def getNumPerfectPackaging(p):
    MOD = 10**9 + 7
    dp0 = [1] * 6  
    dp1 = [0] * 6  

    for i in range(1, len(p)):
        c = [0] * 6
        for s in range(1, 6):
            c[s] = (c[s-1] + dp0[s]) % MOD
        tot = c[5]
        for s in range(1, 6):
            if p[i] > p[i-1]:
                dp1[s] = c[s-1]
            elif p[i] < p[i-1]:
                dp1[s] = (tot - c[s]) % MOD
            else:
                dp1[s] = (tot - dp0[s]) % MOD
        dp0, dp1 = dp1, [0] * 6  

    return sum(dp0[1:6]) % MOD

Amazon ✅

Company: OpenGig
Role: SDE Intern

Batch: 2025, 2026
Salary : 25-30k [Performance based PPO - 12 LPA]

Apply:
https://docs.google.com/forms/d/e/1FAIpQLSeGx23AxYbiop1q0hNRG8fjrbEdZOBGN1-SCD4QeoqMy6JJBw/viewform

TurboHire
https://app.turbohire.co/job/publicjobs/QlwV0cGk5%2F4sXGQL87CSquGyYUisqF6liwYKLSfkgofNgs7nQWtgWg5kxvo1JH22

def magical_string(n, s):
count = 0
expecting_first = True
current_char = ''
for c in s:
if expecting_first:
current_char = c
expecting_first = False
else:
if c != current_char:
count += 2
expecting_first = True
return n - count

Acolite ✅

#include <bits/stdc++.h>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);



/*
* Complete the 'minTimeForRobots' function below.
*
* The function is expected to return a LONG_INTEGER.
* The function accepts following parameters:
*  1. INTEGER the_number_of_robots
*  2. INTEGER_ARRAY rack_capacities
*/


#include<bits/stdc++.h>
using ll = long long;
using namespace std;
bool canAllocate(const vector<int>& capacities, ll k, ll maxLoad) {
    ll robotCount = 1; 
    ll currentLoad = 0; 

    for (int cap : capacities) {
        if (cap > maxLoad) return false; 
        if (currentLoad + cap > maxLoad) {
            robotCount++;
            currentLoad = cap;
            if (robotCount > k) return false;
        } else {
            currentLoad += cap;
        }
    }

    return true;
}


long long minTimeForRobots(int the_number_of_robots, std::vector<int> rack_capacities) {
    ll n = rack_capacities.size();
    if (the_number_of_robots == 0) return -1; 
    if(the_number_of_robots>n)return *max_element(rack_capacities.begin(), rack_capacities.end());
    ll low = *max_element(rack_capacities.begin(), rack_capacities.end());
    ll high = accumulate(rack_capacities.begin(), rack_capacities.end(), 0l);
    ll result = high;

    while (low <= high) {
        ll mid = low + (high - low) / 2;

        if (canAllocate(rack_capacities, the_number_of_robots, mid)) {
            result = mid;
            high = mid - 1; 
        } else {
            low = mid + 1;
        }
    }

    return result;

}

Nvidia ✅

def maximumXorSum(arr1, arr2):
    n = len(arr1)
    MOD = 10**9 + 7
    
    bit_count1 = [0] * 32
    bit_count2 = [0] * 32
    
    for i in range(n):
        for j in range(32):
            bit_count1[j] += (arr1[i] >> j) & 1
            bit_count2[j] += (arr2[i] >> j) & 1
    
    total_sum = 0
    
    for i in range(32):
        ones = (bit_count1[i] * n) + (bit_count2[i] * n) - (2 * bit_count1[i] * bit_count2[i])
        total_sum = (total_sum + (ones * (1 << i))) % MOD
    
    return total_sum

Oracle ✅
Maximum Xor matrix Sum

Nokia is hiring Software Developer

For 2022, 2023 grads
Location: Bangalore

https://fa-evmr-saasfaprod1.fa.ocs.oraclecloud.com/hcmUI/CandidateExperience/en/sites/CX_1/job/12508?src=SNS-102

