Remove Leading Zeros From String in Java
Given a string of digits, remove leading zeros from it.
Illustrations:
Input : 00000123569
Output: 123569
Input: 000012356090
Output: 12356090
Approach: We use the StringBuffer class as Strings are immutable.
- Count leading zeros by iterating string using charAt(i) and checking for 0 at the “i” th indices.
- Converting a string into StringBuffer object as strings are immutable
- Use StringBuffer replace function to remove characters equal to the above count using replace() method.
- Returning string after removing zeros
Example:
Java
import java.util.Arrays;
import java.util.List;
class GFG {
public static String removeZero(String str)
{
int i = 0 ;
while (i < str.length() && str.charAt(i) == '0' )
i++;
StringBuffer sb = new StringBuffer(str);
sb.replace( 0 , i, "" );
return sb.toString();
}
public static void main(String[] args)
{
String str = "00000123569" ;
str = removeZero(str);
System.out.println(str);
}
}
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Time Complexity: O(n), where n is the length of the input string. This is because the program iterates through the input string once to count the leading zeroes and then again to remove them using the StringBuffer replace function.
Auxiliary Space: O(n) , as the program creates a new StringBuffer object with the same length as the input string.
Approach : Using substring() method
- Finding the index of first non zero character
- And then use substring method from that index to length of given string
- Assigning the substring to new string
Example:
Java
import java.util.Arrays;
import java.util.List;
class GFG {
public static void main(String[] args)
{
String str = "00000123569" ;
String newstr = "" ;
int ind = 0 ;
for ( int i = 0 ; i < str.length(); i++) {
char p = str.charAt(i);
if (p != '0' ) {
ind = i;
break ;
}
}
newstr = str.substring(ind, str.length());
System.out.println(newstr);
}
}
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Time Complexity: O(n), where n is the length of the input string ‘str’. The algorithm performs a single loop through the length of the string to find the first non-zero character.
Auxiliary Space Complexity: O(n), where n is the length of the input string ‘str’. The algorithm requires a string variable ‘newstr’ to store the modified string without leading zeros.
Last Updated :
03 Feb, 2023
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