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# Compress the String! Hacker Rank Solution – Queslers

## Problem: Compress the String! Hacker Rank Solution

In this task, we would like for you to appreciate the usefulness of the groupby() function of itertools . To read more about this function .
You are given a string S. Suppose a character ‘c’ occurs consecutively X times in the string. Replace these consecutive occurrences of the character ‘c’ with (X,c)
in the string.
For a better understanding of the problem, check the explanation.

#### Input Format :

A single line of input consisting of the string S.

#### Output Format :

A single line of output consisting of the modified string.

#### Constraints :

All the characters of S denote integers between 0 and 9.

• 1 <= | S | <= 10^4

#### Sample Input :

``````1222311
``````

#### Sample Output :

``````(1, 1) (3, 2) (1, 3) (2, 1)
``````

#### Explanation :

First, the character 1 occurs only once. It is replaced by (1, 1). Then the character 2 occurs three times, and it is replaced by (3, 2) and so on.
Also, note the single space within each compression and between the compressions.

### Compress the String! Hacker Rank Solution in python 2

``````from itertools import groupby
s=map(int,list(raw_input()))
l=[(sum(1 for i in g),k) for k,g in groupby(s)]
print ' '.join(map(str,l))``````

### Compress the String! Hacker Rank Solution in pyhton 3

``````from itertools import groupby
for key, group in groupby(input()):
print('({}, {})'.format(len(list(group)), key), end=" ")
``````

### Compress the String! Hacker Rank Solution in pypy

``````# Enter your code here. Read input from STDIN. Print output to STDOUT
from itertools import groupby

print " ".join(str((len(list(k)),int(i))) for i,k in groupby(raw_input()))``````

### Compress the String! Hacker Rank Solution in pypy 3

``````s=input()
prev_c=''
count=0
Out=()
for c in s :
if c == prev_c:
count+=1
else:
if prev_c!='':
print((count+1,int(prev_c)), end=" ")
prev_c=c
count=0
print((count+1,int(prev_c)),end=" ")``````
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