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  3. Dynamic Programming
  4. LCS Returns
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LCS Returns

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27 Discussions

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  • apkworldsites77
     + 0 comments

    I might be too late but the solution would be to declare it globally. There isn't enough space available in the function segment. For more details visit the portal.

  • Tusenka
     + 0 comments

    What is wrong withthe code:

    memo=[[0]*(len(s2)+1) for _ in range((len(s1)+1))]        
    for i in range(1, len(s1)+1):
        for j in range(1, len(s2)+1):          
            if s1[i-1]==s2[j-1] :          
                memo[i][j]=memo[i-1][j-1]+1
            else:
                memo[i][j]=max(memo[i-1][j], memo[i][j-1])
                                
    _cur=memo[-1][-1]
    suffixes=[[0]*(len(s2)+1) for _ in range((len(s1)+1))] 
    print(memo)      
    for i in range(len(s1)-1, -1, -1):
        for j in range(len(s2)-1, -1, -1):
            print(s1[i], s2[j])
            if s1[i]==s2[j]:   
                suffixes[i][j]=suffixes[i+1][j+1]+1
            else:
                suffixes[i][j]=max(suffixes[i+1][j], suffixes[i][j+1]) 
    count=0
    print(suffixes)
    for i in range(len(s1)+1):
        seen=set()
        for j in range(len(s2)):
            if memo[i][j]+suffixes[i][j+1]+1>_cur:
                count+=1
                seen.add(s2[j])
    return count

  • Tusenka
     + 0 comments

    What is wrong withthe code:

    memo=[[0]*(len(s2)+1) for _ in range((len(s1)+1))]        
    for i in range(1, len(s1)+1):
        for j in range(1, len(s2)+1):          
            if s1[i-1]==s2[j-1] :          
                memo[i][j]=memo[i-1][j-1]+1
            else:
                memo[i][j]=max(memo[i-1][j], memo[i][j-1])
                                
    _cur=memo[-1][-1]
    suffixes=[[0]*(len(s2)+1) for _ in range((len(s1)+1))] 
    print(memo)      
    for i in range(len(s1)-1, -1, -1):
        for j in range(len(s2)-1, -1, -1):
            print(s1[i], s2[j])
            if s1[i]==s2[j]:   
                suffixes[i][j]=suffixes[i+1][j+1]+1
            else:
                suffixes[i][j]=max(suffixes[i+1][j], suffixes[i][j+1]) 
    count=0
    print(suffixes)
    for i in range(len(s1)+1):
        seen=set()
        for j in range(len(s2)):
            if memo[i][j]+suffixes[i][j+1]+1>_cur:
                count+=1
                seen.add(s2[j])
    return count

  • dehaka
     + 0 comments

    quadratic time O(A.size() * B.size())

    int LCSReturns(const string& A, const string& B) {
    int total = 0;
    vector<vector<int>> cache(A.size() + 1, vector<int>(B.size() + 1)), reverse(A.size() + 2, vector<int>(B.size() + 2));
    vector<vector<bool>> insertion(A.size() + 2, vector<bool>(75));
    for (int i = 1; i <= A.size(); i++) {
        for (int j = 1; j <= B.size(); j++) {
            if (A[i - 1] == B[j - 1]) cache[i][j] = cache[i - 1][j - 1] + 1;
            else cache[i][j] = max(cache[i - 1][j], cache[i][j - 1]);
        }
    }
    for (int i = A.size(); i >= 1; i--) {
        for (int j = B.size(); j >= 1; j--) {
            if (A[i - 1] == B[j - 1]) reverse[i][j] = reverse[i + 1][j + 1] + 1;
            else reverse[i][j] = max(reverse[i + 1][j], reverse[i][j + 1]);
        }
    }
    for (int i = 1; i <= A.size() + 1; i++) {
        for (int j = 1; j <= B.size(); j++) {
            if (insertion[i][B[j - 1] - '0']) continue;
            if (cache[i - 1][j - 1] + reverse[i][j + 1] == cache[A.size()][B.size()]) {
                insertion[i][B[j - 1] - '0'] = true;
                total++;
            }
        }
    }
    return total;
}

  • yashparihar729
     + 0 comments

    Here is my solution in java, javascript, python, C ,C++, Csharp HackerRank LCS Returns Problem Solution