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๐ŸšฉMain Group - @SuperExams
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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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Juntran Technologies..!!!
We are looking for DFT & PD Trained Freshers (2021 or earlier passed out graduate) candidates for Hyderabad/Bangalore locations. Interested candidates can drop their resumes to hameed@juntrantech.com

Thanks and Regards,
Syed Hameed
Add Expiry: 31/01/2024
def circles(info):
   result = []

   for circle in info:
       details = [float(data) for data in circle.split()]
       sumRadius = details[2] + details[5]
       diffRadius = abs(details[2] - details[5])
       distance = 0
       if details[0] == details[3] and details[0] == 0:
           distance = abs(details[1] - details[4])
       elif details[1] == details[4] and details[1] == 0:
           distance = abs(details[0] - details[3])

       if distance == sumRadius or distance == diffRadius:
           result.append('Touching')
       elif distance == 0:
           result.append('Concentric')
       elif distance > sumRadius:
           result.append('Disjoint-Outside')
       elif distance < sumRadius and distance > diffRadius:
           result.append('Intersecting')
       elif distance < diffRadius:
           result.append('Disjoint-Inside')
       else:
           print("Operation Failed! data couldn't be handled")
   return result
SELECT
  id,
  title,
  s.quantity AS total_stock,
  r.quantity AS total_returns
  FROM
    products p
      LEFT JOIN stocks s
        ON p.id = s.product_id
      LEFT JOIN ( SELECT product_id, COUNT( * ) AS quantity FROM returns WHERE MONTHNAME( dt ) = 'July' GROUP BY product_id ) r
        ON p.id = r.product_id
  ORDER BY
    id

E-commerce Warehouse stock and Return Report โœ…
int solve(vector<int>& nums) {
    vector<int> s(nums);
    sort(s.begin(), s.end());
    int i = 0, j = nums.size() - 1;
    while (i < nums.size() && nums[i] == s[i]) {
        i++;
    }
    while (j > i && nums[j] == s[j]) {
        j--;
    }
    return j - i + 1;
}

Arrange the heights
Apple โœ…
void solve(string x){
    ll n = x.size();
vector<vector<ll>> dp(n,vector<ll>(n,0));
for(ll i=0;i<n;i++){
    dp[i][i] = 1;
}
for(ll k=1;k<n;k++){
for(ll i=0;i<n-k;i++){
        ll j = i + k;
            if(x[i]==x[j]){
                dp[i][j] = 2 + dp[i+1][j-1];
            } else{
                dp[i][j] = max(dp[i][j-1],dp[i+1][j]);
            }
   }
}
ll ans = 0;
for(ll i=0;i<n;i++){
    for(ll j=0;j<n-1;j++){
        ans = max(ans,dp[i][j]*dp[j+1][n-1]);
      }
   }
cout<<ans;
}

Type According to write function

Game of subsequence reverse

Appleโœ…
#include<bits/stdc++.h>
using namespace std;

void dfs(int node, vector<int>& vis, vector<vector<int>>& adj) {
    vis[node] = 1;
    for(auto it : adj[node]) {
        if(!vis[it]) {
            dfs(it, vis, adj);
        }
    }
}

int main() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> adj(n);
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            int x;
            cin >> x;
            if(x == 1) {
                adj[i].push_back(j);
                adj[j].push_back(i);
            }
        }
    }
    vector<int> vis(n, 0);
    int cc = 0;
    for(int i = 0; i < n; i++) {
        if(!vis[i]) {
            dfs(i, vis, adj);
            cc++;
        }
    }
    cout << cc << endl;
    return 0;
}
int getMinOperations(vector<int> arr) {
      
   int n=arr.size();
    int operations = 0;
    for (int i = 1; i < n - 1; i++)
    {
        if (arr[i] < 0 && arr[i - 1] < 0)
        {
            arr[i] = 1e9;
            operations++;
        }

        else if (arr[i] < 0 && arr[i - 1] > 0 && arr[i + 1] > 0)
        {
            if ((arr[i] + arr[i - 1] < 0) || (arr[i] + arr[i + 1] < 0))
            {
                arr[i] = 1e9;
                operations++;
            }
        }
    }
   return operations; 
}

Make The Array Postive โœ…
def count(a, X):
    a.sort()
    ans = 0
    n = len(a)

    for i in range(n):
        if (i < n // 2):
            ans += max(0, a[i] - X)
        elif (i == n // 2):
            ans += abs(X - a[i])
        else:
            ans += max(0, X - a[i]);

    return ans

IBMโœ