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๐ŸšฉMain Group - @SuperExams
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Ocrolus East Hiring Data Verifier โค๏ธโค๏ธโค๏ธ

Job Role: Data Verifier

Qualifications: Any Graduate

Experience: 0-5 years

Job Type: Full Time

Location: Noida

Salary: 3 to 5 LPA

Skills/Requirements:
1. Review Documents of different formats and quality
2. Correct or Edit any incorrect details captured by the Ocrolus proprietary systems
3. Perform reconciliation of various amounts captured from reviewed documents to ensure high accuracy and identify discrepancies
4. Basic Understanding of financial documents and Terminologies.
5. Understanding basic accounting concepts would be an added advantage.

Date of offline Interview: 20th August, 2024

Time : 11.00 AM โ€“ 2.00 PM

Address: Ocrolus East Private Limited, A-61, 2nd Floor, ASF Building, Sector 63, Noida- 201301
Company โ€“ Payinstacard Private Limited
Role โ€“ Artificial Intelligence (AI) Intern
Exp. โ€“ Fresher
Apply Here โ€“ https://internshala.com/internship/details/part-time-artificial-intelligence-ai-internship-in-hyderabad-at-payinstacard-private-limited1723443866?utm_source=cp_link&referral=web_share

Company โ€“ Bilateral Solutions Private Limited
Role โ€“ Machine Learning Internship
Exp. โ€“ Fresher
Apply Here โ€“ https://internshala.com/internship/details/machine-learning-internship-in-mandi-at-bilateral-solutions-private-limited1723470090?utm_source=cp_link&referral=web_share

Company โ€“ EcoRatings
Role โ€“ Associate Data Scientist
Exp. โ€“ 0-2 yrs
Apply Here โ€“ https://www.linkedin.com/jobs/view/3996441253
๐Ÿ‘1
#include <string>
#include <algorithm>
string solution(const string& a, const string& b) {
string result = "";
    int lenA = a.length();
    int lenB = b.length();
    int maxLength = max(lenA, lenB);
    for (int i = 0; i < maxLength; ++i) {
        int digitA = (i < lenA) ? (a[lenA - 1 - i] - '0') : 0;
        int digitB = (i < lenB) ? (b[lenB - 1 - i] - '0') : 0;
        int sum = digitA + digitB;
        result = to_string(sum)+result;
    }
    return result;
}


Databrick โœ…
int solution(int initialHealth, vector<int>& deltas) { 
    int currentHealth = initialHealth;

    for (int delta : deltas) {
        currentHealth += delta;

        if (currentHealth < 0) {
            currentHealth = 0;
        } else if (currentHealth > 100) {
            currentHealth = 100;
        }
    }

    return currentHealth;
}


Databrick โœ…
#include <iostream>
#include <string>
#include <unordered_map>

using namespace std;

int findFirstUnique(string s) {
    unordered_map<char, int> charCount;

    for (char c : s) {
        charCount[c]++;
    }

    for (int i = 0; i < s.length(); i++) {
        if (charCount[s[i]] == 1) {
            return i + 1;
        }
    }

    return -1;
}


Amazon (intern) โœ…
int getDistinctPairs(vector<int> stocksProfit, int target)

    int n = stocksProfit.size();
  unordered_map<int, int> m;
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (m.find(target - stocksProfit[i]) != m.end()) {
      count += m[target - stocksProfit[i]];
    }
    m[stocksProfit[i]]++;
  }
  return count;
}


Amazon(intern)โœ…
int solve(int m, int w) 
{
    if (m >= 1 && w >= 1)
    {
        int ways = m * (w * (w - 1)) / 2 + w * (m * (m - 1)) / 2;
        return ways;
    }
    else
    {
        return 0;
    }


IBMโœ…
vector<vector<int>> findOverlappingTimes(vector<vector<int>>& intervals) {
    if (intervals.empty()) {
        return {};
    }

    sort(intervals.begin(), intervals.end());

    vector<vector<int>> merged;
    merged.push_back(intervals[0]);

    for (int i = 1; i < intervals.size(); i++) {
        if (merged.back()[1] >= intervals[i][0]) {
            merged.back()[1] = max(merged.back()[1], intervals[i][1]);
        } else {
            merged.push_back(intervals[i]);
        }
    }

    return merged;
}


Amazon (intern) โœ…
๐Ÿ‘1
def solve1(number):
    digits = [int(d) for d in str(number)]
    sum_digits = sum(digits)
    max_digit = max(digits)
    return (sum_digits - max_digit) == max_digit

def solve2(X, Y):
    total_savings = 0
    for number in range(X, Y + 1):
        if solve1(number):
            total_savings += number
    return total_savings

X, Y = map(int, input().split())
print(solve2(X, Y))


MasterCard โœ…
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
#include <bits/stdc++.h>
using namespace std;

int stockLounge(vector<int> onHand, vector<int> supplier, int demand) {
   
    sort(onHand.begin(), onHand.end());
    sort(supplier.begin(), supplier.end());

    int totalSupplierUsed = 0;
    int n = onHand.size(), m = supplier.size();
   
  
    for (int i = 0; i < n; i++) {
        demand -= min(demand, 1);
    }

  
    for (int i = 0; i < m; i++) {
        if (demand > 0) {
            int canTake = min(supplier[i], demand);
            totalSupplierUsed += canTake;
            demand -= canTake;
        }
    }
    return totalSupplierUsed;
}


BNY Mellon โœ