MAQ Hiring Software Engineer

https://forms.office.com/pages/responsepage.aspx?id=0o3Z5JmR5UK6i9o-dj7eLq0Di02SxiRKvrcmuY04xrFUQTlQM0tZNEpaQktGT09OUjJZOEhHSzNaNC4u

int main() {
    int n, x;
    cin >> n >> x;
    vector<int> c(n);
    for(int i = 0; i < n; i++) {
        cin >> c[i];
    }

    int m = 0, s = 0, i = 0;
    for(int e = 0; e < n; e++) {
        s += c[e];
        while(s > x) {
            s -= c[i++];
        }
        m = max(m, e - i + 1);
    }

    cout << m << endl;
    return 0;
}

Mercedes ✅

def maximize_profit(prices):
    if len(prices) < 2:
        return "Not Possible" 
    min_price = prices[0]
    max_profit = 0
    for price in prices[1:]:
        max_profit = max(max_profit, price - min_price)
        min_price = min(min_price, price)
    if max_profit > 0:
        return "Possible\n{}".format(max_profit)
    else:
        return "Not Possible"
n = int(input())
prices = list(map(int, input().split()))
print(maximize_profit(prices))

Maximize Profit ✅

# include <bits/stdc++.h>
using namespace std;
#define ll long long
signed main(){
    ll n; cin>>n;
    vector<ll>a(n);
    for(ll i=0;i<n;i++) cin>>a[i];
    ll x=0;
    for(ll i=0;i<n;i++) x^=a[i];
    ll ans=LONG_MAX;
    for(ll i=0;i<n;i++)  ans=min(ans,x^a[i]);
    cout<<ans<<endl;
    return 0;
}

Minimum Xor ✅

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✅