ibequa very simple and short solution is just to subtract ascii codes of the letters.
zpf7879 Exactly! Substract ascii codes in the symmetrical positions.
Schuetzl That's what I did, worked like a charm, and was very simple.
aeshwer thks..very good logic ..made it very simple
rahulvic7 can u explain the logic behind this??
aayushis How did you code it? can you elaborate please
ashunimbz [deleted]ashunimbz #include<bits/stdc++.h> using namespace std; int main() { int n ; cin>>n ; for(int k = 0 ; k< n ; ++k) { string s1 ; cin>>s1 ; int i = 0 ; int j = s1.length() - 1 ; int ans = 0 ; while(i <= j ) { int max1 = max((int)s1[i],(int)s1[j]) ; int min1 = min ((int)s1[i],(int)s1[j]) ; ans += (max1 - min1) ; i++; j--; } cout<<ans<<"\n" ; } return 0; }
keshav_thakur998 while(i <= j ) { //int max1 = max((int)s1[i],(int)s1[j]) ; //int min1 = min ((int)s1[i],(int)s1[j]) ;
//ans += (max1 - min1) ; ans+=abs(s1[i]-s1[j]); i++; j--; }hiddencoder you are great
Kanudo_10 [deleted]
mackdelton def theLoveLetterMystery(s): out = 0 for i in range(len(s) // 2): out += abs(ord(s[i]) - ord(s[len(s) - i - 1])) return out
nimish_mangal910 Yeah exactly just subtract the ascii values of the reverse string and normal string and take the absolute value of it and then divide by 2. The code is given below:-
t=int(input()) for i in range (t): s=list(input()) rev=s[::-1] changes=0 for i in range (len(s)): changes+=abs(ord(rev[i])-ord(s[i])) print(changes//2)
arastogi1997 Yeah, from both ends, whichever is smaller, it get reduced to its counterPart on symmetrically oppsite side.
for(int i=0,j=c.length-1; i<c.length/2; i++,j--) count+= Math.abs(c[i]-c[j]); return count;
mchaudh41 I am new at programming. Can any one explain how the for loop is working in the above solution?
macurvello First, two variables are declared, i and j, and assigned values 0 and c.length - 1 respectively. for( int i = 0 // declares variable i and assigns it the value 0 , j = c.length - 1; // declares variable j and assigns it the value c.length - 1
c.length is the length of the character array. You don't have to declare 'int' again to declare variable j if you just use a comma in Java. Then comes the statement that will be evaluated before each loop. If true, the iteration continues, if not, stop. i < c.length/2; // At the beginning of each loop, check if variable i is less than the length of the character arry divided by two (the middle* of the array)
Next are the commands that are to be executed at the end of each loop, separated by commas. i++, j--) // Increment i, decrement j
Inside each loop, Math.abs(c[i]-c[j]) is added to the value of the variable 'count'. Math.abs is a method of the Java language that returns the absolute value of a number, i.e. if negative, return the positive counterpart, else return the number itself. So what is being calculated is the absolute value of the difference between the numerical value of the characters in the position i and j of the array.
count+= Math.abs(c[i]-c[j]); // Add to 'count' the absolute value of the difference between the numerical value of the characters in the position i and j of the array.
- the 'middle' of the array is actually between two positions if the size is even. In this case, c.length / 2 returns the highest of those two positions The same effect can be achieved if you just use i < j instead, without having to work out rounding mechanics and if you need to use < or <=, or if you have to subtract one or whatever.
imhktc Oneliner in python
def theLoveLetterMystery(s): return sum([abs(ord(s[i])-ord(s[len(s)-1-i])) for i in range(len(s)//2)])
DR3am3r yes exactly, i did the same .
arun_prashanth Please explain it in detail.
sarathy_v_krish1 C++ solution :
int theLoveLetterMystery(string s) { int sum=0; for (int i=0, j=s.size()-1;i<j;i++, j--) sum+=abs(s[i]-s[j]); return sum; }
yli131 public class Solution {
public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); for(int ii = 0; ii < n; ii++){ String s = sc.next(); int count = 0; // use two pointers; int i = 0, j = s.length() - 1; while(i < j){ count += Math.abs(s.charAt(i) - s.charAt(j)); i++; j--; } System.out.println(count); } }}
KeyurPotdar [deleted]KeyurPotdar [deleted]
positivedeist_07 +1 for your edit :D
hooni why is the 2 pointer solution better than the 1 pointer solution?
jay39 And this is why I don't use Java.
