dhtml18:
Omo, the kind of codes flying about here na be am o

#include <iostream>

int len_str(char s[]) {
    int i = 0;
    while (1) {
        if(s[i] == NULL)
            break;
        i++;
    }
    return i;
}

bool cmp_char(char s[], char c[]) {

    if(len_str(s) != len_str(c)) { // different lengths
        return false;
    }

    for(int i = 0; i < len_str(s); i++) {
        if(s[i] != c[i])
            return false;
    }
    return true;
}

bool find(char word[], char sentence[]) {
    int word_len = len_str(word);
    int sen_len = len_str(sentence);

    if(word_len > sen_len) {
        std::cout << "warn: word to find is greater than sentence" << std::endl;
        return false;
    }

    char temp[word_len+1];

    for(int i = 0; i <= sen_len - word_len; i++) {

        for(int j = 0; j < word_len; j++) {
            temp[j] = sentence[i+j];
        }
        // terminate temp word
        temp[word_len] = NULL;

        if(cmp_char(word, temp)) {
            std::cout << "From position " << i << " to " << i + word_len << std::endl;
            return true;
        }
    }

    return false;
}

int main() {
    char sentence[] = "the boy is good";
    char word[] = "boy";

    if(find(word, sentence)) {
        std::cout << "Found" << std::endl;
    } else {
        std::cout << "Not Found" << std::endl;

    }
    return 0;
}

CryptoCoinr:
I'm more of an amateur than a newbie (sort of) but it took longer than I expected, and more iterations than I anticipated (still have them in Sublime Text), but I finally did it. This is in JavaScript:

[-- former presentation of code --]

You can test it yourself with your browser (Chrome; any browser really but the shortcut is for Chrome) if you're using a PC.

- Press Ctrl + Shift + J in any open tab
- Copy and Paste the above code into the console and hit the Return/Enter key
- Type find("whatever word you want to find", "whatever sentence you want the word to be found in" ) into the console and hit the Return/Enter key.

function find(word, text)
{
  let index = 0;
  let lastIndex = 0;
  let startIndex = 0;
  let placeholder = "";

  while (startIndex < text.length && placeholder !== word)
  {
    lastIndex = startIndex + (word.length - 1);

    if (word[index] === text[startIndex] && word[word.length - 1] === text[lastIndex])
    {
      for (startIndex; startIndex <= lastIndex; startIndex++)
      {
        if (word[index] === text[startIndex])
        {
          placeholder+= word[index];
          index++;
        }
        else
        {
          index = 0;
          placeholder = "";
          startIndex = lastIndex;
        }
      }
    }
    else
    {
      startIndex++;
    }
  }

  return (placeholder === word) ? 1 : 0;
}

def length(s):
    l=0
    while s:
        l+=1
    return l

def findphrase(p,s):
    sl = length(s)
    pl = length(p)
    fpl = 0 #found phrase length
    cpi = 0 #current phrase index
    csi = 0 #current sentence index
    while sl:
        if fpl == pl: return 1
        elif p[cpi] == s[csi]:
            fpl += 1, csi += 1, cpi += 1, sl - 1
        else:
            fpl = 0, cpi = 0, sl - 1

phrase = input('enter phrase: ')
sentence = input('enter sentence: ')
print(findphrase(phrase, sentence))

words' :: String -> [String]
words' "" = []
words' s = transform s "" []
  where
    transform "" cur acc = (if cur /= "" then cur:acc else acc)
    transform (x:xs) cur acc =
      case x of
        ' ' -> transform xs "" (if cur /= "" then cur:acc else acc)
        _   -> transform xs (cur ++ [x]) acc

finder :: String -> [String] -> Int
finder _ []                = 0
finder needle (x:haystack) = if x == needle then 1 else (finder needle haystack)

find :: String -> String -> Int
find needle haystack = finder needle (words haystack)

greatface:
Implemented in python 3

def length(s):
    l=0
    while s:
        l+=1
    return l

def findphrase(p,s):
    sl = length(s)
    pl = length(p)
    fpl = 0 #found phrase length
    cpi = 0 #current phrase index
    csi = 0 #current sentence index
    while sl:
        if fpl == pl: return 1
        elif p[cpi] == s[csi]:
            fpl += 1, csi += 1, cpi += 1, sl - 1
        else:
            fpl = 0, cpi = 0, sl - 1

phrase = input('enter phrase: ')
sentence = input('enter sentence: ')
print(findphrase(phrase, sentence))

def lenght(x):
    count = 0;
    for r in x:
        count +=1;
    return count;

def findString(word,Sentence):
    count = 0;
    xlength = lenght(word);
    ySplit = Sentence.split();
    ySplit_len = lenght(ySplit);
    while count <= ySplit_len:
        if count == ySplit_len:
            break;
        if x in ySplit:
            return word
        else:
            return "'{}' not Found".format(word.upper());
            
        count += 1;
x = input("Enter the Word : ";
y = input("Enter the Sentence containing the '{}' : ".format(x.upper()));

shittu123:

def lenght(x):
    count = 0;
    for r in x:
        count +=1;
    return count;

def findString(word,Sentence):
    count = 0;
    xlength = lenght(word);
    ySplit = Sentence.split();
    ySplit_len = lenght(ySplit);
    while count <= ySplit_len:
        if count == ySplit_len:
            break;
        if x in ySplit:
            return word
        else:
            return "'{}' not Found".format(word.upper());
            
        count += 1;
x = input("Enter the Word : ";
y = input("Enter the Sentence containing the '{}' : ".format(x.upper()));

edicied:
I no even understand the question self Op are you saying in a String for example "The boy is here" we should check for if the words "The" "BOy" or "Here" is there?

