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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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Role: Software Engineer Intern

Batch eligible: Only 2023 passouts

Link: https://bit.ly/3dxVqst


๐.๐’. ๐†๐จ ๐ฐ๐ข๐ญ๐ก ๐ซ๐ž๐Ÿ๐ž๐ซ๐ซ๐š๐ฅ, ๐ข๐Ÿ ๐ฉ๐จ๐ฌ๐ฌ๐ข๐›๐ฅ๐ž
๐Ÿ‘1
int getMaximumLength(string lotterryID, string winnerID, int k)
{
    int ans = 0;
    int i = 0;
    while (i < min(lotterryID.size(), winnerID.size()))
    {
        if (lotterryID[i] == winnerID[i])
        {
            ans++;
            i++;
        }
        else if (((lotterryID[i] - 96) % 26) + 1 == winnerID[i] - 96)
        {
            ans++;
            i++;
        }
        else if (((lotterryID[i] - 96) % 26) - 1 == winnerID[i] - 96)
        {
            ans++;
            i++;
        }
        else
        {
            i++;
        }
    }
    return ans;
}

Lottery Winner โœ…(C++)
โค1
def longestCommonSubsequence(self, text1: str, text2: str)
        #Tabulation Approach
        if len(text1)==0 or len(text2)==0:
            return 0
        rows,columns=(len(text2)+1,len(text1)+1)
        dp = [[0 for j in range(columns)] for i in range(rows)]
        for col in range(1,columns):
            for row in range(1,rows):
                if text1[col-1]==text2[row-1]:
                    dp[row][col]=1+dp[row-1][col-1]
                else:
                    dp[row][col]=max(dp[row-1][col],dp[row][col-1])
        return dp[rows-1][columns-1]

Lottery Winner โœ…(Python 3)
โค1
int minimumKeypadClickCount(string input)
{
    char right[9][3] = {
        {'a', 'j', 's'},
        {'b', 'o', 't'},
        {'c', 'p', 'u'},
        {'d', 'k', 'v'},
        {'h', 'm', 'z'},
        {'g', 'l', '@'},
        {'e', 'n', 'w'},
        {'f', 'q', 'x'},
        {'i', 'r', 'y'}};
    int n, c = 0, ans = 0;
    n = input.length();
    for (int i = 0; i < n; i++)
    {
        int d = input[i] - 97;
        d = d % 3;
        c += (d + 1);
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 9; j++)
        {
            for (int k = 0; k < 3; k++)
            {
                if (right[j][k] == input[i])
                    ans += (k + 1);
            }
        }
    }
    return ans;
}

Amazon โœ…
โค1