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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