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

pair<vector<int>, int> elimination_game(int n) {
    deque<int> s(n);
    iota(s.begin(), s.end(), 1);
    vector<int> result;
    bool l2r = true;

    while (s.size() > 1) {
        int size = s.size();
        for (int i = 0; i < size; ++i) {
            if ((i % 2 == 0 && l2r) || (i % 2 == 1 && !l2r)) {
                result.push_back(s[i]);
            } else {
                s.push_back(s[i]);
            }
        }
        s.erase(s.begin(), s.begin() + size);
        l2r = !l2r;
    }

    return {result, s.front()};
}

Kickdrum ✅

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

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while(t--) {
        int n, k;
        cin >> n >> k;
        vector<int> a(n);
        for(auto &x : a) cin >> x;
        int ans = accumulate(a.begin(), a.end(), 0);
        for(int i = 0; i + k < n; i++) {
            int mn = *min_element(a.begin() + i, a.begin() + i + k + 1);
            int sum = 0;
            for(int j = 0; j < n; j++) {
                if(j >= i && j < i + k + 1) sum += mn;
                else sum += a[j];
            }
            ans = min(ans, sum);
        }
        cout << ans << '\n';
    }
    return 0;
}

Kickdrum ✅

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int N, M;
    cin >> N >> M; 

    vector<int> pouchCount(N + 1, 0); 
    vector<int> output; 

    for (int i = 0; i < M; ++i) {
        int candies;
        cin >> candies; 

        if (candies > N) {
            candies = N;
        }
        
        pouchCount[candies]++; 

        bool canPackBasket = true;
        for (int j = 1; j <= N; ++j) {
            if (pouchCount[j] == 0) {
                canPackBasket = false; 
                break;
            }
        }

        if (canPackBasket) {
            output.push_back(1);
            for (int j = 1; j <= N; ++j) {
                pouchCount[j]--;
            }
        } else {
            output.push_back(0); 
        }
    }

    for (size_t i = 0; i < output.size(); ++i) {
        cout << output[i] << (i == output.size() - 1 ? "" : " "); 
    }
    cout << endl;

    return 0;
}

Amagi (intern) ✅

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

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int &x : a) cin >> x;

    long long sum = accumulate(a.begin(), a.end(), 0LL), pre = 0;
    int idx = 0, min_diff = INT_MAX;

    for (int i = 0; i < n; ++i) {
        pre += a[i];
        long long avg1 = pre / (i + 1);
        long long avg2 = (sum - pre) / max(1, n - i - 1);
        int diff = abs(avg1 - avg2);

        if (diff < min_diff) {
            min_diff = diff;
            idx = i;
        }
    }
    cout << idx+1 << endl;
}

Amagi (Intern) ✅

Exciting opportunity for Data Scientists in Gurgaon! Join the Quant Trading firm to work on parsing and analysing large data sets, discovering new data sources, and building predictive machine learning models.

Responsibilities include:
- Exploring diverse data sets
- Developing a framework to monitor critical data sources
- Implementing machine learning techniques and predictive modelling.

Requirements:
- Technical degree in Computer Science or related field from IIT with 8+ GPA
- Experience in statistical analysis
- Experience with Machine Learning, Development of ML Models
- Proficiency in Linux, Bash, Python, and SQL Database

Work in a dynamic and rewarding environment as a Data Scientist. Please reach out to me at anshulgoel@aquissearch.com for more details.

Commerce IQ is hiring Software Engineer (UI)

For 2021, 2022, 2023 grads
Location: Bangalore

https://job-boards.greenhouse.io/commerceiq/jobs/6242977003?gh_src=6c2d12b73us

def solve(matrix):
    rows = len(matrix)
    cols = len(matrix[0]) if rows > 0 else 0
    border_row = '*' * (cols + 2)
    bordered_matrix = [border_row]
    for row in matrix:
        bordered_matrix.append('*' + ''.join(row) + '*')
    bordered_matrix.append(border_row)
    return bordered_matrix

Fico ✅

def getRegions(ip_addresses):
    def isValidIP(ip):
        parts = ip.split('.')
        if len(parts) != 4:
            return False
        for part in parts:
            if not part.isdigit():
                return False
            if not 0 <= int(part) <= 255:
                return False
        return True

    def getRegion(ip):
        first_octet = int(ip.split('.')[0])
        if 0 <= first_octet <= 127:
            return 1
        elif 128 <= first_octet <= 191:
            return 2
        elif 192 <= first_octet <= 223:
            return 3
        elif 224 <= first_octet <= 239:
            return 4
        elif 240 <= first_octet <= 255:
            return 5
        else:
            return -1

    regions = []
    for ip in ip_addresses:
        if isValidIP(ip):
            regions.append(getRegion(ip))
        else:
            regions.append(-1)
    return regions

Location Detection
HPE ✅

Company Name : Groww
Batch : 2023/2022 passouts
Role : Web Developer

Link : https://www.linkedin.com/jobs/view/4064817342

Company Name : Swiggy
Role : Machine Learning Engineer 1
Batch : 2024/2023 passouts

Link :
https://www.linkedin.com/jobs/view/4064814309