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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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class node{
  public:
  long ind;
  char c;
  int t;
  node()
  {
   
  }
  node(long ind,char c,int t)
  {
    this->ind=ind;
    this->c=c;
    this->t=t;
  }
};
bool cmp(node &a,node &b)
{
  if(a.ind<b.ind)return true;
  else return false;
}
vector<string>fun(int n,vector<vector<string>>messages)
{
  vector<node>v(n);
  for(int i=0;i<n;i++)
  {
    long ind=stol(messages[i][0]);
    char c=messages[i][1][0];
    int t=i;
    node temp(ind,c,t);
    v[i]=temp;
  }
  sort(v.begin(),v.end(),cmp);
  vector<pair<int,string>>check;
  vector<string>ans;
  for(int i=0;i<n;i++)
  {
    if(v[i].c=='*')
    {
      int j=i+1;
      while(j<n&&v[j].c!='*'&&v[j-1].ind+1==v[j].ind)j++;
      if(v[j].c=='*')
      {
        int mx=0;
        string temp="";
        for(int k=i+1;k<j;k++)temp.push_back(v[k].c);
        for(int k=i;k<=j;k++)mx=max(mx,v[k].t);
        check.push_back({mx,temp});
      }
    }
  }
  sort(check.begin(),check.end());
  for(int i=0;i<check.size();i++)ans.push_back(check[i].second);
  return ans;
}

String Parse โœ…
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SELECT
  c.name AS name,
  COUNT(w.customer_id) AS warehouses,
  MIN(w.volume) AS min_volume,
  MAX(w.volume) AS max_volume,
  SUM(w.volume) AS total_volume
FROM
  customers c
INNER JOIN
  warehouses w ON c.id = w.customer_id
WHERE
  w.is_active = 1
GROUP BY
  c.name
ORDER BY
  c.name ASC;

IBM โœ…
def getBitDistances(n):
    binary = bin(n)[2:]
    return max(len(binary) - 1 - i for i, bit in enumerate(binary) if bit == '1')

def getTopKBitDistances(numbers, k):
    bit_distances = [(num, getBitDistances(num)) for num in numbers]
    sorted_numbers = sorted(bit_distances, key=lambda x: (-x[1], -x[0]))
    return [num for num, _ in sorted_numbers[:k]]


IBMโœ…
class Solution {
    public int solution(int[][] peanuts, int[][] carrots) {
        int N = peanuts.length;
        int M = peanuts[0].length;

        int[][] dp = new int[N][M];

        dp[0][0] = peanuts[0][0] - carrots[0][0];

        for (int j = 1; j < M; j++) {
            dp[0][j] = dp[0][j - 1] + peanuts[0][j] - carrots[0][j];
        }

        for (int i = 1; i < N; i++) {
            dp[i][0] = dp[i - 1][0] + peanuts[i][0] - carrots[i][0];
        }

        for (int i = 1; i < N; i++) {
            for (int j = 1; j < M; j++) {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + peanuts[i][j] - carrots[i][j];
            }
        }

        return dp[N - 1][M - 1];
    }

Atlan fellowship โœ…
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