# Challenge :

Your job is to find the given number N.

# Input :

You will be given a string in the following form:

dN ± Y = X


Where:

d, N, X, Y are all numbers with d , X , Y being integers (not decimals).


# Output :

Solve the equation and output the value of N rounded. Round up if decimal is greater than or equal to 0.5 and round down if less than 0.5

# Examples :

Input                  Output
N + 1 = 3        ---> 2
N - 1 = 5        ---> 6
2N + 1 = 3        ---> 1
2N + 1 = 4        ---> 2


# Winning criteria :

This is so the shortest code in each language wins.

# Notes :

• Answer must be rounded it cannot be a float
• Whitespace do not matter. You can have every element in the string separated by white space if you want or if you don't want leave it as it is.
• You do not need to check for invalid input. i.e. All inputs will be strings in the given format.

• Comments are not for extended discussion; this conversation has been moved to chat. – Mego Apr 19 '18 at 2:31
• @Mego : I have (hopefully) made it more clear. Can you kindly take a look and open it? – Muhammad Salman Apr 19 '18 at 7:08
• @MuhammadSalman I'll leave both the original and add the alternative [-128,127] range answer then. (And I will delete some of my answers now, thanks for answering them.) – Kevin Cruijssen May 25 '18 at 12:26
• Can d , X , and Y be negative? what is the maximum value that needs to be handled? – Taylor Scott May 25 '18 at 13:23
• @TaylorScott : If you want you can implement that. – Muhammad Salman May 25 '18 at 13:30

# Python 2, 8874 70 bytes

-14 bytes thanks to Chas Brown.
-4 bytes thanks to Sunny Patel.

d,w,Y,_,X=input().split()
print round((int(X)-int(w+Y))/float(d[:-1]))


Try it online!

• @ovs : He he broke it. If I remove even a single space bam. It blows up. This is whitespace sensitive. – Muhammad Salman Apr 16 '18 at 18:54
• @MuhammadSalman This was the case before. This solution requires at least one space between the different parts. If this is not allowed I will delete my answer. – ovs Apr 16 '18 at 19:02
• @ovs: Na it is perfectly alright. – Muhammad Salman Apr 16 '18 at 19:04

under the assumption (according to comments) that we deal only with digits

# JavaScript (Node.js), 30 bytes

*thanks for @Rick Hitchcock for pointing out i can use 1 instead of space (reduces 13 bytes

x=>(x[3]+x[5]-x[9])/-x[0]+.5|0


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# Python 2, 64 63 bytes

Using eval

d,w,Y,_,X=input().split()
print-eval('('+w+Y+'-'+X+')/'+d[:-1])


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Saved 1 byte thanks to Keyu Gan

# Convex, 19 bytes

S-®-?\d+"ô:d~\-\/mo


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# Wolfram Language (Mathematica), 54 46 bytes

Round[ToExpression[""<>(#/."="->"-")]~Root~1]&


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# JavaScript (Node.js), 44 bytes

x=>([m,a,r]=x.split(/N|=/),r-a)/(+m||1)+.5|0


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Or, for only single-digit numbers:

# JavaScript (Node.js), 36 bytes

([m,,o,a,,r])=>-(o+a-r)/(+m||1)+.5|0


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# C (clang), 133 bytes

N,d,y,x,a[9],o;f(*s){sscanf(s," %[0-9N] %c %d = %d",a,&o,&y,&x);d=!sscanf(a,"%d",&d)?1:d;N=(x-(o>43?-y:y))*2/d;printf("%d",N/2+N%2);}


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This handles all spaces except between d and N.

[Removed wrong answer that rounds down]

• some spaces can be removed – l4m2 May 26 '18 at 1:41
• 124 bytes – ceilingcat Nov 23 '18 at 19:43

# Befunge-93 (FBBI), 34 26 bytes

&~$~90p0&X&-2*\/-:2%\2/+.@  Try it online! If we didn’t need to bother with rounding correctly, it could be much smaller: &~$~90p0&X&-\/-.@


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-\/-

• I think this fails to round, and instead truncates. For example, it returns 0 for 3N+1=3 when it should round 2/3 up to 1 – MildlyMilquetoast Apr 17 '18 at 21:45
• Fixed <padding> – Jo King Apr 18 '18 at 7:16
• For rounding, you can double the numerator before dividing, add the quotient to itself mod 2, and then divide the sum of that by 2, instead of doing the multiply by 10 thing – MildlyMilquetoast Apr 18 '18 at 21:11
• np, always happy to help – MildlyMilquetoast Apr 18 '18 at 23:40

# Java 10, 111 94 (or 92) bytes

s->{var a=s.split("N|=");return(int)((new Float(a[2])-new Float(a[1]))/new Float(a[0])+.5);}


Input is in the format dN±Y=X (without any spaces) instead of dN ± Y = X.

