You are given two strings, remove the characters from the first string which are present in the second string.
Algorithm: Let first input string be “test string” and the string which has characters to be removed from first string be “mask”
1: Initialize:
res_ind = 0 /* index to keep track of processing of each character in i/p string */
ip_ind = 0 /* index to keep track of processing of each character in the resultant string */
2: Construct count array from mask_str. Count array would be:
(We can use Boolean array here instead of int count array because we don’t need count, we need to know only if character is present in mask string)
count[‘a’] = 1
count[‘k’] = 1
count[’m’] = 1
count[’s’] = 1
3: Process each character of the input string and if count of that character is 0 then only add the character to the resultant string.
str = “tet tringng” // ’s’ has been removed because ’s’ was present in mask_str but we we have got two extra characters “ng”
ip_ind = 11
res_ind = 9
4: Put a ‘′ at the end of the string if required.
Java code sample is :
public class RemoveFirstFromSecond {
public static void main(String\[\] args) {
String first = "this is a test";
String second = "mask";
boolean bit\[\] = new boolean\[256\]; //boolean are defaulted to false
//Set the bits on bit\[\] for the mask string
for(int i=0;i<second.length();i++){
bit\[second.charAt(i)\] = true;
}
//This is not good, we can use StringBuffer
//char\[\] is just for the proof of concept
char\[\] result = new char\[first.length()\];
for(int i=0,j=0;i<first.length();i++){
if(bit\[first.charAt(i)\]==false){
result\[j\] = first.charAt(i);
j++;
}
}
String resultstr = new String(result);
System.out.println("Result - " + resultstr);
}
}//:~