ibequa very simple and short solution is just to subtract ascii codes of the letters.
zpf7879 Exactly! Substract ascii codes in the symmetrical positions.
vishalbty Full solution in python can be found here.
mandirasarkarma1 can u tell me about that enumerate part please?
Schuetzl That's what I did, worked like a charm, and was very simple.
aeshwer thks..very good logic ..made it very simple
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?
poloju_shashidh1 time complexity
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);
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; }
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