Check if a number starts with another number or not

Given two numbers A and B where (A > B), the task is to check if B is a prefix of A or not. Print "Yes" if it is a prefix Else print "No".

Examples:

Input: A = 12345, B = 12
Output: Yes

Input: A = 12345, B = 345
Output: No

Approach:

Convert the given numbers A and B to strings str1 and str2 respectively.
Traverse both the strings from the start of the strings.
While traversing the strings, if at any index characters from str1 and str2 are unequal then print "No".
Else print "Yes".

Below is the implementation of the above approach:

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# 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>

Output

Yes

Time Complexity: O(n2), where n2 is the size of string s2
Auxiliary Space: O(1), as no extra space is required

Using in-built function: Using inbuilt function std::boost::algorithm::starts_with(), it can be checked whether any string contains prefix of another string or not.