input string = aabbbccccaaa, maintain the insertion order and output should be a2b3c4a3

Question

Sindhu Nagabhushan · Accepted Answer

//input string = aabbbccccaaa, maintain the insertion order   and output should be a2b3c4a3

public class Test {
	
	public static void process(String mystring) {
		StringBuilder sb = new StringBuilder();
		char[] mychar = mystring.toCharArray();
		int count = 1;
		for(int i=0; i< (mychar.length)-1;i++) {
			System.out.println("mychar[p1]: "+mychar[i]+" mychar[p2]: "+mychar[i+1]);
			if(mychar[i] == mychar[i+1]) {
				count = count+1;
				System.out.println("count: "+count);
			}
			else {
				sb.append(mychar[i]);
				sb.append(count);
				count = 1;
			}
			if(i==(mychar.length)-2) {
				sb.append(mychar[i+1]);
				sb.append(count);
			}
		}
		System.out.println(sb.toString());
	}
	
	public static void main(String[] args) {	
        process("aabbbccccaaa");
		process("ggggyyynnnkkkkkk");
		process("aabbcccdef");
		process("abcdefghj");
	}
	
}

jeena · Answer

public class Stringoupt {

	int count =0;
	void outputString(String str){
		char st = str.charAt(0);
		for(int i=0;i

Anonymous · Answer

a = "aabbbccccaaa"

if not a:
  print ("")
if len(a) == 1:
  print (a+"1")

start, end = 0, 1
new_string = a[0]
count = 1
while end < len(a):
  if a[start] == a[end]:
    count += 1
    end += 1
  else:
    new_string += str(count)
    count = 1
    start, end = end, end+1
    new_string += a[start]
new_string += str(count)
print (new_string)

Anonymous · Answer

words = 'aaaabahhhhhaaa'

count = 1
answer = ''
last_char = None

for i, char in enumerate(words):
                if last_char != None:
                                if char == last_char:
                                                count = count + 1
                                else:
                                                answer = '%s%s%d' % (answer, last_char, count)
                                                count = 1

                if i == (len(words) - 1):
                                print('hi')
                                answer = '%s%s%d' % (answer, last_char, count)

                last_char = char

print(answer)

input string = aabbbccccaaa, maintain the insertion order and output should be a2b3c4a3 - in C# · Answer

using System;
using System.Collections.Generic;

namespace Letter__Count_in_String
{
    class Program
    {
        static void Main(string[] args)
        {
            string original = "aabbbccccaaa";
            List result = new List();

            char[] arrOriginal = original.ToCharArray();
            int count = 1;
            for (int i = 0; i < arrOriginal.Length; i++)
            {
                if ((i < arrOriginal.Length - 1) && (arrOriginal[i] == arrOriginal[i + 1]))
                {
                    count++;
                }
                else
                {
                    result.Add(arrOriginal[i].ToString() + count.ToString());
                    count = 1;
                }
            }
            foreach(var a in result)
            {
                Console.Write(a);
            }
        }
    }
}

Anonymous · Answer

def compressString(arr):
    count=1
    op=""
    for i in range(len(arr)):
        if (i+1)

Anonymous · Answer

def compressString(arr):
    count=1
    op=""
    for i in range(len(arr)):
        if (i+1)

Rameshkumar · Answer

a = "aaabbbaaaa"

i=0
j=0
fnl_result =""
for item in a:
	if i== 0:
		temp = item
		j = j+1
	elif item == temp:
		j = j+1
	elif item!=temp:
		fnl_result = fnl_result+str(temp)+str(j)
		j=1
	temp = item
	i=i+1
fnl_result = fnl_result+str(temp)+str(j)
print fnl_result

#output = a3b3a4

Amazon

Amazon Interview Question

Interview Answers

Want the inside scoop on your own company?

Bowls

Followed companies

Job searches