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# Check Whether a number is Duck Number or not

• Difficulty Level : Basic
• Last Updated : 23 Mar, 2023

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.

Recommended Practice

Implementation:

## C++

 `// C++ Program to check whether` `// a number is Duck Number or not.` `#include ` `using` `namespace` `std;`   `// Function to check whether` `// given number is duck number or not.` `bool` `check_duck(string num)` `{` `    ``// Ignore leading 0s` `    ``int` `i = 0, n = num.length();` `    ``while` `(i < n && num[i] == ``'0'``)` `        ``i++;`   `    ``// Check remaining digits` `    ``while` `(i < n) {` `        ``if` `(num[i] == ``'0'``)` `            ``return` `true``;` `        ``i++;` `    ``}`   `    ``return` `false``;` `}`   `// Driver Method` `int` `main(``void``)` `{` `    ``string num = ``"1023"``;` `    ``if` `(check_duck(num))` `        ``cout << ``"It is a duck number\n"``;` `    ``else` `        ``cout << ``"It is not a duck number\n"``;`   `    ``return` `0;` `}`

## Java

 `// Java Program to check whether a` `// number is Duck Number or not.`   `import` `java.io.*;` `class` `GFG {`   `    ``// Function to check whether` `    ``// the given number is duck number or not.` `    ``static` `boolean` `check_duck(String num)` `    ``{` `        ``// Ignore leading 0s` `        ``int` `i = ``0``, n = num.length();` `        ``while` `(i < n && num.charAt(i) == ``'0'``)` `            ``i++;`   `        ``// Check remaining digits` `        ``while` `(i < n) {` `            ``if` `(num.charAt(i) == ``'0'``)` `                ``return` `true``;` `            ``i++;` `        ``}`   `        ``return` `false``;` `    ``}`   `    ``// Driver Method` `    ``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

 `# Python program to check whether a number is Duck Number or not.`   `# Function to check whether ` `# the given number is duck number or not.` `def` `check_duck(num) :`   `    ``# Length of the number(number of digits)` `    ``n ``=` `len``(num) ` `   `  `    ``# Ignore leading 0s` `    ``i ``=` `0` `    ``while` `(i < n ``and` `num[i] ``=``=` `'0'``) :` `        ``i ``=` `i ``+` `1`   `    ``# Check remaining digits` `    ``while` `(i < n) : ` `        ``if` `(num[i] ``=``=` `"0"``) :` `            ``return` `True` `        ``i ``=` `i ``+` `1`   `    ``return` `False` `    `    `# Driver Method` `num1 ``=` `"1023"` `if``(check_duck(num1)) :` `    ``print` `"It is a duck number"` `else` `:` `    ``print` `"It is not a duck number"`   `# Write Python3 code here`

## C#

 `// C# Program to check whether a ` `// number is Duck Number or not.` `using` `System;`   `class` `GFG {` `    `  `    ``// Function to check whether ` `    ``// the given number is duck` `    ``// number or not.` `    ``static` `bool` `check_duck( String num)` `    ``{` `        `  `        ``// Ignore leading 0s` `        ``int` `i = 0, n = num.Length;` `        ``while` `(i < n && num[i] == ``'0'``)` `            ``i++;`   `        ``// Check remaining digits` `        ``while` `(i < n) {` `            ``if` `(num[i] == ``'0'``)` `                ``return` `true``;` `            ``i++;` `        ``}`   `        ``return` `false``;` `    ``}` `    `    `    ``// Driver Method` `    ``public` `static` `void` `Main()` `    ``{` `        `  `        ``String num1 = ``"1023"``; ` `        `  `        ``// checking number1` `        ``if``( check_duck(num1))` `            ``Console.Write(``"It is a "` `                      ``+ ``"duck number"``);` `        ``else` `            ``Console.Write(``"It is not "` `                    ``+ ``"a duck number"``);` `    ``}` `}`   `// This code is contributed by` `// nitin mittal.`

## Javascript

 ``

Output

`It is a duck number`

Time Complexity: O(n) where n is length of string.
Auxiliary Space: O(1)

Approach 2:String manipulation

1. Take the input number as an integer n.
2. Convert the integer n to a string s using the to_string function in C++.
3. Find the length of the string s and initialize a boolean variable hasZero to false.
4. 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.
5. If any character is equal to ‘0’, set hasZero to true and break out of the loop.
6. Check if the first character of the string s is not equal to ‘0’.
7. 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 ` `#include ` `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;` `}`

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.

This article is contributed by Nikita Tiwari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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