Short Palindrome Discussions | Algorithms | HackerRank

Short Palindrome

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  • + 0 comments

    standard dynamic programming algo, O(n), but uses a lot of memory (couple hundred MBs). pass all test

    long pairPalindrome(int i, int j, const vector<vector<int>>& count) {
        if (i >= j) return 0;
        long sum = 0;
        for (int k=0; k < 26; k++) sum = (sum + ((count[j][k] - count[i-1][k]) * (count[j][k] - count[i-1][k] - 1) / 2)) % 1000000007;
        return sum;
    }
    
    long shortPalindrome(const string& s) {
        vector<int> TEMP1(26), TEMP2(26);
        vector<vector<int>> count = {TEMP1}, mostRecent = {TEMP2};
        for (int i=1; i <= s.size(); i++) {
            TEMP1[s[i-1]-'a']++;
            count.push_back(TEMP1);
            mostRecent.push_back(TEMP2);
            TEMP2[s[i-1]-'a'] = i;
        }
        long total = 0;
        vector<long> result(s.size()+1);
        vector<vector<int>> cache(s.size()+1, vector<int>(26));
        for (int i=1; i <= s.size(); i++) {
            int index = mostRecent[i][s[i-1]-'a'];
            if (index == 0) continue;
            result[i] = result[index] + (count[index][s[i-1]-'a']-1) * pairPalindrome(index, i-1, count) + pairPalindrome(index+1, i-1, count);
            for (int k=0; k < 26; k++) {
                result[i] = (result[i] + (long)cache[index][k] * (count[i-1][k] - count[index-1][k])) % 1000000007;
                cache[i][k] = (cache[index][k] + (long)(count[index][s[i-1]-'a']-1)*(count[i-1][k] - count[index-1][k])
                              + count[i-1][k] - count[index][k]) % 1000000007;
            }
            total = (total + result[i]) % 1000000007;
        }
        return total;
    }
    
  • + 0 comments
    def shortPalindrome(s):
      arr1 = [0]*26
      arr2 = [[0]*26 for i in range(26)]
      arr3 = [0]*26
      ans = 0
      for i in range(len(s)):
          idx = ord(s[i]) - ord('a')
          ans += arr3[idx]
          for j in range(26):
              arr3[j] += arr2[j][idx]
          for j in range(26):
              arr2[j][idx] += arr1[j]
          arr1[idx] += 1
      return ans % (10**9+7)
    
  • + 0 comments

    C++ (more at https://github.com/IhorVodko/Hackerrank_solutions , feel free to give a star :) )

    static size_t const alphabetSize{26};
    static size_t const aASCII{97};
    static size_t const modulo{static_cast<size_t>(std::pow(10, 9)) + 7};
    
    size_t shortPalindrome(
        std::string const & str
    ){
        using namespace std;
        size_t tuplesCount = 0;
        auto char1{vector<size_t>(alphabetSize, 0)}; 
        auto chars12{vector<size_t>(alphabetSize * alphabetSize, 0)}; 
        auto chars123{vector<size_t>(alphabetSize, 0)};
        size_t chNum{0};
        size_t chCopy{0};
        for(auto const & ch: str){
            chNum = static_cast<size_t>(ch) - aASCII;
            chCopy = chNum;
            tuplesCount = (tuplesCount % modulo) + chars123.at(chCopy);
            for(auto i = 0; i < alphabetSize; ++i){
                chars123.at(i) += chars12.at(chCopy);
                chars12.at(chCopy) += char1.at(i);
                chCopy += 26;
            }
            ++char1.at(chNum);
        } 
        return tuplesCount % modulo;
    }
    
  • + 0 comments
    const alphabet = "abcdefghijklmnopqrstuvwxyz".split("")
    const mod = (10 ** 9 + 7)
    function shortPalindrome(s: string): number {
        let total = 0;
        
        // i
        const firstCharCount: Record<string, number> = {};
        // ij
        const charPairCount: Record<string, number> = {};
        // ijj
        const charDoublePairCount: Record<string, number> = {};
        
        for(const currChar of s){
            for(const char of alphabet){
                const ij = char + currChar; 
                const ijj = char + currChar + currChar; 
                const ijji = currChar + char + char;
                
                total += charDoublePairCount[ijji] || 0;
                total = total % mod;
                
                if(!charDoublePairCount[ijj])
                    charDoublePairCount[ijj] = 0
                charDoublePairCount[ijj] += charPairCount[ij] || 0;
                charDoublePairCount[ijj] = charDoublePairCount[ijj] % mod;
                
                if(!charPairCount[ij])
                    charPairCount[ij] = 0
                charPairCount[ij] += firstCharCount[char] || 0;
                charPairCount[ij] = charPairCount[ij] % mod;
            }
            if(!firstCharCount[currChar])
                firstCharCount[currChar] = 0;
            firstCharCount[currChar] += 1;
            firstCharCount[currChar] = firstCharCount[currChar] % mod;
        }
        return total;
    }
    
  • + 0 comments

    my solution

    def shortPalindrome(s):
        charCount = {}
        pairCount = {}
        pairHash = {}
        totalCount = 0
        MOD = 10**9 + 7
    
        for char in s:
            charIndex = ord(char) - ord('a')
            pairIndex = charIndex
            totalCount = (totalCount + pairCount.get(charIndex, 0)) % MOD
    
            for pairChar in range(26):
                pairCount[pairChar] = (pairCount.get(pairChar, 0) + pairHash.get(pairIndex, 0)) % MOD
                pairHash[pairIndex] = (pairHash.get(pairIndex, 0) + charCount.get(pairChar, 0)) % MOD
                pairIndex += 26
    
            charCount[charIndex] = charCount.get(charIndex, 0) + 1
    
        return totalCount