Compress the String
Ninja has been given a program to do basic string compression. For a character that is consecutively repeated more than once, he needs to replace the consecutive duplicate occurrences with the count of repetitions.

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

Question

Compress the String

Ninja has been given a program to do basic string compression. For a character that is consecutively repeated more than once, he needs to replace the consecutive duplicate occurrences with the count of repetitions.

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

Accepted Answer

This was nothing but the Run Length Encoding Algorithm which I was familiar with. So I coded it preety fast.

Algorithm :

1) Pick the first character from the source string.
2) Append the picked character to the destination string.
3) Count the number of subsequent occurrences of the picked character and append the count to the destination
string.
4) Pick the next character and repeat steps (2),(3) and (4) if the end of the string is NOT reached.

TC : O(N), where N=length of the string
SC : O(1)

Accepted Answer

Greedy 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’.

Space Complexity: O(1)Explanation:

O(1)

We only need constant space. So it’s O(1).

Time Complexity: O(n)Explanation:

O(N) where N is the length of the input string.

Though we are using two loops to iterate through the string, our outer loop is only iterating through the different characters present in the string while the inner loop is iterating to count the repetition of the current character in the outer loop. Hence, each character of the string is only traversed once, so our time complexity is O(N).

Java (SE 1.8)


/*
	Time Complexity: O(N)
	Space Complexity: O(1)
	
	Where N is the length of the input string.
*/
public class Solution {
	public static String compressTheString(String s) {

		// Declare an integer variable n to store the length of the input string
		int n = s.length();

		// Declare a string variable compressString to hold the final compressed
		// string.
		String compressString = "";

		// Declare an integer variable localCount to keep the count of repetitions
		// of the current character of the string.
		int localCount = 0;

		// Declare a character type variable currentCharacter to store the current
		// character of the string.
		char currentCharacter;

		// Declare two integer variables i and j as two loop counters.
		int i = 0;
		int j = 0;
		for (i = 0; i < n; ++i) {

			// Set currentCharacter to s[i] denoting the current character
			currentCharacter = s.charAt(i);
			localCount = 0;
			for (j = i; j < n; ++j) {

				// If our currentCharacter is equal to s[j]
				if (currentCharacter == s.charAt(j))
					localCount++;
				else
					break;
			}

			// If our localCount is equal to 1 denoting a single occurrence of the
			// currentCharacter
			if (localCount == 1) {
				compressString = compressString + currentCharacter;
			} else {
				compressString = compressString + currentCharacter + String.valueOf(localCount);
			}

			/*
			 * 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.
			 */
			i = j - 1;
		}

		// Return compressed string
		return compressString;
	}

}

Python (3.5)

'''

    Time Complexity: O(N)

    Space Complexity: O(1)



    Where N is the length of the input string.

'''



def compressTheString(s):

	

	n = len(s)



	# Declare empty string.

	answer = ""

	

	# Initialize current count of any character in string.

	currentCharCount = 1



	# Add first letter of string to answer.

	answer += s[0]

	

	for i in range(1, n):



		# If the current letter is same as previous.

		if s[i] == s[i - 1]:

			currentCharCount += 1

			

		else:

			if currentCharCount > 1:



				# Add count of letter in answer.

				answer += str(currentCharCount)

				currentCharCount = 1

				

			answer += s[i]



	if currentCharCount > 1:

		answer += str(currentCharCount)



	# Return answer.

	return answer

C++ (g++ 5.4)

/*
    Time Complexity: O(N)
    Space Complexity: O(1)

    Where N is the length of the input string.
*/

string compressTheString(string &s)
{
    // Declare an integer variable ‘n’ to store the length of the input string
    int n = s.size();

    // Declare a string variable ‘compressString’ to hold the final compressed string.
    string compressString = "";

    // Declare an integer variable ‘localCount’ to keep the count of repetitions of the current character of the string.
    int localCount = 0;

    // Declare a character type variable ‘currentCharacter’ to store the current character of the string.
    char currentCharacter;

    // Declare two integer variables ‘i’ and ‘j’ as two loop counters.
    int i = 0;
    int j = 0;
    for (i = 0; i < n; ++i)
    {
        // Set ‘currentCharacter’ to ‘s[i]’ denoting the current character
        currentCharacter = s[i];
        localCount = 0;
        for (j = i; j < n; ++j)
        {
            // If our ‘currentCharacter’ is equal to ‘s[j]’
            if(currentCharacter == s[j])
                localCount++;
            else 
                break;
        }

        // If our ‘localCount’ is equal to 1 denoting a single occurrence of the ‘currentCharacter’
        if(localCount == 1)
        {
            compressString = compressString + currentCharacter;
        }
        else
        {
            compressString = compressString + currentCharacter + to_string(localCount);
        }

        /* 
        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.
        */
        i = j - 1;
    } 

    // Return compressed string
    return compressString;   
}

Accepted Answer

import java.util.LinkedHashMap; public class Main { public static void main(String[] args) { String s="iyyer"; String s1=""; char [] ch=s.toCharArray(); for(char i:ch){ System.out.print(i+" "); } LinkedHashMap L=new LinkedHashMap(); for(char i:ch){ if(L.containsKey(i)) { L.put(i,L.get(i)+1); } else { L.put(i,1); } } for(char i:L.keySet()) { s1=s1+i+L.get(i); } System.out.println(s1); } }