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# Remove leading zeros from a Number given as a string

• Last Updated : 05 Oct, 2022

Given numeric string str, the task is to remove all the leading zeros from a given string. If the string contains only zeros, then print a single “0”.

Examples:

Input: str = “0001234”
Output: 1234
Explanation:
Removal of leading substring “000” modifies the string to “1234”.
Hence, the final answer is “1234”.

Input: str = “00000000”
Output:
Explanation:
There are no numbers except 0

Naive Approach:
The simplest approach to solve the problem is to traverse the string up to the first non-zero character present in the string and store the remaining string starting from that index as the answer. If the entire string is traversed, it means all the characters in the string are ‘0’. For this case, store “0” as the answer. Print the final answer.

Below is the implementation of the above idea:

## C++

 `// C++ Program to remove all the leading``// zeros from a given numeric string``#include ` `using` `namespace` `std;` `// Function to remove the leading zeros``string removeLeadingZeros(string num)``{``    ``// traverse the entire string``    ``for` `(``int` `i = 0; i < num.length(); i++) {` `        ``// check for the first non-zero character``        ``if` `(num[i] != ``'0'``) {``            ``// return the remaining string``            ``string res = num.substr(i);``            ``return` `res;``        ``}``    ``}` `    ``// If the entire string is traversed``    ``// that means it didn't have a single``    ``// non-zero character, hence return "0"``    ``return` `"0"``;``}` `// Driver Code``int` `main()``{``    ``string num = ``"1023"``;``    ``cout << removeLeadingZeros(num) << endl;` `    ``num = ``"00123"``;``    ``cout << removeLeadingZeros(num) << endl;``}` `// This code is contributed by phasing17`

## Java

 `// Java Program to remove all the leading``//zeros from a given numeric string``import` `java.io.*;` `class` `GFG``{``    ` `    ``// Function to remove the leading zeros``    ``static` `String removeLeadingZeros(String num)``    ``{``        ``//traverse the entire string``        ``for``(``int` `i=``0``;i

## Python3

 `# Python Program to remove all the leading``# zeros from a given numeric string` `# Function to remove the leading zeros``def` `removeLeadingZeros(num):` `    ``# traverse the entire string``    ``for` `i ``in` `range``(``len``(num)):` `        ``# check for the first non-zero character``        ``if` `num[i] !``=` `'0'``:``            ``# return the remaining string``            ``res ``=` `num[i::];``            ``return` `res;``        ` `    ``# If the entire string is traversed``    ``# that means it didn't have a single``    ``# non-zero character, hence return "0"``    ``return` `"0"``;` `# Driver Code``num ``=` `"1023"``;``print``(removeLeadingZeros(num));` `num ``=` `"00123"``;``print``(removeLeadingZeros(num));` `# This code is contributed by phasing17`

## C#

 `// C# Program to remove all the leading``//zeros from a given numeric string``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `  ``// Function to remove the leading zeros``  ``static` `string` `removeLeadingZeros(``string` `num)``  ``{``    ``// traverse the entire string``    ``for``(``int` `i = 0; i < num.Length; i++){` `      ``// check for the first non-zero character``      ``if``(num[i] != ``'0'``)``      ``{``        ``// return the remaining string``        ``string` `res = num.Substring(i);``        ``return` `res;``      ``}``    ``}` `    ``// If the entire string is traversed``    ``// that means it didn't have a single``    ``// non-zero character, hence return "0"``    ``return` `"0"``;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``string` `num = ``"1023"``;``    ``Console.WriteLine(removeLeadingZeros(num));` `    ``num = ``"00123"``;``    ``Console.WriteLine(removeLeadingZeros(num));``  ``}``}` `// This code is contributed by phasing17`

