Last Updated: 28 Nov, 2020

# Compress the String

Moderate

## Problem statement

#### Example:

If a string has 'x' repeated 5 times, replace this "xxxxx" with "x5".

The string is compressed only when the repeated character count is more than 1.

#### Note :

The consecutive count of every character in the input string is less than or equal to 9.
##### Input Format:
The first line contains an integer 'T' which denotes the number of test cases or queries to be run.

The first line of each test case contains one string ‘S’ denoting the input string that needs to be compressed.
##### Output Format:
For each case, we need to print a string representing the compressed string.

The output of each test case will be printed in a separate line.
##### Note:
You do not need to input or print anything, and it has already been taken care of. Just implement the given function.
##### Constraints:
1 <= T <= 5
1 <= |S| <= 5000

Where |S| is the size of the string.

Time Limit: 1 sec

## Approaches

### 01 Approach

Here, we can simply traverse the string and run two loops where the outer loop will hold the unique characters and the inner loop will count the consecutive repetitions of that character. Once, we encounter a new character, our inner loop breaks and now our outer loop will start again from the new character. In this process, while getting the repetitions of each character, we can append the character and its count of repetitions depending on the condition that the repetition should be greater than 1.

## Algorithm:

• Declare an integer variable ‘n’ to store the length of the input string
• Declare a string variable ‘compressString’ to hold the final compressed string.
• Declare an integer variable ‘localCount’ to keep the count of repetitions of the current character of the string.
• Declare a character type variable ‘currentCharacter’ to store the current character of the string.
• Declare two integer variables ‘i’ and ‘j’ as two loop counters.
• Run a loop from ‘i’ = ‘0’ to ‘n’.
• Set ‘currentCharacter’ to ‘s[i]’ denoting the current character
• Set ‘’localCount’ to ‘0’
• Run a loop from ‘j’ = ‘i’ to ‘n’
• If our ‘currentCharacter’ is equal to ‘s[j]’
• Increase ‘localCount’ by 1
• Otherwise, we have encountered a different character and we need to break the loop
• If our ‘localCount’ is equal to 1 denoting a single occurrence of the ‘currentCharacter’
• Append only that character to the ‘compressString’
• Otherwise, if ‘localCount’ is greater than 1
• Append the character and ‘localCount’ to ‘compressString’.
• Set ‘i’ equal to ‘j - 1’ because ‘j’ has already encountered a different character. Hence, ‘i’ needs to be set one step back so that on the next increment of ‘i’, it will start from that different character.
• Return ‘compressString’.