₦airaland Forum

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

no

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.

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

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



But here, it is a phrase

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"

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.

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

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

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

presence mean "boy" not "aboyoy"

the two word are different

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;
}

NICE WORK BUT return 1

when matched to "the aboyu is a man"

justanALIAS:
Dnt get

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"

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?

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"

[quote author=justanALIAS post=59077579]I dont think so but i will try. By the way, the above code is buggy; see d fixed version:


package main

import (
"fmt"
)
func length(dstring string) int { // i made my own len() method
var counter int
for range dstring {
counter++
}
return counter
}
func main() {
mainstr := "i am a bon boni"
substr := "i am"
fmt.Println(check(mainstr, substr))
}
func check(mainstr string, substr string) int{
var status bool = false
for i, _ := range mainstr {
if mainstr[i] == substr[0] {
if length(mainstr[i:]) >= length(substr) {
last := len(substr)
di := i + last
var rangeof string = mainstr[i:di]
if substr == rangeof{
status = true
} else {
continue
}
} else {
continue
}
}
}
if status {
return 1
} else {
return 0
}
}

[/quote




NICE WORK BUT your code

returns 1 when "i am" matched against "i ammy"

closed testing result winners will be announced once done

justanALIAS:
I dont think so but i will try. By the way, the above code is buggy; see d fixed version:


package main

import (
"fmt"
)
func length(dstring string) int { // i made my own len() method
var counter int
for range dstring {
counter++
}
return counter
}
func main() {
mainstr := "i am a bon boni"
substr := "i am"
fmt.Println(check(mainstr, substr))
}
func check(mainstr string, substr string) int{
var status bool = false
for i, _ := range mainstr {
if mainstr[i] == substr[0] {
if length(mainstr[i:]) >= length(substr) {
last := len(substr)
di := i + last
var rangeof string = mainstr[i:di]
if substr == rangeof{
status = true
} else {
continue
}
} else {
continue
}
}
}
if status {
return 1
} else {
return 0
}
}

ok

justanALIAS:
package main

import (
"fmt"
)
// i made my own len() method
func length(dstring string) int {
var counter int
for range dstring {
counter++
}
return counter
}
func main() {
mainstr := "i am a bon boi"
substr := "boi"
fmt.Println(check(mainstr, substr))
}
func check(mainstr string, substr string) int {
var status bool = false
for i, _ := range mainstr {
if mainstr[i] == substr[0] {
last := length(substr)
di := i + last
if substr == mainstr[i:di] {
status = true
} else {
continue
}
}
}
if status {
return 1
} else {
return 0
}
}

range
is inbuilt operator in golang can you scrap it

finally we are improving

pcguru1:
Sorry just seeing this, I will post mine the goal is no inbuilt function and yet people still using contains and strpos this is why Nigerians fail algorithm test, even the interview we conducted we spelt it out loud and yet still the same thing.

nawa the tire me ooo

mustaphagreens:
Please check the attached images. Nairaland clearly does not allow for indentation.

Language: Python
Version: 3.0
Using the in operator
def findString(x,y):
if x in y:
return 1
else:
return 0

#an instance of findString
print(findString("boy","I am a boy" ))



Done!!!
Check out the code below or in the second image
#findString without the in operator

def findString(x,y):
i=0
p=""
while i<=len(y)-1:
if (y[i]!=" " and p!=x):
p+=y[i]
i+=1
elif p==x:
return 1
else:
i+=1
p=""
if p==x:
return 1
else:
return 0

#client program invoking findString
x=input("enter target word:" )
y=input("enter phrase or sentence:" )
print(findString(x,y))

wow that's great nice indentation in your code

But can you rewrite the code so that if Len()

will not be there

the essence of this contest is to make you understand what happen beneath the standard libraries and function

anyway you will do great things in python keep on trying you got the heart of a coded indeed

python user without the help of in operator rewrite you code

golang hard ooo

dhtml18:
And of course I will be umpire. . .

