Remove leading zeros from a Number given as a string - GeeksforGeeks

10 Jun, 2021
Given numeric string str, the task is to remove all the leading zeros from a given string. If the string containing 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: 0 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. 
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 “$” means the end of the string. 
Use the inbuilt replaceAll() method of the String class which accepts two parameters, a Regular Expression, and a Replacement String.
To remove the leading zeros, pass a Regex as the first parameter and empty string as the second parameter.
This method replaces the matched value with the given string.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ Program to implement// the above approach#include <iostream>#include <regex>using namespace std; // Function to remove all leading// zeros from a a given stringvoid removeLeadingZeros(string str){    // Regex to remove leading    // zeros from a string    const regex pattern("^0+(?!$)");     // Replaces the matched    // value with given string    str = regex_replace(str, pattern, "");    cout << str;} // Driver Codeint main(){    string str = "0001234";    removeLeadingZeros(str);    return 0;}
// Java Program to implement// the above approachimport java.util.regex.*;class GFG{     // Function to remove all leading    // zeros from a a given string    public static void removeLeadingZeros(String str)    {         // Regex to remove leading        // zeros from a string        String regex = "^0+(?!$)";         // Replaces the matched        // value with given string        str = str.replaceAll(regex, "");         System.out.println(str);    }     // Driver Code    public static void main(String args[])    {        String str = "0001234";         removeLeadingZeros(str);    }}
# Python3 Program to implement# the above approachimport re # Function to remove all leading# zeros from a a given stringdef removeLeadingZeros(str):     # Regex to remove leading    # zeros from a string    regex = "^0+(?!$)"     # Replaces the matched    # value with given string    str = re.sub(regex, "", str)     print(str) # Driver Codestr = "0001234"removeLeadingZeros(str) # This code is contributed by avanitrachhadiya2155
// C# program to implement// the above approachusing System;using System.Text.RegularExpressions; class GFG{ // Function to remove all leading// zeros from a a given stringpublic static void removeLeadingZeros(string str){         // Regex to remove leading    // zeros from a string    string regex = "^0+(?!$)";     // Replaces the matched    // value with given string    str = Regex.Replace(str, regex, "");     Console.WriteLine(str);} // Driver Codepublic static void Main(string[] args){    string str = "0001234";     removeLeadingZeros(str);}} // This code is contributed by ukasp
<script>// Javascript Program to implement// the above approach // Function to remove all leading    // zeros from a a given stringfunction removeLeadingZeros(str){    // Regex to remove leading        // zeros from a string        const regex = new RegExp("^0+(?!$)",'g');          // Replaces the matched        // value with given string        str = str.replaceAll(regex, "");          document.write(str);} // Driver Codelet str = "0001234";  removeLeadingZeros(str); // This code is contributed by unknown2108</script>
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.
avanitrachhadiya2155
yuvraj_chandra
ukasp
unknown2108
java-StringBuffer
regular-expression
Mathematical
Pattern Searching
Strings
Strings
Mathematical
Pattern Searching
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