Remove leading zeros from a Number given as a string

Given a 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

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:

  1. 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.

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

Below is the implementation of the above approach:

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to implement
// the above approach
import 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);
    }
}

chevron_right


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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up


If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.