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Hello Developers,
We are a startup from Kolkata, we are currently building a very innovative and exciting prototype in the gaming sector. Currently we need a developer with skills and knowledge in machine and deep learning and Artificial intelligence, like llm and nlp. The project is a paid project, and the developer MUST BE FROM KOLKATA, as we will need to guide and interact the person on site about the project.

The key roles will be -
. Understanding and building the required code and project.
. Being punctual and reporting any problems and following instructions.

If interested for this exciting opportunity please dm and send resume or you work to
+91 9330860396. You can also have a chance to work with us full time in future.

The project is contractual work and the pavkage is around โ‚น15k-20k
๐Ÿ‘2
#include <iostream>
#include <string>
#include <unordered_map>

using namespace std;

long getkRepValue(string user_history, long k) {
    long n = user_history.size();
    unordered_map<char, long> count;
    long left = 0, right = 0, ans = 0;

    while (right < n) {
        count[user_history[right]]++;
        while (count[user_history[right]] >= k && left <= right) {
            ans += n - right;
            count[user_history[left]]--;
            left++;
        }
        right++;
    }

    return ans;


Machine learning Amazon โœ…
int solve(vector<int> tc) {
    int n = tc.size();
    vector<int> p(n), s(n);

    p[0] = tc[0];
    for (int i = 1; i < n; ++i) {
        p[i] = p[i - 1] + tc[i];
    }

    s[n - 1] = tc[n - 1];
    for (int i = n - 2; i >= 0; --i) {
        s[i] = s[i + 1] + tc[i];
    }

    int m = max(p[0], s[0]);
    for (int i = 1; i < n; ++i) {
        m = max(m, max(p[i], s[i]));
    }

    return m;
}

Amazon โœ…
long Count_sol (int N, vector<int> A, vector<int> B) {
   // Write your code here
   vector<long>dp(MAXN,0);
   unordered_map<int, int>f;
   unordered_map<int, int>g;
   for(int i=0;i<N;i++)
   {
    f[A[i]]++;
    g[B[i]]++;
   }
   long ans=0;
   for(int i=100000;i>=2;--i)
   {
    int c=0;
    int d=0;
    int s=0;
    for(int j=i;j<=100000;j+=i)
    {
        c+=f[j];
        d+=g[j];
        s+=dp[j];
    }
    long pw=(long)c*d;
    pw=pw-s;
    ans+=pw;
    dp[i]=pw;
   }
   return ans;
}

Count the Pairs โœ…
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