print Yes
if result:
print("Yes")
else:
print("No")
# Driver code
if __name__=='__main__':
# Given numbers
A = 12345
B = 12
# Function call
checkprefix(A, B)
# This code is contributed by virusbuddah_
C#
// C# program for the above approach
using System.Threading;
using System.Globalization;
using System;
class GFG{
// Function to check if B is a
// prefix of A or not
static void checkprefix(int A, int B)
{
// Convert numbers into Strings
string s1 = A.ToString();
string s2 = B.ToString();
bool result;
// Check if s2 is a prefix of s1
// or not using starts_with() function
result = s1.StartsWith(s2, false,
CultureInfo.InvariantCulture);
// If result is true, print "Yes"
if (result)
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
// Driver code
static void Main()
{
// Given numbers
int A = 12345, B = 12;
// Function call
checkprefix(A, B);
}
}
// This code is contributed by divyeshrabadiya07
Javascript
<script>
// JavaScript program for the above approach
// Function to check if B is a
// prefix of A or not
function checkprefix(A, B) {
// Convert numbers into Strings
var s1 = A.toString();
var s2 = B.toString();
var result;
// Check if s2 is a prefix of s1
// or not using starts_with() function
result = s1.startsWith(s2);
// If result is true, print "Yes"
if (result) {
document.write("Yes");
} else {
document.write("No");
}
}
// Driver code
// Given numbers
var A = 12345,
B = 12;
// Function call
checkprefix(A, B);
</script>
Producción:
Yes
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)