Position: HR Fresher
Location: Phase 8B, Mohali, Punjab
Interview: Face-to-face only
Requirements: Training or internship in HR.
How to Apply: Send CV to sanjot@brucode.com or call/WhatsApp 7009008101.

Company – Unified Mentor Private Limited
Role – data analyst intern
Exp. – Fresher
Apply Here – https://www.linkedin.com/jobs/view/3907324906

Company – Acencore Technologies
Role – AI & ML Internship
Exp. – Fresher
Apply Here – https://internshala.com/internship/details/work-from-home-ai-ml-internship-at-acencore-technologies1713779302?utm_source=cp_link&referral=web_share

Company – INI8 Labs Private Limited
Role – Data Engineering Internship
Exp. – Fresher
Apply Here – https://internshala.com/internship/details/data-engineering-internship-in-bangalore-at-ini8-labs-private-limited1713853327?utm_source=cp_link&referral=web_share

Hashedln hiring for 2024 Batch

https://forms.office.com/pages/responsepage.aspx?id=8UXaNizdH02vE1q-RrmZIbWv986yhRlFvG4VkjASqEZURExLQU45WjFNNUZQTDFQWVpFWkxHR0hBNy4u

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

int findMinimumLengthSubarray(vector<int> arr, int k) {
    int n = arr.size();
    unordered_map<int, int> freq;
    int distinct = 0;
    int minLength = INT_MAX;
    int left = 0;
   
    for (int right = 0; right < n; ++right) {
        if (freq[arr[right]] == 0) {
            distinct++;
        }
        freq[arr[right]]++;
       
        while (distinct >= k) {
            minLength = min(minLength, right - left + 1);
            freq[arr[left]]--;
            if (freq[arr[left]] == 0) {
                distinct--;
            }
            left++;
        }
    }
   
    return minLength == INT_MAX ? -1 : minLength;
}

Minimum Sub Array ✅

int minimalFilps(string bits)
{
    int n = bits.length();
 
    int flipsFromLeft[n];
    int flipsFromRight[n];
    int flips = 0;
    for (int i = 0; i < n; i++) {
        if (bits[i] == '0')
            flips++;        
        flipsFromLeft[i] = flips;
    }
 
    flips = 0;
    for (int i = n - 1; i >= 0; i--) {
        if (bits[i] == '1')
            flips++;        
        flipsFromRight[i] = flips;
    }
 
    int minFlips = INT_MAX;
    for (int i = 1; i < n; i++) {
        if (flipsFromLeft[i - 1] + flipsFromRight[i] < minFlips)
            minFlips = flipsFromLeft[i - 1] + flipsFromRight[i];
    }
 
    return minFlips;
}

IBM✅
Password string

vector<string> areAlmostEquivalent(const vector<string>& s, const vector<string>& t) {
    vector<string> result;
    for (int i = 0; i < s.size(); ++i) {
        vector<int> s_count(26, 0);
        vector<int> t_count(26, 0);
        for (char c : s[i]) {
            s_count[c - 'a']++;
        }
        for (char c : t[i]) {
            t_count[c - 'a']++;
        }
        bool valid = true;
        for (int j = 0; j < 26; ++j) {
            if (abs(s_count[j] - t_count[j]) > 3) {
                valid = false;
                break;
            }
        }
        if (valid) {
            result.push_back("YES");
        } else {
            result.push_back("NO");
        }
    }

    return result;
}

IBM✅

def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True

def is_googly(n):
    digit_sum = sum(int(digit) for digit in str(n))
   
    if is_prime(digit_sum):
        return "GOOGLY"
    else:
        return "NOT GOOGLY"

N = int(input())

print(is_googly(N))

Googly Number ✅

from itertools  import combinations
n=int(input())
a=list(map(int,input().split()))
res=0
if n>1:
  for i in combinations(a,2):
    if sum(i)%60==0:
        res+=1
  if res! =0:
      print(res%((10**9)+7))
if res==0 or n<2:
       print("NO HOURS")


Hours Count ✅

#include<stdio.h>
#include<stdlib.h>

#define MOD 1000000007

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int main() {
    int N;
    scanf("%d", &N);

    if (N <= 1) {
        printf("NO HOURS\n");
        return 0;
    }

    int *A = (int*)malloc(N * sizeof(int));

    for (int i = 0; i < N; i++) {
        scanf("%d", &A[i]);
    }

    qsort(A, N, sizeof(int), compare);

    int count = 0;

    for (int i = 0; i < N - 1; i++) {
        for (int j = i + 1; j < N; j++) {
            if ((A[i] + A[j]) % 60 == 0) {
                count = (count + 1) % MOD;
            }
        }
    }

    if (count > 0) {
        printf("%d\n", count);
    } else {
        printf("NO HOURS\n");
    }

    free(A);

    return 0;
}

Hours Count ✅