## Javascript

 `// JS Program to remove all the leading``// zeros from a given numeric string` `// Function to remove the leading zeros``function` `removeLeadingZeros(num)``{``    ``// traverse the entire string``    ``for` `(``var` `i = 0; i < num.length; i++) {` `        ``// check for the first non-zero character``        ``if` `(num.charAt(i) != ``'0'``) {``            ``// return the remaining string``            ``let res = num.substr(i);``            ``return` `res;``        ``}``    ``}` `    ``// If the entire string is traversed``    ``// that means it didn't have a single``    ``// non-zero character, hence return "0"``    ``return` `"0"``;``}` `// Driver Code``let num = ``"1023"``;``console.log(removeLeadingZeros(num));` `num = ``"00123"``;``console.log(removeLeadingZeros(num));`  `// This code is contributed by phasing17`

Output

```1023
123```

Time Complexity: O(N)
Auxiliary Space: O(N)

Space-Efficient Approach:
Follow the steps below to solve the problem in constant space using Regular Expression

• Create a Regular Expression as given below to remove the leading zeros

regex = “^0+(?!\$)”
where:
^0+ match one or more zeros from the beginning of the string.
(?!\$) is a negative look-ahead expression, where “\$” means the end of the string.

• Use the inbuilt replaceAll() method of the String class which accepts two parameters, a Regular Expression, and a Replacement String.
• To remove the leading zeros, pass a Regex as the first parameter and empty string as the second parameter.
• This method replaces the matched value with the given string.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``#include ``using` `namespace` `std;` `// Function to remove all leading``// zeros from a given string``void` `removeLeadingZeros(string str)``{``    ``// Regex to remove leading``    ``// zeros from a string``    ``const` `regex pattern(``"^0+(?!\$)"``);` `    ``// Replaces the matched``    ``// value with given string``    ``str = regex_replace(str, pattern, ``""``);``    ``cout << str;``}` `// Driver Code``int` `main()``{``    ``string str = ``"0001234"``;``    ``removeLeadingZeros(str);``    ``return` `0;``}`

## Java

 `// Java Program to implement``// the above approach``import` `java.util.regex.*;``class` `GFG``{` `    ``// Function to remove all leading``    ``// zeros from a given string``    ``public` `static` `void` `removeLeadingZeros(String str)``    ``{` `        ``// Regex to remove leading``        ``// zeros from a string``        ``String regex = ``"^0+(?!\$)"``;` `        ``// Replaces the matched``        ``// value with given string``        ``str = str.replaceAll(regex, ``""``);` `        ``System.out.println(str);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``String str = ``"0001234"``;` `        ``removeLeadingZeros(str);``    ``}``}`

## Python3

 `# Python3 Program to implement``# the above approach``import` `re` `# Function to remove all leading``# zeros from a given string``def` `removeLeadingZeros(``str``):` `    ``# Regex to remove leading``    ``# zeros from a string``    ``regex ``=` `"^0+(?!\$)"` `    ``# Replaces the matched``    ``# value with given string``    ``str` `=` `re.sub(regex, "", ``str``)` `    ``print``(``str``)` `# Driver Code``str` `=` `"0001234"``removeLeadingZeros(``str``)` `# This code is contributed by avanitrachhadiya2155`

## C#

 `// C# program to implement``// the above approach``using` `System;``using` `System.Text.RegularExpressions;` `class` `GFG{` `// Function to remove all leading``// zeros from a given string``public` `static` `void` `removeLeadingZeros(``string` `str)``{``    ` `    ``// Regex to remove leading``    ``// zeros from a string``    ``string` `regex = ``"^0+(?!\$)"``;` `    ``// Replaces the matched``    ``// value with given string``    ``str = Regex.Replace(str, regex, ``""``);` `    ``Console.WriteLine(str);``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``string` `str = ``"0001234"``;` `    ``removeLeadingZeros(str);``}``}` `// This code is contributed by ukasp`

## Javascript

 ``

Output

`1234`

Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1)

Java-specific approach: Refer to this article for the Java-specific approach using StringBuffer.

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