nobe0716 one line with lambda
s = raw_input() print sum(map(lambda x: abs(ord(s[x]) - ord(s[len(s) - x - 1])), range(len(s)/2)))
Norman99 LOL I did almost the same.
for i in range(int(input())): print((lambda w: sum(abs(ord(w[i]) - ord(w[len(w)-i-1])) for i in range(len(w)//2)))(input()))
hiddencoder lol answer is great brother
jmorgensen IMO, comprehensions/generators are more readable:
print(sum(abs(ord(s[x]) - ord(s[len(s) - x - 1])) for x in range(len(s)//2)))
And, though speed isn't usually important, for fun:
s = list(map(ord, s)) print(sum(map(abs, (s[x] - s[-x - 1] for x in range(len(s)//2)))))
This would be even faster due to reducing function calls and eliminating global namespace searches in the innermost loop.
sandninja I tried my best in C#:
int numberOfTest = int.Parse(Console.ReadLine()); for (int t = 0; t < numberOfTest; t++) { var s = Console.ReadLine().ToArray(); var ops = 0; // Ignore the character in the middle, if any (it will automatically be excluded by the condition in the for-loop) for (int i = 0, j = s.Length - 1; i < (s.Length / 2) && j >= (s.Length / 2); i++, j--) ops += Math.Abs(s[j] - s[i]); Console.WriteLine(ops.ToString()); } }
mohanraj13 cool!
gmoshe27 That's pretty much the same result that I came up with. One little tweak you can make in your for loop to make it tighter is that you only need to know that i is not greater than j. For words with an odd number of characters, when i == j, that's your midpoint, so you don't have to do the math. For words with even number charactesr, i < j is the middle two characters. In your next loop, i will be greater than j, so you break. Here's the result:
for(int i = 0, j = s.Length - 1; i < j; i++, j--) { ops += Math.Abs(s[j] - s[i]); }
MarsFreewill wow, that's an excellent answer.THX.
eugen3 Thank you! A very useful suggestion.
Nambo I did something similar except I created a variable that is reverse of the input.
string tc = Console.ReadLine(); char[] tcArray = tc.ToCharArray(); char[] tcArrayReverse = tc.ToCharArray(); Array.Reverse(tcArrayReverse); for(int j = 0; j < (tcArray.Length/2); j++) { counter += Math.Abs(tcArray[j] - tcArrayReverse[j]); } Console.WriteLine(counter);
Spatzist [deleted]
ykurmangaliyev :)
string s = Console.ReadLine(); Console.WriteLine(Enumerable.Range(0, s.Length / 2) .Sum(i => Math.Abs(s[i] - s[s.Length - i - 1])));
batman25663 for ( int i = 0; i < s.length()/2; i++ ) { min_op += abs(s[s.length() - 1 - i] - s[i]); }
JuanDavidCanti My Java Solution
static int theLoveLetterMystery(String s) { int oper=0, n=s.length(); for(int i=0,j=n-1 ; i<n/2 ; i++,j--) oper+=Math.abs(s.charAt(i)-s.charAt(j)); return oper; }
alphasingh One-liner in python
return sum([abs(ord(s[i])-ord(s[-i-1])) for i in range(len(s)//2)])
indrasenareddyg static int theLoveLetterMystery(String s) {
int count = 0; int i = 0; int j = s.length() -1; while(i < j){ int start = s.charAt(i); int end = s.charAt(j); int diff = Math.abs(start - end); count = count + diff; i++; j--; } return count; }
dhawal_1304 n = int(raw_input()) for _ in range(n): count=0 s = raw_input() if s == s[::-1]: count="0" else: for i in range(len(s)//2): j = len(s) - i - 1 if ord(s[i]) > ord(s[j]): count+= ord(s[i]) - ord(s[j]) elif ord(s[i]) < ord(s[j]): count+= ord(s[j]) - ord(s[i]) else: next print count
mattgurney This one was easy, only a few minutes required
function processData(input) { const inputLines = input.split('\n') const words = inputLines.slice(1) for (let word of words){ var steps = 0 for (let i = 0; i < word.length / 2 ; i++){ steps = steps + Math.abs(word[i].charCodeAt(0) - word[word.length -i -1 ].charCodeAt(0)) } console.log(steps) } }
runcy Here's a O(n/2) solution in python:
T = int(input().strip()) for _ in range(T): str_ = str(input().strip()) ops = 0 for i in range(len(str_) // 2): ops += abs(ord(str_[i]) - ord(str_[-(i+1)])) print(ops)
permenko actually you cannot have n/2 solution coz big-O not allow constants. so this is O(n) solution
Sort 391 Discussions, By:
Please Login in order to post a comment