Token Metrics Hiring Crypto Full-Stack Developer Intern (Remote)
Batch: 2024 & 2025 pass outs
NOTE: unpaid
https://jobs.lever.co/tokenmetrics/a78f1b4b-371e-403e-90cd-7d10741a7046

Software Engineer, Internship at Palantir Technologies

Batch 2025
https://www.hackerrank.com/apply/99120

Dear candidate, Genpact is hiring for AML/KYC (Financial crime & compliance)

Please find the below criteria

1) candidates must have basic knowledge of AML/KYC along with good communication skills written/ verbal as well

2) 0-2 years of experience / fresher/graduate/ MBA fresher also apply.

3) Noida location

4) UK shift and 5 days working from office Please send your resume Priyanka.Kharbanda@Genpact.com

Company Name: SjInnovation

🛄 Job Title: Software Engineer Intern

✍🏻 YOE: 2023 and 2024 grads

➡️ Apply: https://career.sjinnovation.com/jobDetails/194

Please do share in your college grps and in case you are applying please react on this post:) 😇❤️

Company Name: Adobe

🛄 Job Title: Software Development Engineer

✍🏻 YOE: 2022 and 2023 grads



➡️ Apply: https://careers.adobe.com/us/en/job/ADOBUSR145089EXTERNALENUS/Software-Development-Engineer

Try to referral

Please do share in your college grps and in case you are applying please react on this post:) 😇❤️

void solve(int tc)
{
        int n,k;
        string s;
        cin >> k >> s;
        n = s.size();
        string str="";
        string find=s.substr(k,n-1);
        int i=0,j=k-1;
        while(i<j)
            str+=s[i++],str+=s[j--];
        if(i==j) str+=s[j];
        reverse(all(str));
        string res = str + find;
        cout << res << '\n';
    }

String Reversal ✅

Company Name: Zinnia

🛄 Job Title: Software Engineer Intern

✍🏻 YOE: 2025 grads only


➡️ Apply: https://boards.greenhouse.io/zinnia/jobs/4301004006?gh_src=5b8360806us

Please do share in your college grps and in case you are applying please react on this post:) 😇❤️

ll dp[102][2][902];
const int mod = 1e9 + 7;
ll solve(int i, bool tight, int sum_so_far, int Sum, string number, int len) {
    if (i == len) {
        if (sum_so_far <= Sum)
            return 1;
        else
            return 0;
    }

    ll &ans = dp[i][tight][sum_so_far];
    if (~ans) return ans;

    ans = 0;
    bool ntight;
    int nsum_so_far;
    for (char currdigit = '0'; currdigit <= '9'; currdigit++) {
        if (!tight && currdigit > number[i]) break;
        ntight = tight || currdigit < number[i];
        nsum_so_far = sum_so_far + (currdigit - '0');
        ans += solve(i + 1, ntight, nsum_so_far, Sum, number, len);
        ans %= mod;
    }
    return ans;
}
void op() {
    string num1, num2;
    int mxmSum, mnmSum;
    cin >> num1 >> num2 >> mnmSum >> mxmSum;
    mnmSum--;
    ll res1, res2, fin1 = 0;

    if(num1 != "0") {
        for(int i = num1.size() - 1; i >= 0; i--){
            if(num1[i] != '0'){
                num1[i]--;
                break;
            }else num1[i] = '9';
        }
        if(num1[0] == '0'){
            reverse(num1.begin(), num1.end());
            num1.pop_back();
            reverse(num1.begin(), num1.end());
        }
        ::memset(dp, -1, sizeof dp);
        res1 = solve(0, 0, 0, mnmSum, num1, num1.size());
        ::memset(dp, -1, sizeof dp);
        res2 = solve(0, 0, 0, mxmSum, num1, num1.size());
        fin1 = res2 - res1;
        fin1 %= mod;
    }

    ::memset(dp, -1, sizeof dp);
    res1 = solve(0, 0, 0, mnmSum, num2, num2.size());
    ::memset(dp, -1, sizeof dp);
    res2 = solve(0, 0, 0, mxmSum, num2, num2.size());
    ll fin2 = res2 - res1;
    fin2 %= mod;

    cout << (fin2 - fin1) % mod << "\n";
}

Digit sums ✅

EY is hiring for Freshers into Researchanalyst role 📢

Candidates with experience in secondarymarketresearch and thirdpartyduediligence are applicable for this role only.

📍 Location: Noida
📌 Work mode: Work from Office

✅ Interested candidates can share their CVs at himani.mittal@in.ey.com along with their below details.

Name:
current location:
preferred location:
Current CTC:
Expected CTC:
Current Company:
Official Notice period:

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

int main() {
    int n;cin>>n;
    vector<int>v(n);
    int bud;cin>>bud;
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    int diff=INT_MAX;
    pair<int,int>p;
    for(int i=0;i<(n-1);i++){
        int left=bud-v[i];
        if(left<=0)
        {
             int curr_diff=bud-v[i]-v[i+1];
            if(abs(curr_diff)<diff){
                diff=abs(curr_diff);
                p.first=v[i];
                p.second=v[i+1];
               
            }
            continue;
        }
        if(v[n-1]<=left){
            int curr_diff=bud-v[i]-v[n-1];
            if(abs(curr_diff)<diff){
                diff=abs(curr_diff);
                p.first=v[i];
                p.second=v[n-1];
               
            }
            continue;
        }
        int idx=lower_bound(v.begin()+i+1,v.end(),bud-v[i])-v.begin();
        if(idx==(i+1)){
            int curr_diff=bud-v[i]-v[idx];
            if(abs(curr_diff)<diff){
                diff=abs(curr_diff);
                p.first=v[i];
                p.second=v[idx];
               
            }
        }
        else{
            int curr_diff1=abs(bud-v[i]-v[idx]);
            int curr_diff2=abs(bud-v[i]-v[idx-1]);
            if(curr_diff2<curr_diff1){
                if(curr_diff2<diff){
                diff=curr_diff2;
                p.first=v[i];
                p.second=v[idx-1];
                }
            }
            else{
                if(curr_diff1<diff){
                diff=curr_diff1;
                p.first=v[i];
                p.second=v[idx];
                }
            }
           
        }
       
    }
    cout<<p.first<<endl;
    cout<<p.second<<endl;
   
return 0;
}

Budget issues ✅

https://www.testgorilla.com/job-board/jaffle/frontend-developer-vue-3-nuxt/

Instant exam link ✅
Solution uploaded ✅

class UserMainCode(object):
    @classmethod
    def musicTeacher(cls, input1):
        n = len(input1)
        dp = [i for i in range(n + 1)]
        for i in range(1, n + 1):
            cnt = [0] * 26
            for j in range(i, 0, -1):
                d = ord(input1[j - 1]) - ord('a')
                cnt[d] += 1
                val = cnt[d]
                f = 0
                for k in range(26):
                    if cnt[k] != 0 and cnt[k] != val:
                        f = 1
                        break
                if f == 0:
                    dp[i] = min(dp[i], 1 + dp[j - 1])
        return dp[n]

Music Teacher ✅