Remove Consecutive Characters

Problem

Given a string S. For each index i(1<=i<=N-1), erase it if s[i] is equal to s[i-1] in the string.

Examples:

Example 1:

Input:
S = aabb
Output:  ab 
Explanation: 'a' at 2nd position is appearing 2nd time consecutively. Similiar explanation for b at 4th position.

Example 2:

Input:
S = aabaa
Output:  aba
Explanation: 'a' at 2nd position is appearing 2nd time consecutively. 'a' at fifth position is appearing 2nd time consecutively.

Your task:

You dont need to read input or print anything. Complete the function removeConsecutiveCharacter() which accepts a string as input parameter and returns modified string.

• Expected Time Complexity: $O(|S|)$
• Expected Auxiliary Space: $O(|S|)$

Constraints:

• $1<=|S|<=10^5$

All characters are lowercase alphabets.

Solution

Python

def removeConsecutiveCharacter(self, S):
    if len(S) == 0:
        return ""
    result = []
    result.append(S[0])
    for i in range(1, len(S)):
        if S[i] != S[i - 1]:
            result.append(S[i])
    return ''.join(result)

Java

public String removeConsecutiveCharacter(String S){
    if (S == null || S.length() == 0) {
        return "";
    }
    StringBuilder result = new StringBuilder();
    result.append(S.charAt(0));  
    for (int i = 1; i < S.length(); i++) {
        char currentChar = S.charAt(i);
        char previousChar = result.charAt(result.length() - 1);  
        if (currentChar != previousChar) {
            result.append(currentChar);
        }
    }
    return result.toString();
}

C++

string removeConsecutiveCharacter(string S) {
    if (S.empty()) {
        return "";
    }
    string result;
    result.push_back(S[0]);  
    for (int i = 1; i < S.length(); i++) {
        if (S[i] != result.back()) {
            result.push_back(S[i]);
        }
    }
    return result;
}

C

char* removeConsecutiveCharacter(const char* S) {
    if (S == NULL || strlen(S) == 0) {
        char* result = (char*) malloc(sizeof(char));
        result[0] = '\0';
        return result;
    }
    int len = strlen(S);
    char* result = (char*) malloc((len + 1) * sizeof(char)); 
    if (result == NULL) {
        return NULL; 
    }
    int resultIndex = 0;
    result[resultIndex++] = S[0]; 
    for (int i = 1; i < len; i++) {
        if (S[i] != result[resultIndex - 1]) {
            result[resultIndex++] = S[i];
        }
    }
    result[resultIndex] = '\0'; 
    return result;
}

• Time Complexity: $O(|S|)$
• Auxiliary Space: $O(|S|)$