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2 more bytes could be saved if the numbers are never outside the [-128, 127] range, in which case the last and one of the two first Float can be changed to Byte.
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Explanation:

s->{                             // Method with String parameter and integer return-type
var a=s.split("N|=");          //  And split by "N" or "=" to a String-array
return(int)((new Float(a[2])   //  Return the third number (after the equal sign),
-new Float(a[1])) //   minus the second number (including leading +/-),
/new Float(a[0])   //   divided by the first number
+.5);}                  //  Rounded by using (int)(R + 0.5)


# Yabasic, 142 bytes

An anonymous function that takes input as a string and outputs to the console.

Line Input""s$dim n$(5)
k=token(s$,n$())
n$=n$(1)
If n$="N"n$=("1N")
?Int((Val(n$(5))-Val(n$(2)+"1")*Val(n$(3)))/Val(Left$(n$,Len(n$)-1))+.5)


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# PHP, 45 bytes

<?=round((($s=$argn)[3].$s[5]-$s[9])/-$s[0]);  Run with echo '<input>' | php -nF <filename> or Try it online. This requires that d always be given, and that the string is the same length each time, e.g.: • N + 1 = 3 won't work, but 1N + 1 = 3 will work. • 1N +1 =3 won't work, but 1N + 1 = 3 will work. # J, 26 Bytes <.({.%~{:-1&{)".>0 2 4{cut  Note that the result for N is floored (not rounded), so for the last test case it returns 1 instead of 2. If this is unacceptable I will change the answer. Input should be of the form 'd N +-Y = X' ### Explanation:  cut '10 N +2 = 23' NB. Split on whitespace ┌──┬─┬──┬─┬──┐ │10│N│+2│=│23│ └──┴─┴──┴─┴──┘ 0 2 4{cut '10 N +2 = 23' NB. Choose relevant portions ┌──┬──┬──┐ │10│+2│23│ └──┴──┴──┘ ".>0 2 4{cut '10 N +2 = 23' NB. Eval each 10 2 23 ({.%~{:-1&{)".>0 2 4{cut'10 N +2 = 23' NB. Compute (x-y)/d 2.1 <.({.%~{:-1&{)".>0 2 4{cut'10 N +2 = 23' NB. Floor 2  ## C# (149 Bytes) s=>(int)Math.Round(((s=s.Replace(" ",""))[s.Length-1]-48-((s.Contains("-")?-1:1)*(s.Split(new[]{'+','-'})[1][0]-48)))/(decimal)(s[0]=='N'?1:s[0]-48))  Craptacular bloated C# solution. Guess I could've ported the Java solution but I try not to look at others' stuff before posting mine, so here we are. Try it Online # Ruby, 66 bytes ->s{a,_,b,_,c=s.split.map &:to_f;((s[?+]?c-b:c+b)/(a+0**a)).round}  Try it online! # awk, 48 bytes awk -F'[ N]' '{printf("%.f",($6-$4)/($1?$1:1))}'  Use cases: echo "N + 1 = 3" | awk -F'[ N]' '{printf("%.f",($6-$4)/($1?$1:1))}' echo "N - 1 = 5" | awk -F'[ N]' '{printf("%.f",($6-$4)/($1?$1:1))}' echo "2N + 1 = 3" | awk -F'[ N]' '{printf("%.f",($6-$4)/($1?$1:1))}' echo "2N + 1 = 4" | awk -F'[ N]' '{printf("%.f",($6-$4)/($1?$1:1))}'  Extra case: echo "3N + 1 = 11" | awk -F'[ N]' '{printf("%.f",($6-$4)/($1?$1:1))}'  Output: 3 ## PHP, 83 Bytes Try it online Try it online2 For some reason round does not round to the closest integer on tio run, on this version the solution has some extra bytes (87 Bytes) Code <?=round(($c=explode(" ",$argv))[4]-($c[1].$c[2]))/($c[0]==N?1:substr($c[0],0,-1));  Explanation //ecuation structure xN sign c = m //array structure like ["xN","sign", "c", "=", "m"] round( //round the result ( //first element of the ecuation$c=explode(" ",$argv))[4] //explode the string and get the 4th value - //substract, php accepts (1-(-1)) so, //it can ve solved like this ($c[1].$c[2]) //concat the sign with c ) / //divide by x ($c[0]==N?1:substr($c[0],0,-1) //if the first element xN it's N //divide by 1, else remove N and divide );  Question About negative numbers, my code should work for an equation like -xN +- c = m, but when you are considering a x = -1, the input will be -N +- c = m or -1N +- c = m? # R, 90 bytes -round(eval(parse(t=gsub("=","-",gsub("\\dN","",s<-scan(,"")))))/(gsub("N.*$","",s):1)[1])


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Assumes d is present as per the spec. Probably not the shortest option - I am sure there are better regex.