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:

- 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:

## Java

`// 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); ` ` ` `} ` `} ` |

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*filter_none*

**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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