CareerCup

Let str[] denote the given string. I am creating an array arr[] which stores the words as I read them.
i=0,k=0,count=0
while(str[i++]!='\0')
do
if(str[i]!=' ')
arr[k++]=str[i]
count++
else
if(count==1)
arr is one of the ans and continue finding for others
k=0
empty arr again

The question is : are we looking for unique words of 1 length? or not?

Assume we have the answer and we are taking care of this in every approach.

First Approach:
- Split by " " String[] token = input.split(" ") , go word by word and if it is of length 1 count++;
// it uses O(nr of words) additional space

SecondApproach:
- go element by element and if it is surround it by " " or if it is at the beginning of the string and followed by " " or it at the end of a string and preceded by " "
- if so count++;
// this one uses O(1) space

Third approach:
-use pattern matching:

Pattern p = Pattern.compile("\\w+");
  Matcher m = p.matcher(input);
  while(m.find()){
     if(m.group().length()==1) count++;
  }

public int singleAlphebetCounter(String s)
{
int counter = 0;

for(String temp:s.split(" "))
if(temp.length()==1)
counter++;

return counter;
}

assign each space char as operation +
each char as a value = 1
we need to convert the string into a numerical expression where a multiple char numerical value = 0
When a multiple characters string is found, its value would be str[0]-str[1], ignoring remaining characters in building expression
So the string "I am a king" will become
1+(1-1)+1+(1-1) = 2 //replace space with + and while scanning a multichar string, just scan first two chars
Complexity = O(n)

My Algorithm as following:
1) increase the count if the character is between A-Z or a-z (we can expand the list as needed).
2) for the rest characters, reset the counter.
3) Before reset the counter, we check and see if counter == 1

while (str != 0 && str[pos] != '\0')
   {      
      if ((str[pos] >= 'A' && str[pos] <= 'Z') || (str[pos] >='a' && str[pos] <= 'z'))
      {
        counter++;
      }
      else
      {
         if (counter == 1)
         {
            printf("found:%c\n", str[pos-1]);
         }  
            
         counter = 0;
      }   
      pos++;         
   }

if (isspace(*(a+1)))
{
count++;
}

if (isspace(*(a+length-2)))
{
count++;
}

for (i=1;*a;i++)
{
if (isspace(*(a-1)) && isspace(*(a+1)))
{
count++;
}
a++;
}

How about this??

int main()
{
    char str[]="i am a king";
    char *str2=strtok(str," ");
    
    int count=0;
    
    while(str2!=NULL)
    {
                    if(strlen(str2)==1)
                                      count++;
                    str2=strtok(NULL," ");
    }
    
    cout<<"count is"<<count;
    
    getch();
    return 1;
}

int main()
{
char *str = "I am a M a c";
char *tmp_str1 = NULL;
int count = 0;

tmp_str1 = str;

while (tmp_str1)
{
if ((tmp_str1[1] == '\0') || (tmp_str1[1] == ' '))
{
count++;
}
tmp_str1 = strchr (tmp_str1,' ');
if (tmp_str1)
{
tmp_str1++;
}
else
{
break;
}
}

printf ("%d", count);

}

#include<iostream>
#include<stdio.h>
#include<conio.h>
using namespace std;
int alpha(char *a)
{
int count=0;
if(*(a+1)==' ');
count++;
while(*a)
{
if(*(a-1)==' '&&*(a+1)=='\0')
count++;
if(*(a+1)==' '&&*(a-1)==' ')
count++;
a++;
}
return count;

}
int main()
{
char a[20];
gets(a);
cout<<alpha(a);

getch();
return 0;
}

int countSingleLetterWord(const char *in)
{
int count,i;
i=0;count=0;
if(*in=='\0')
{
return 0;
}
do
{
if(*in==' ' || *in=='\0')
{
if(i==1)
{
count++;
}
i=0;
}
else
{
i++;
}
}while(*in++);
return count;
}

#include<stdio.h>
#include<conio.h>

int main()
{
char arr[20]="Am i a king?";
int i=0,k=0,count=0;

while(i<strlen(arr))
{
if( arr[i]==' ')
{
if((i-k)==1)
{
count+=1;
}
k=i+1;
}
i++;
}
printf("\Number of single letter words in a sentence:%d \n ",count);
return 0;
}

