Sherlock and Cost Discussions | Algorithms | HackerRank
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Sherlock and Cost

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  • nidanabi1414
     + 0 comments

    Bottom up DP

    int cost(vector<int> B) {
vector<vector<int>>dp(B.size(),vector<int>(2,0));
int n=dp.size()-1;
for(int i=1;i<B.size();i++){
    dp[i][0]=max(dp[i-1][0],dp[i-1][1]+abs(B[i-1]-1));
    dp[i][1]=max(dp[i-1][0]+abs(B[i]-1),dp[i-1][1]+abs(B[i]-B[i-1]));
}
return max(dp[n][0],dp[n][1]);

}

  • yashparihar729
     + 0 comments

    Here is my solution in java, javascript, python, C, C++, Csharp HackerRank Sherlock and Cost Problem Solution

  • ekaramath
     + 2 comments

    I see this as a convex maximization problem over a closed bounded set which means the solution is on the boundary (e.g., think of maximizing abs(x) for a <= x <= b. This means A[i] = 1 or B[i]. The rest is simple DP.

    def cost(B):
    # Write your code here
    Smax = 0
    Smin = 0
    for index, value in enumerate(B):
        if index == 0:
            continue
        smin = max([Smin, Smax + abs(B[index - 1] - 1)])
        Smax = max([Smin + abs(1 - B[index]), Smax + abs(B[index] - B[index - 1])])
        Smin = smin
    return max([Smin, Smax])

  • mineman1012221
     + 0 comments

    Here is the solution of Sherlock and Cost Click Here

  • thecodingsoluti2
     + 0 comments

    Here is problem solution - https://programs.programmingoneonone.com/2021/07/hackerrank-sherlock-and-cost-problem-solution.html

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