Check Whether a number is Duck Number or not
A Duck number is a positive number which has zeroes present in it, For example 3210, 8050896, 70709 are all Duck numbers. Please note that a numbers with only leading 0s is not considered as Duck Number. For example, numbers like 035 or 0012 are not considered as Duck Numbers. A number like 01203 is considered as Duck because there is a non-leading 0 present in it.
Examples :
Input : 707069
Output : It is a duck number.
Explanation: 707069 does not contains zeros at the beginning.
Input : 02364
Output : It is not a duck number.
Explanation: in 02364 there is a zero at the beginning of the number.
Implementation:
C++
Java
import java.io.*;
class GFG {
static boolean check_duck(String num)
{
int i = 0 , n = num.length();
while (i < n && num.charAt(i) == '0' )
i++;
while (i < n) {
if (num.charAt(i) == '0' )
return true ;
i++;
}
return false ;
}
public static void main(String args[]) throws IOException
{
String num = "1023" ;
if (check_duck(num))
System.out.println( "It is a duck number" );
else
System.out.println( "It is not a duck number" );
}
}
|
Python
def check_duck(num) :
n = len (num)
i = 0
while (i < n and num[i] = = '0' ) :
i = i + 1
while (i < n) :
if (num[i] = = "0" ) :
return True
i = i + 1
return False
num1 = "1023"
if (check_duck(num1)) :
print "It is a duck number"
else :
print "It is not a duck number"
|
C#
using System;
class GFG {
static bool check_duck( String num)
{
int i = 0, n = num.Length;
while (i < n && num[i] == '0' )
i++;
while (i < n) {
if (num[i] == '0' )
return true ;
i++;
}
return false ;
}
public static void Main()
{
String num1 = "1023" ;
if ( check_duck(num1))
Console.Write( "It is a "
+ "duck number" );
else
Console.Write( "It is not "
+ "a duck number" );
}
}
|
Javascript
<script>
function check_duck(num)
{
let i = 0, n = num.length;
while (i < n && num[i] == '0' )
i++;
while (i < n)
{
if (num[i] == '0' )
return true ;
i++;
}
return false ;
}
let num = "1023" ;
if (check_duck(num))
document.write( "It is a duck number" );
else
document.write( "It is not a duck number" );
</script>
|
Output
It is a duck number
Time Complexity: O(n) where n is length of string.
Auxiliary Space: O(1)
Approach 2:String manipulation
- Take the input number as an integer n.
- Convert the integer n to a string s using the to_string function in C++.
- Find the length of the string s and initialize a boolean variable hasZero to false.
- Iterate over the characters of the string s starting from the second character (i.e., the first non-zero digit) to the end of the string.
- If any character is equal to ‘0’, set hasZero to true and break out of the loop.
- Check if the first character of the string s is not equal to ‘0’.
- If both conditions in steps 4 and 5 are satisfied, return true (i.e., the number is a Duck Number); otherwise, return false (i.e., the number is not a Duck Number).
C++
#include <iostream>
#include <string>
using namespace std;
bool isDuckNumber( int n) {
string s = to_string(n);
int len = s.length();
bool hasZero = false ;
for ( int i = 1; i < len; i++) {
if (s[i] == '0' ) {
hasZero = true ;
break ;
}
}
return (hasZero && s[0] != '0' );
}
int main() {
int n = 1023;
if (isDuckNumber(n)) {
cout << "It is a Duck Number" << endl;
} else {
cout << "It is not a Duck Number" << endl;
}
return 0;
}
|
Java
import java.util.*;
public class Main {
public static boolean isDuckNumber( int n) {
String s = Integer.toString(n);
int len = s.length();
boolean hasZero = false ;
for ( int i = 1 ; i < len; i++) {
if (s.charAt(i) == '0' ) {
hasZero = true ;
break ;
}
}
return (hasZero && s.charAt( 0 ) != '0' );
}
public static void main(String[] args) {
int n = 1023 ;
if (isDuckNumber(n)) {
System.out.println( "It is a Duck Number" );
} else {
System.out.println( "It is not a Duck Number" );
}
}
}
|
Python3
def isDuckNumber(n):
s = str (n)
len_s = len (s)
hasZero = False
for i in range ( 1 , len_s):
if s[i] = = '0' :
hasZero = True
break
return (hasZero and s[ 0 ] ! = '0' )
if __name__ = = '__main__' :
n = 1023
if isDuckNumber(n):
print ( "It is a Duck Number" )
else :
print ( "It is not a Duck Number" )
|
C#
using System;
public class GFG{
public static bool isDuckNumber( int n){
string s = n.ToString();
int len = s.Length;
bool hasZero = false ;
for ( int i = 1; i < len; i++){
if (s[i] == '0' ){
hasZero = true ;
break ;
}
}
return (hasZero && s[0] != '0' );
}
public static void Main( string [] args){
int n = 1023;
if (isDuckNumber(n)){
Console.WriteLine( "It is a Duck Number" );
}
else {
Console.WriteLine( "It is not a Duck Number" );
}
}
}
|
Javascript
function isDuckNumber(n) {
const s = n.toString();
const len = s.length;
let hasZero = false ;
for (let i = 1; i < len; i++) {
if (s[i] === '0' ) {
hasZero = true ;
break ;
}
}
return hasZero && s[0] !== '0' ;
}
const n = 1023;
if (isDuckNumber(n)) {
console.log( "It is a Duck Number" );
} else {
console.log( "It is not a Duck Number" );
}
|
Output
It is a Duck Number
Time Complexity: O(n)
Auxiliary Space: O(n)
The time complexity of this algorithm is O(n), where n is the number of digits in the input integer. This is because we need to iterate over all the digits in the integer to check if there is a zero after the first non-zero digit.
The space complexity of this algorithm is also O(n), because we need to store the string representation of the input integer. The length of this string is equal to the number of digits in the integer, which is at most log10(n) + 1.
In general, string manipulation algorithms have higher space complexity compared to their integer-based counterparts because they require additional space to store the string representation of the input. However, they can sometimes be more intuitive to understand and implement.
Last Updated :
09 Aug, 2023
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