#include<stdio.h>
#include<conio.h>

int main()
{
char arr[50]="i am a MAC ? a HOST";
int i=0,k=0,count=0;

while(i<strlen(arr))
{
if( arr[i]==' ')
{
printf("\n %d %c",i,arr[i-1]);
if( (65<=arr[i-1]&&arr[i-1]<=90) || ( 97<=arr[i-1] && arr[i-1]<=122 ) )
{
printf(" \t HI ");
printf("%d %d \n",i,k);
if((i-k)==1)
{
count+=1;
}
k=i+1;
}
else
{
k=i+1;
}
}
i++;
}
printf("\n Number of single letter words in a sentence:%d \n ",count);
return 0;
}

My implementation:

#include <stdio.h>
unsigned int countsingleword(char*p)
{
        unsigned int count, count1;
        if(NULL==p)
                return 0;
        for(count=0, count1=0;*p;p++)
        {
                if(' '==*p)
                {
                        if(1==count1)
                                count++;
                        count1=0;
                }else
                {
                        count1++;
                }
        }
        if(1==count1)
                count++;
        return count;
}

int main()
{
        char p[]="I am a humble human! I am ";
        printf("string %s has %d single word\n", p, countsingleword(p));
}

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int no_single_letter_words(char str[])
{
int length=0,check,count=0,i=0;
while(i<strlen(str))
{
if(str[i]!=' ')
length++;
else
{
check=length;
length=0;
if(check==1)
count++;
}
i++;
}
if(length==1)count++; //cheking if the last word consists of a single letter
return count;
}
int main()
{
char str[]="I am a King";
int answer=no_single_letter_words(str);
printf("%2d\n",answer);
}

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main ()
{
char *p="I am a king";
int n=0,i;
for(i=0; i<strlen(p);i++)
{
if((i==1) || (i==(strlen(p)-2)))
if(p[i]==' ')
n++;

if((p[i]==' ') && (p[i+2]==' '))
n++;
}

printf(" Number of single letter word in the sentence is %d \n", n);

}

What about double space in case we will have , Pallavi ?
Like : char *p="I am king r t";

/* This is not a compact code ... Just an approach */
#include<stdio.h>
void wordcount(char *str)
{
	char *temp = str;
	int num =0,count = 0;
    while(*temp!= '\0')
	{
		if(*temp == ' ' && num == 1) 
		{
			count++;
			num = 0;
		}
		else
		{
		num++;
	
		}
		if(*temp == ' ' )

		{
			num = 0;
		}
			temp++;


	}
			printf("the count is %d",count);

}
int main()
{
char *str = "I am a king";
wordcount(str);
return 0;
}

#include<stdio.h>

int main()
{
	char c[] = "t t h to the j enfo j o k";
	char *cp = c;
	int count = 0;
	if(c[1] == ' ')/*checks 1st character is single*/
		count++;
	while((*cp) != '\0') {
		if((*cp) == ' ') {
			if( *(cp+= 2) == ' ') {
				count++;
				cp-=2;/* i used to modify the pointer, so it missed some blanks*/
			}
		}
		cp++;
	}
	if(c[strlen(c) - 2] == ' ')/*checks last character is single*/
		count++;
	printf("count = %d\n",count);
	return 0;
}

int FindSingleWord(const char* s)
{
	int size=strlen(s);
	int count=0;
	int wordsum=0;
	if(s==NULL)
		return 0;
	for(int i=0;i<size;i++)
	{
		if(*(s+i)==' ')
		{
			if(wordsum==1)
				count++;
		    wordsum=0;
		}
		else
		{
			wordsum++;
		}
      
	}
	return count;
}

#include <stdio.h>

int main() {
  char *s = "I am A king";
  int count = 0;
  int base = 0;
  int iter = 0;
  while (*s != '\0') {
    if (*s == ' ') {
       if ((iter - base) == 1) {
          count ++;
       }
       base = iter+1;
    }
    iter ++;
    s ++ ;
  }

  printf("%d\n", count);
}

int find1s(char* s)
{
	bool before=1;
	int size =strlen(s);
	int cnt=0;
	for(int i=0;i<size;i++)
	{
		if(*(s+i)!=' ')
		{
			if((*(s+i+1)==' ')||(*(s+i+1)==0))
			{
				if(before==1)
				{
				cnt++;
				i=i+1;
				}
				else
				{
					
				}
			}
			else
			{
				before=0;
			}
		}
		else
		{
			before=1;
		}
	}
	return cnt;

}