the troll master

Welete:
#python 3.x
def wordsearch(x, sent):
for x in sent:
return 1
break
else: return 0

x = input('word: ')
sent = input('sentence: ')
wordsearch(x, sent)

in operator a membership operator only in python I consider it inbuilt function

Welete:
nairaland is messing with my indentation..

no fancy operator

Jaddo19:
Using Php
.
.
<?Php
If (strpos("is This What You Want","What"){
Echo true;
}else{
Echo (int)false;
}
?>

great but no inbuilt function or library and fancy operator

https://msdn.microsoft.com/en-us/library/2hxce09y(v=vs.90).aspx?_e_pi_=7%2CPAGE_ID10%2C1207831521

Desyner:
Clarify please. Custom functions that don't call library function?

just a custom function but no library or inbuilt function

Jaddo19:
I Really Dont Understand D Question Fully, Can U Explain Better Pls

I hope you understand it now

3days for the contest to run

just a simple but powerful function to improve our coding skills


beginners onlys

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


contest is closing on 3rd of this month

Note : you are not allowed to use any inbuilt library or functions just from scratch code every thing




some time similar to str.find()

in python


any language is welcomed but c and c ++ Will be more interesting






pros are not welcomed

just New coders




SINCE WE DO NOT HAVE A WINNER

I WILL LIKE TO Honor THIS GREAT PROGRAMMERS

1.justanALIAS (golang)

2.jidez007 (c++)


please drop your numbers for a little appreciation









SOLUTION IN C language

#include <stdio.h>



//FUNCTION THAT CHECKS if a phrase exists in sentence

int found(char* phrase,char *sentence)
{
    //one line function to check lenght
    int len(char*item){int t=0;while(item[t] !='\0')t++;return t;}

     // ckecking length
    int sent_len = len(sentence);
    int word_len = len(phrase);

    //variables for keeping records and also indexing through string array
    int c=0,j=0,chk=0;

    //for loop to loop through sentence
    for(c=0;c<sent_len;c++)
    {


        //check if phrase[0] is equal to sentence[c]
    	if(phrase[j]==sentence[c])
    	{
    	    /*if sentence[c-1] is not  a whitespace ,and also j==0 and c not ==0
    	    that mean that the phrase is just similar word like 'to' and 'toy' to avoid it such scenario this if statement is must
                */

    	    if(sentence[c-1] !=32 & j==0 & c !=0)
    	    {
    	        break;
    	    }

    	    //if above evaculation is false then this one increment this record variables
    	    else
            {
                chk+=1;
                j+=1;
            }


		}

		/*
		while matching word like 'toma' against 'toys' once it match 'm' to 'y'
            it will reset record variable to null and it means not matched

            */
		else
		{
			chk=0;
			j=0;
		}


        //if chk record variable is equal to phrase lenght then break loop because a match found
		if((chk==word_len)&(sentence[c+1]==32))
		{

			break;
		}



    }

    if(chk==len(phrase))
        return 1;
	else
        return 0;

}


int main()
{

    char r[]="boy are goo";
    char t[] = "d boy you boy are goo d";
    printf("%d ",found(r,t));
    return 0;
}

orimion:
web?

although it is not enough but c and web Google it

php and Python and c and c++ I haven't found anything this Lang can't do if somebody can use c to program an os what else do I need again that it cannot offer

hahaha that one go be tori master dhtml18

seun act Now all we go force you to act

python my wife

problem with Africans make small money and forget the future seun is lost hope on programming section because it is not generating enough traffic but no problem we will manage it anyhow we see it but mind you one day you will need our help to move your website to the next level na that day you go hear version 3


tutorial on git loading for Newbies

stay tuned for the topic

seun do something now is time for change

seun why are u mute to this issues is it because you don't have competitors just tell me if naijaloaded is still a forum website like nairaland that will over look this section for a very long period

serious what attracts me to nairaland was the programming section and I know alot of young devs that are here but disappointed with how things are going here

all we are asking is for a mod that can delete useless post and ban spammers

is that much please

I can't die with all this my knowledge on c,cpp,python,php,css,etc

we need to share and evolve all together

you too can learn from others and improve nairaland and don't tell me you love nairaland as it is now I just discovered that you hardly code since last year if you do code at all