1. IsAnagram
2. Count unique substring of length k

Question

1. IsAnagram
2. Count unique substring of length k

Kemo · Accepted Answer

class Solution {
    public static void main(String[] args) {
	System.out.println(isAnagram("kemo", "okme"));

	}

    public static boolean isAnagram(String s, String t) {
        String str =s.toLowerCase();
		String str2 = t.toLowerCase();
		if(str.length() != str2.length()) {
			return false;
		}
		//HashMap hasmap=  new HashMap();
        LinkedList ch =  new LinkedList();
		for (int i = 0; i 0){
            return false;
        }
		return true;		
	}
}

Anonymous · Answer

Your "Anagram" program is wrong and you are using HaspMap (costly data structure) for wrong purpose

I dont get what is your second program

Raktim · Answer

import java.util.Arrays;

public class Anagram {
	
	
	public static void main(String[] args) {
		System.out.println(isAnagram("silent","lentsi"));

	}

	private static boolean isAnagram(String string1, String string2) {
		if(string1.length() != string2.length()){
			return false;
		}
		else{
			 char[] temp1 = string1.toCharArray();
			 Arrays.sort(temp1);
			 char[] temp2 = string2.toCharArray();
			 Arrays.sort(temp2);
			 
			 for(int i=0;i

Wei · Answer

unique substring of length k in python:
================================
#!/usr/bin/python
import argparse

p = argparse.ArgumentParser()
p.add_argument('-s')
p.add_argument('-l', type=int)
args = p.parse_args()
entered_str = args.s
entered_len = args.l
print 'Entered string is:', entered_str
print 'Entered length is:', entered_len

#out_raw = list(entered_str)
#print 'The raw char list is:', out_raw
out_raw_k = entered_str[:entered_len]
print 'The shortened raw char list is:', out_raw_k
print 'Unique substring with length of 1:', set(out_raw_k)
print 'Unique substring with length of 1 count:', len(set(out_raw_k))
out = [0, len(set(out_raw_k))]
i = 2 
while i <= entered_len:
  chunk = (len(out_raw_k)/i)*i
  sub_list = [ out_raw_k[y:y+i] for y in range(0, chunk, i) ]
  print 'Unique substring with length of', i, ':', set(sub_list)
  print 'Unique substring with length of', i, 'count:', len(set(sub_list))
  out.append(len(set(sub_list)))
  i += 1

print 'Total unique substring count:', sum(out)

=======Output from UNIX============
$ ./test -s phenomenon -l 3
Entered string is: phenomenon
Entered length is: 3
Starting position: 0 substring count: 3 list: ['phe', 'nom', 'eno']
Starting position: 1 substring count: 3 list: ['hen', 'ome', 'non']
Starting position: 2 substring count: 2 list: ['eno', 'men']
Starting position: 3 substring count: 2 list: ['nom', 'eno']
Starting position: 4 substring count: 2 list: ['ome', 'non']
Starting position: 5 substring count: 1 list: ['men']
Starting position: 6 substring count: 1 list: ['eno']
Starting position: 7 substring count: 1 list: ['non']
Total substring list: ['phe', 'nom', 'eno', 'hen', 'ome', 'non', 'eno', 'men', 'nom', 'eno', 'ome', 'non', 'men', 'eno', 'non']
Total unique substring count: 7 list: set(['nom', 'non', 'ome', 'men', 'phe', 'eno', 'hen'])

Wei · Answer

Use Python for anagram:

#!/usr/bin/python
import argparse

p = argparse.ArgumentParser()
p.add_argument('-source')
p.add_argument('-comp')
args = p.parse_args()
str_01 = args.source
str_02 = args.comp

str_01_list = sorted(list(str_01))
str_02_list = sorted(list(str_02))

if str_01_list == str_02_list:
  print 'Source string:', str_01, 'is anagram of the comparing string:', str_02
else:
  print 'Source string:', str_01, 'is NOT anagram of the comparing string:', str_02

Wei · Answer

unique substring of length k in python:
================================
#!/usr/bin/python
import argparse

p = argparse.ArgumentParser()
p.add_argument('-s')
p.add_argument('-l', type=int)
args = p.parse_args()
entered_str = args.s
entered_len = args.l
print 'Entered string is:', entered_str
print 'Entered length is:', entered_len

#out_raw = list(entered_str)
#print 'The raw char list is:', out_raw
out_raw_k = entered_str[:entered_len]
print 'The shortened raw char list is:', out_raw_k
print 'Unique substring with length of 1:', set(out_raw_k)
print 'Unique substring with length of 1 count:', len(set(out_raw_k))
out = [0, len(set(out_raw_k))]
i = 2 
while i <= entered_len:
  chunk = (len(out_raw_k)/i)*i
  sub_list = [ out_raw_k[y:y+i] for y in range(0, chunk, i) ]
  print 'Unique substring with length of', i, ':', set(sub_list)
  print 'Unique substring with length of', i, 'count:', len(set(sub_list))
  out.append(len(set(sub_list)))
  i += 1

print 'Total unique substring count:', sum(out)

=======Output from UNIX============
$ ./test -s phenomenon -l 3
Entered string is: phenomenon
Entered length is: 3
Starting position: 0 substring count: 3 list: ['phe', 'nom', 'eno']
Starting position: 1 substring count: 3 list: ['hen', 'ome', 'non']
Starting position: 2 substring count: 2 list: ['eno', 'men']
Starting position: 3 substring count: 2 list: ['nom', 'eno']
Starting position: 4 substring count: 2 list: ['ome', 'non']
Starting position: 5 substring count: 1 list: ['men']
Starting position: 6 substring count: 1 list: ['eno']
Starting position: 7 substring count: 1 list: ['non']
Total substring list: ['phe', 'nom', 'eno', 'hen', 'ome', 'non', 'eno', 'men', 'nom', 'eno', 'ome', 'non', 'men', 'eno', 'non']
Total unique substring count: 7 list: set(['nom', 'non', 'ome', 'men', 'phe', 'eno', 'hen'])

Kemo · Answer

import java.util.HashMap;

public class IsAnagram {

	public static void main(String[] args) {
	System.out.println(isAnagram("kkemo", "okkme"));

	}

	private static boolean isAnagram(String string, String string2) {
		String str =string.toLowerCase();
		String str2 = string2.toLowerCase();
		if(str.length() != str2.length()) {
			return false;
		}
		HashMap hasmap=  new HashMap();
		for (int i = 0; i < string.length(); i++) {
			hasmap.put(str.charAt(i),i+1);
		}
		for (int i = 0; i < hasmap.size(); i++) {
			if(hasmap.get(str2.charAt(i)) == null) {
				return false;
			}
		}
		return true;		
	}

}

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