justanALIAS:

Dnt get

silento:



the program will able to match a phrase

like "under it" if matched to " my bed is under it but am cool"


it will return 1 to show that match was found


but 0 when "under" matched to"my bed is underneath it"

silento:



the program will able to match a phrase

like "under it" if matched to " my bed is under it but am cool"


it will return 1 to show that match was found


but 0 when "under" matched to"my bed is underneath it"

jidez007:
Solution in C++

Got c++ exam tomorrow so I may as well write in c++

#include <iostream>

int len_str(char s[]) {
    int i = 0;
    while (1) {
        if(s[i] == NULL)
            break;
        i++;
    }
    return i;
}

bool cmp_char(char s[], char c[]) {

    if(len_str(s) != len_str(c)) { // different lengths
        return false;
    }

    for(int i = 0; i < len_str(s); i++) {
        if(s[i] != c[i])
            return false;
    }
    return true;
}

bool find(char word[], char sentence[]) {
    int word_len = len_str(word);
    int sen_len = len_str(sentence);

    if(word_len > sen_len) {
        std::cout << "warn: word to find is greater than sentence" << std::endl;
        return false;
    }

    char temp[word_len+1];

    for(int i = 0; i <= sen_len - word_len; i++) {

        for(int j = 0; j < word_len; j++) {
            temp[j] = sentence[i+j];
        }
        // terminate temp word
        temp[word_len] = NULL;

        if(cmp_char(word, temp)) {
            std::cout << "From position " << i << " to " << i + word_len << std::endl;
            return true;
        }
    }

    return false;
}

int main() {
    char sentence[] = "the boy is good";
    char word[] = "boy";

    if(find(word, sentence)) {
        std::cout << "Found" << std::endl;
    } else {
        std::cout << "Not Found" << std::endl;

    }
    return 0;
}

justanALIAS:

Oooo ... I thought as long as the string is present..

CryptoCoinr:


I noticed this but I don't think you clarified it in the OP so I'm working on a fix for my solution now.

silento:

so the function will do the following

when given a two parameters the first will be a word and second will be a sentence

it should be able to detect that the word is in the sentence example

function find("boy","obi is a boy"
{
code........

return 0 or 1
}

the function will return 0 or 1
0 if word not found in the sentence
1 if word is found in the sentence

silento:


the program is expected to return 1 in this case:

"i am" => to "i am a boy"


and 0 in this case :

"i am" => "i ammy is a boy"

silento:



i don't to clarify it because it is so obvious

in English is you spell boy as aboy is not a boy again

orimion:
The original question says the first parameter is a word



But here, it is a phrase

CryptoCoinr:


Um, no it isn't. As orimion said above, your question was unclear and even the inbuilt functions of various languages, or JS at least, finds "words" whether they are fragments on solitary.

silento:



that's true but that why regular expressions sell like water

silento:


if you are on a computer

open up your browser


and press ctrl+f now type "word" and click the button for whole words


did you browser highlighted "words" to be "word" if it didn't then i think you got your answer

we are here to learn you will only find examples all over the internet with this bug of matching "boy" to be same with "boym"

silento:
so the function will do the following

when given a two parameters the first will be a word and second will be a sentence

it should be able to detect that the word is in the sentence example

orimion:


I don't think you understand me

I was making the distinction between a word and a phrase

in the question spec, the example given was

"boy" & "i am a boy" should return 1
(i.e word and sentence)

but the part i quoted earlier,

"i am" & "i am a boy"
(i.e phrase and sentence)

I pointed this out because in my solution, i split the sentence into a list of words first (as per the question) then check if the word is in the list
obviously it's not going to work in the second situation (phrase & sentence)

PS. as for the challenge, chrome doesn't have "match whole word" and so highlighted "words" for "word"
PPS. did you even try my haskell solution

silento:
I hope you guys learned a lot from this tiny problem

if some company paid you to wrote a code for them like this one and a after you code has this bug kiss it goodbye I know alot of dudes on nl that will know that matching any occurrence is a bug


if you think I cheated you guys go to stackoverflow and you will see that 86% of there code has this bug while alot of devs there have solved it which others stick with inbuilt buggy function


please check the first page and drop your number if your name is mentioned

silento:
I hope you guys learned a lot from this tiny problem

if some company paid you to wrote a code for them like this one and a after you code has this bug kiss it goodbye I know alot of dudes on nl that will know that matching any occurrence is a bug


if you think I cheated you guys go to stackoverflow and you will see that 86% of there code has this bug while alot of devs there have solved it which others stick with inbuilt buggy function