Given a string s containing only lower case English letters and the ‘?’ character, convert all the ‘?’ characters into lower case letters such that the final string does not contain any consecutive repeating characters. You cannot modify the non ‘?’ characters. It is guaranteed that there are no consecutive repeating characters in the given string except for ‘?’.

Return the final string after all the conversions (possibly zero) have been made. If there is more than one solution, return any of them. It can be shown that an answer is always possible with the given constraints.

Example 1:
Input: s = “?zs”
Output: “azs”
Explanation: There are 25 solutions for this problem. From “azs” to “yzs”, all are valid. Only “z” is an invalid modification as the string will consist of consecutive repeating characters in “zzs”.

Example 2:
Input: s = “ubv?w”
Output: “ubvaw”
Explanation: There are 24 solutions for this problem. Only “v” and “w” are invalid modifications as the strings will consist of consecutive repeating characters in “ubvvw” and “ubvww”.

Example 3:
Input: s = “j?qg??b”
Output: “jaqgacb”

Example 4:
Input: s = “??yw?ipkj?”
Output: “acywaipkja”

Constraints:
1 <= s.length <= 100
s contains only lower case English letters and ‘?’.

Hints:
Processing string from left to right, whenever you get a ‘?’ character, check left character and right character, and select a character not equal to either of them
Do take care to compare with replaced occurrence of ‘?’ when checking the left

String Algorithm to Avoid Consecutive Repeating Characters by Replacing the ? Character

All the characters are lowercases, thus if we are to replace the ? character so that to avoid consecutive repeating characters, we can use the following function to return the first character that is not the same as previous and next character.

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private:
    char diff(char a, char b) {
        for (auto ans = 'a'; ans <= 'z'; ++ ans) {
            if ((ans != a) && (ans != b)) {
                return ans;
            }
        }
        return ' ';
    }

private:
    char diff(char a, char b) {
        for (auto ans = 'a'; ans <= 'z'; ++ ans) {
            if ((ans != a) && (ans != b)) {
                return ans;
            }
        }
        return ' ';
    }

Then the algorithm works like this: we process the string from left to right, when we meet a ‘?’ to replace – we replace it with a character (above function) that is neither previous and next character. Please note that when we replace it with a character – we have to updated the previous character.

The following implementation store the new string separately – O(N) time and O(N) space complexity.

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class Solution {
public:
    string modifyString(string s) {
        string ans;
        char prev = ' ';
        for (int i = 0; i + 1 < s.size(); ++ i) {
            if (s[i] == '?') {
                ans += diff(prev, s[i + 1]);
                prev = ans.back();
            } else {
                ans += s[i];
                prev = s[i];
            }
        }    
        if (s.back() == '?') {
            ans += diff(prev, ' ');
        } else {
            ans += s.back();
        }
        return ans;
    }
};

class Solution {
public:
    string modifyString(string s) {
        string ans;
        char prev = ' ';
        for (int i = 0; i + 1 < s.size(); ++ i) {
            if (s[i] == '?') {
                ans += diff(prev, s[i + 1]);
                prev = ans.back();
            } else {
                ans += s[i];
                prev = s[i];
            }
        }    
        if (s.back() == '?') {
            ans += diff(prev, ' ');
        } else {
            ans += s.back();
        }
        return ans;
    }
};

This can also be implementing by replacing the characters in place:

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class Solution {
public:
    string modifyString(string s) {
        char prev = ' ';
        for (int i = 0; i + 1 < s.size(); ++ i) {
            if (s[i] == '?') {
                s[i] = diff(prev, s[i + 1]);
                prev = s[i];
            } else {
                prev = s[i];
            }
        }    
        if (s.back() == '?') {
            s.back() = diff(prev, ' ');
        } 
        return s;
    }
}

class Solution {
public:
    string modifyString(string s) {
        char prev = ' ';
        for (int i = 0; i + 1 < s.size(); ++ i) {
            if (s[i] == '?') {
                s[i] = diff(prev, s[i + 1]);
                prev = s[i];
            } else {
                prev = s[i];
            }
        }    
        if (s.back() == '?') {
            s.back() = diff(prev, ' ');
        } 
        return s;
    }
}

We can also use the index to get the previous character:

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class Solution {
public:
    string modifyString(string s) {
        for (int i = 0; i + 1 < s.size(); ++ i) {
            if (s[i] == '?') {
                s[i] = diff(i > 0 ? s[i - 1] : ' ', s[i + 1]);
            } 
        }    
        if (s.back() == '?') {
            s.back() = diff(s.size() > 1 ? s[s.size() - 2] : ' ', ' ');
        } 
        return s;
    }

class Solution {
public:
    string modifyString(string s) {
        for (int i = 0; i + 1 < s.size(); ++ i) {
            if (s[i] == '?') {
                s[i] = diff(i > 0 ? s[i - 1] : ' ', s[i + 1]);
            } 
        }    
        if (s.back() == '?') {
            s.back() = diff(s.size() > 1 ? s[s.size() - 2] : ' ', ' ');
        } 
        return s;
    }

This implementation runs O(N) time and O(1) space complexity.

–EOF (The Ultimate Computing & Technology Blog) —