Interview Question

Java Senior Software Engineer/Lead Position Interview

-Bangalore

Applied Materials

Input is a string like "AAAAABBCCAA" and it should print "5A2B2C2A". 5 being the continuous number of occurance for character 'A'. Same is with other characters also.

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Interview Answers

3 Answers

0

let string:String = "AAAAABBBVVVCVCAA" let characters = Array(string) var counter:Int = 1 var newArray:[String] = [String]() let lastCount = characters.count - 1 for count in 0...(characters.count - 2) { if characters[count] == characters[count + 1] { counter = counter + 1; }else { newArray.append("\(counter)\(characters[count])") counter = 1 } if lastCount == count + 1 { newArray.append("\(counter)\(characters[count])") } } print(newArray)

Wasim on

0

public static void main(String[] args) { String s = "AAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBCC"; Set set = new HashSet(); char b = 0; int count = 1; String newString = ""; for (int i = 0; i 0) { b = s.charAt(i - 1); } char c = s.charAt(i); if (!set.add(c)) { count++; } else { if (b != 0) { newString = newString + b + count; count = 1; } } } newString = newString + b + count; System.out.println(newString); }

Anonymous on

0

public static void main(String[] args) { String s = "AAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBCC"; Set set = new HashSet(); char b = 0; int count = 1; String newString = ""; for (int i = 0; i 0) { b = s.charAt(i - 1); } char c = s.charAt(i); if (!set.add(c)) { count++; } else { if (b != 0) { newString = newString + b + count; count = 1; } } } newString = newString + b + count; System.out.println(newString); }

Anonymous on

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