CareerCup

simple kmp

#include<stdio.h>
#include<string.h>

int occur(char[],char[]);

int occur(char sent[],char pattern[])
{

	int count=0;

	for(int i=0,j=i;sent[i]!='\0';)
	{
		if(sent[i+j]==pattern[j]&&pattern[j]!='\0')
		{
			j++;
			
			
		}
	   else if(j==strlen(pattern))
		{
			count++;
			i=i+j;
			j=0;
		}
		else
		{
			i++;
			j=0;
			
		}
		
	}
	return count;
}

int main()
{
	char sent[] = "aabaabaaabbbabababddfggaabbbasab";
	char pattern[] = "aba";
	int result = occur(sent,pattern);
	printf("\nNo of Occurrences --- > %d",result);
	return 0;
}

public static void countOccurence(String str, String pattern)
    {
    int n = 0, i = 0;
    while (i < str.length()
            && (i = str.indexOf(pattern, i)) != -1) {
        ++n;
        ++i;
    }
    System.out.println("n: " + n);
   }

It should be O(n) time complexity I think.

public class FindPatternOccurrences {

	public static void main(String[] args) {
		String str = "aabababcfdc";
		String pattern = "ab";
		matchPattern(str, pattern);
	}

	public static void matchPattern(String str, String pattern) {

		int count = 0;
		for (int i = 0; i < str.length() - 1; i++) {
			String str1 = str.substring(i, i + 2);
			if (str1.equals(pattern))
				count++;
		}
		System.out.println("NUmber of occurrences= " + count);
	}

}

This prints the number of occurrences of the pattern:

import java.util.Collections;
import java.util.HashMap;
import java.util.Map;

public class SubString {
public static void main(String[] args) {
	String str= "abbbacctlkababcdjlkcccabcdaaabbbabcd";
	String temp = "abcd";
	Map<String,Integer> map = new HashMap<>();
	int len = str.length();
	System.out.println(len);
	int templen = temp.length();
	for(int i=0;i<len-templen+1;i++){
		if(map.containsKey(str.substring(i,i+templen)))
		map.put(str.substring(i,i+templen), map.get(str.substring(i,i+templen))+1);
		else
			map.put(str.substring(i,i+templen), 1);	
	}
	System.out.println(map.get(temp));
	
	
}
}