INDMoney is looking for web intern role

Batch eligible: 2024 passouts only


https://docs.google.com/forms/d/e/1FAIpQLSerXmlfS7hOpg6UuKW6OU-AWT17U76eRbcnP9yyiq-4XDIFdQ/viewform

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

Company: Ciena

Role: Software Engineer

Location: Gurugram, India

Batch eligible: 2024 passouts

Apply Link: https://careers.ciena.com/us/en/job/CIENUSR023021ENUS/Software-Engineer-NewGrad

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 ✅

Company Name: Upstox

Role: SDE Intern

Batch eligible: 2024 grads only

Apply: https://jobs.lever.co/upstox/e15b9350-848e-4840-808b-94d557e35d3b

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 ✅

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

Apple✅

#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;
}

Walmart Round 2 Result Out check out now

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 ✅

Walk Drive
Only Batch Eligible 2022/2023

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

IBM✅

Hi All,

We are looking for MTech Computer science/Electronics 2023/2022 passouts to work with our Cloud Team in Bangalore. We are doing a hiring event on 30th Jan 2024 to hire folks . If interested please share your resume to aayushar@qti.qualcomm.com

EY is on the lookout for Chartered Accountants (with 0-1 year of experience) to join Indirect Tax team.

Candidates should be based in Pune or willing to relocate.

If you're seeking an opportunity that offers growth, learning, and a chance to be part of a renowned global organization, we'd like to connect with you.

You may email your resume at shweta.patni@in.ey.com and take a step towards a rewarding career with EY!

Hello Freshers,

Mirrar is hiring for the below Internship roles :
1.Frontend Intern
2.Backend Intern
3.Testing Intern

If you are interested in being a part of a B2B fashion tech organization with focus on Augmented Reality, Kindly share your resume to  jeevitha.m@mirrar.com

long long solution(const vector<int> &no_adjacent, const vector<int> &one_adjacent, const vector<int> &both_adjacent) {
    const int n = no_adjacent.size();
    vector<long long> dp = {no_adjacent[0], one_adjacent[0]};

    for (int i = 2; i < n; ++i) {
        dp = {
            max(dp[0] + one_adjacent[i - 1], dp[1] + no_adjacent[i - 1]),
            max(dp[0] + both_adjacent[i - 1], dp[1] + one_adjacent[i - 1])
        };
    }

    return max(dp[0] + one_adjacent[n - 1], dp[1] + no_adjacent[n - 1]);
}


Efficient Deployments✅

from functools import cmp_to_key

def romanToInt(s):
    rtoi = {'I': 1, 'V': 5, 'X': 10, 'L': 50}
    length = len(s)
    value = 0
   
    for i in range(length-1):
        if (rtoi[s[i]] >= rtoi[s[i+1]]):
            value += rtoi[s[i]]
        else:
            value -= rtoi[s[i]]
    value += rtoi[s[length-1]]

    return value

def compare(name1, name2):
    namelst1 = name1.split()
    namelst2 = name2.split()

    if (namelst1[0] < namelst2[0]):
        return -1
    elif (namelst1[0] > namelst2[0]):
        return 1
    else:
        roman1 = romanToInt(namelst1[1])
        roman2 = romanToInt(namelst2[1])
        if (roman1 < roman2):
            return -1
        elif (roman1 > roman2):
            return 1
    return 0

def func(names):
    return sorted(names, key=cmp_to_key(compare))

Sort Roman numeral✅

import bisect
import functools
def jobScheduling(pickUp, drop, tip):
jobs = sorted(zip(pickup, drop, tip), key=lambda v: v[1])
print(jobs)
dp = [[0, 0]]
for s, e, p in jobs:
     i = bisect.bisect(dp, [s + 1]) - 1
     if dp[i][1] + e-s+p > dp[-1][1]:
         dp.append([e, dp[i][1] + e-s+p])
return dp[-1][1]

Maximum Earnings ✅