Remove three consecutive duplicates from string

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Given a string, you have to remove the three consecutive duplicates from the string. If no three are consecutive then output the string as it is.

Examples: 

Input : aabbbaccddddc
Output :ccdc
Input :aabbaccddc
Output :aabbaccddc

Explanation : 

We insert the characters of string one by one to vector and keep on checking the size of vector. If the size of vector is greater than 2, then we will check whether the last 3 characters of the string are same or not. If the characters are same 

Then we will move three steps backwards in array using resize() else not. 

Implementation:

ccdc

Time Complexity: O(n)
Auxiliary Space: O(n)

This article is contributed by Roshni Agarwal.  

Approach#2:using re

Algorithm

1. Use the re library to search for any sequence of 3 consecutive identical characters in the input string.
2. Replace any matches with an empty string.
3. Repeat step 1 and 2 until no matches are found.

ccdc

Time Complexity: O(n^2), where n is the length of the input string. This is because we could potentially iterate over the entire string multiple times (once for each match).
Space Complexity: O(n), where n is the length of the input string. This is because we need to store the input string and any intermediate regular expression matches..

Approach: Using Dynamic Programming

• Define a dynamic programming desk, dp, with a size of n (the period of the enter string). Each access dp[i] will constitute the minimal variety of characters required to form the string with out 3 consecutive duplicates, considering the substring up to index i (inclusive).
• Initialize dp[0] as 1, because the minimum wide variety of characters required to form a string of length 1 is 1.
  Iterate from index 1 to n-1 to fill the dynamic programming table:                                                                                      a. If the current person, str[i], isn’t like the previous person, str[i-1], then we will append str[i] to the result without introducing three consecutive duplicates. Therefore, dp[i] = dp[i-1]   1.                                                                               b. If the modern-day character, str[i], is the same as the preceding man or woman, str[i-1], we need to don’t forget  instances:
• If the previous two characters, str[i-1] and str[i-2], also are the same, we can not append str[i] to the result as it might introduce 3 consecutive duplicates. In this case, dp[i] = dp[i-1].
• If the previous two characters, str[i-1] and str[i-2], are distinct, we can append str[i] to the end result with out introducing three consecutive duplicates. In this example, dp[i] = dp[i-2]   1.
  The final answer can be the string fashioned through considering the characters up to index n-1 of the enter string, aside from any characters which have a corresponding dp[i] price more than or identical to 3.
   

Input: aabbbaccddddc
Output: aaaccdc
Input: aabbaccddc
Output: aabbaccddc

Time Complexity: O(n), where n is the length of the input String.

Auxiliary Space: O(n), dp table requires O(n) space to store the count of consecutive duplicates. In worst case, the space complexity will be near about O(n), where n is the length of the input String.

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Article Tags :

• DSA
• Misc
• Strings

Practice Tags :

• Misc
• Strings