String Python C++

557. Reverse Words in a String III

Reverse Words in a String III - LeetCode

Python is very easy

class Solution:
    def reverseWords(self, s: str) -> str:
        ss = s.split()
        ss = [c[::-1] for c in ss]
        return ' '.join(ss)

C++ similar idea and much longer as we have to write our own split and reverse code.

class Solution {
public:
    string reverse(string s){
        if(s.empty())return s;
        string res = "";
        while(!s.empty()){
            res += s.back();
            s.pop_back();
        }
        return res;
    }
    vector<string> split(string s){
        vector<string> res;
        if(s.empty())return res;
        string temp = "";
        for(int i=0;i<s.length();++i){
            if(s[i]==' '){
                res.push_back(temp);
                temp="";
            }
            else temp+=s[i];
        }
        if(!temp.empty())res.push_back(temp);
        return res;
    }
    string reverseWords(string s) {
        vector<string> ss = split(s);
        string res = "";
        for(auto&s:ss){
            res += reverse(s);
            res += ' ';
        }
        res.pop_back();
        return res;
    }
};

Actually, C++ can use a more efficient way to solve this problem.

class Solution {
public:
    string reverseWords(string s) {
        auto size = s.size();
        int from = 0;
        for(int i = 1; i < size; ++i) {
            if(s[i] == ' '){
                reverse(begin(s)+from, begin(s)+i);
                from = i+1;
            }
        }
        reverse(begin(s)+from, end(s));
        return s;
    }
};

1047. Remove All Adjacent Duplicates In String

https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/

class Solution {
public:
    string removeDuplicates(string S) {
        //old_len = S.length();
        while(true){
            string new_S = "";
            int i=0;
            while(i<S.length()){
                if(i+1<S.length() && S[i+1]==S[i]){
                    i += 2;
                }
                else{
                    new_S += S[i];
                    ++i;
                }
            }
            if (new_S.length()==S.length())return new_S;
            S = new_S;
        }
        return "";
        
    }
};

Better one from lee215

https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/discuss/294893/JavaC++Python-Two-Pointers-and-Stack-Solution

string removeDuplicates(string s) {
        int i = 0, n = s.length();
        for (int j = 0; j < n; ++j, ++i) {
            s[i] = s[j];
            if (i > 0 && s[i - 1] == s[i]) // count = 2
                i -= 2;
        }
        return s.substr(0, i);
    }

Or we can use stack

class Solution {
public:
    string removeDuplicates(string S) {
        string res = "";
        for(auto&c:S){
            if(res.size() && c==res.back())res.pop_back();
            else res.push_back(c);
        }
        return res;
    }
};

422. Valid Word Square

class Solution {
public:
    bool validWordSquare(vector<string>& words) {
      for(int i = 0; i < words.size(); ++i) {
            for(int j = 0; j < words[i].size(); ++j)
                if(j >= words.size()|| i>=words[j].size() || words[j][i] != words[i][j])
                   return false;
        }
        return true;
    }
};

Python

class Solution:
    def validWordSquare(self, words: List[str]) -> bool:
        return words== [''.join(c) for c in itertools.zip_longest(*words,fillvalue='')]

387. First Unique Character in a String

class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char,int> m;
        for (auto &c:s)m[c]++;
        for(int i=0;i<s.length();++i){
            if(m[s[i]]==1)return i; 
        }
        return -1;
        
    }
};

392. Is Subsequence

https://leetcode.com/problems/is-subsequence/

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if(s.empty())return true;
        if(t.empty())return false;
        int j=0;
        for(int i=0;i<t.size()&&j<s.size();++i)
            if (t[i]==s[j])++j;   
        return j==s.size();
    }
};

Binary search

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if(s.empty())return true;
        if(t.empty())return false;
        vector<vector<int>> d(256);
        for(int i=0;i<t.length();++i)d[t[i]].push_back(i);
        int from = -1;
        for(auto &c:s){
            auto it = upper_bound(d[c].begin(), d[c].end(), from);
            if (it==d[c].end())return false;
            from = *it;
        }
        return true;
    }

125. Valid Palindrome

https://leetcode.com/problems/valid-palindrome/

class Solution {
public:
    bool isPalindrome(string s) {
        string t="";
        for(auto &c:s){
            if (isalnum(c)){
                if(isalpha(c))t.push_back(tolower(c));
                else t.push_back(c);
            }
        }
        
        /*
        string tt=t;
        reverse(t.begin(),t.end());
        return tt==t;
        */
        for (int i=0;i<t.size()/2;++i){
            if(t[i]!=t[t.size()-i-1])return false;
        }
        return true;
    }
};

Or do it like this

class Solution {
public:
    bool isPalindrome(string s) {
       for (int i = 0, j = s.size() - 1; i < j; i++, j--) { // Move 2 pointers from each end until they collide
        while (!isalnum(s[i])  && i < j) i++; // Increment left pointer if not alphanumeric
        while (!isalnum(s[j])  && i < j) j--; // Decrement right pointer if no alphanumeric
        if (toupper(s[i]) != toupper(s[j])) return false; // Exit and return error if not match
    }
    
    return true;
    }
};

1048. Longest String Chain

https://leetcode.com/problems/longest-string-chain/

https://leetcode.com/problems/longest-string-chain/discuss/294890/C%2B%2BJavaPython-DP-Solution

class Solution {
public:
    int longestStrChain(vector<string>& words) {
        unordered_map<string, int> dp;
        for(auto &word:words) dp[word] = 1;
        sort(words.begin(), words.end(), [](string &a, string &b){
            return a.length()<b.length();
        });
        for(auto &word:words){
            for(int j=0; j <word.length();++j){
                string new_word = word.substr(0,j)+word.substr(j+1);
                if(dp.count(new_word))dp[word] = max(dp[word], dp[new_word]+1);
            }
        }
        int res = 0;
        for(auto& d:dp)res=max(res, d.second);
        return res;
    }
};

Python code

class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        dp = {}
        for word in sorted(words, key=len):
            dp[word] = max(dp.get(word[:i]+word[i+1:], 0)+1 for i in range(len(word)))
        return max(dp.values())

451. Sort Characters By Frequency

32 ms

class Solution:
    def frequencySort(self, s: str) -> str:
        return ''.join(c*freq for c, freq in collections.Counter(s).most_common())

16 ms

class Solution {
public:
    string frequencySort(string s) {
        unordered_map<char,int> freq;
        vector<string> bucket(s.size()+1, "");
        string res;
        
        //count frequency of each character
        for(char c:s) freq[c]++;
        //put character into frequency bucket
        for(auto& it:freq) {
            int n = it.second;
            char c = it.first;
            bucket[n].append(n, c);
        }
        //form descending sorted string
        for(int i=s.size(); i>0; i--) {
            if(!bucket[i].empty())
                res.append(bucket[i]);
        }
        return res;
    }
};