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

int solution(vector<int>& A, vector<int>& B, int X, int Y) {
    int N = A.size();
    vector<int> dpA(N), dpB(N);

    // Initial condition: Start at position 0 on either line
    dpA[0] = A[0];
    dpB[0] = B[0];

    // Fill the dp arrays
    for (int i = 1; i < N; ++i) {
        dpA[i] = min(dpA[i - 1] + A[i], dpB[i - 1] + Y + A[i]);
        dpB[i] = min(dpB[i - 1] + B[i], dpA[i - 1] + X + B[i]);
    }

    // The result is the minimum time to complete the car on either line at the last position
    return min(dpA[N - 1], dpB[N - 1]);
}

int main() {
    int N, X, Y;
    cin >> N >> X >> Y;
    vector<int> A(N), B(N);
    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }
    for (int i = 0; i < N; ++i) {
        cin >> B[i];
    }
    cout << solution(A, B, X, Y) << endl;
    return 0;
}
Ms task2✅

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

ll solve(vector<int>& s, int k) {
  
    sort(s.begin(), s.end());
    int n = s.size();
    vector<ll> ps(n+1, 0);

    for (int i = 0; i < n; ++i) {
        ps[i + 1] = ps[i] + s[i];
    }

    ll minEff = LLONG_MAX;

    for (int i = 0; i <= n - k; ++i) {
        int mIdx = i + k / 2;
        int x = s[mIdx];

        ll lSum = (mIdx - i) * x - (ps[mIdx] - ps[i]);
        // cout << lSum <<" ";
        ll rSum = (ps[i + k] - ps[mIdx + 1]) - (i + k - mIdx - 1) * x;
        // cout << rSum <<" ";
        
        ll eff = lSum + rSum;
        minEff = min(minEff, eff);
    }

    return minEff;
}

Atlassian
findminimumeffort✅

takeUforward | SWE Interns and FTE

https://www.linkedin.com/posts/takeuforward_hiring-wearehiring-activity-7228273874994442240-Fd3c

MX is hiring Front-end Developer

For 2021, 2022 grads
Location: Chennai

https://mx.wd1.myworkdayjobs.com/MX-IND/job/Chennai-Tamil-Nadu-India/Software-Engineer---Frontend-Developer-Experience_R1517?source=LinkedIn

Full the Form: https://forms.gle/7mSaBWyMme7XGzW76
OR
Send your updated resume to: parnavi.srivastava@yendigital.com

🚀Vertex Infosoft Solutions Pvt. Ltd is urgently looking for B.tech (CS/IT) candidates with:

Excellent english communication skills 🗣️ for the 'Associate Engineer' position in the Technical Support department.

Interested candidates can share their CVs 📃at
careers@vertexinfosoft.com

def is_valid_expression(s):
    valid_chars = set("0123456789+-*/ ")
    if any(c not in valid_chars for c in s):
        return False
    
    prev_char = ''
    for i, c in enumerate(s.strip()):
        if c in '+-*/':
            if prev_char in '+-*/' or i == 0 or i == len(s.strip()) - 1:
                return False
        prev_char = c
    
    return True

def evaluate_expression(s):
    try:
        result = eval(s)
        if isinstance(result, int):
            return result
        else:
            return -1
    except ZeroDivisionError:
        return -1
    except Exception:
        return -1

def integer_calculator(s):
    s = s.strip()
    
    if not is_valid_expression(s):
        return -1
    
    result = evaluate_expression(s)
    
    return result

Jaguar ✅

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

struct node {
    int h[26];
    int n;     
}

node null;
vector<node> trie; 
void addContact(string &name) {
    int act = 0; 
    for (char c : name) {
        int p = c - 'a'; 
        if (!trie[act].h[p]) { 
            trie[act].h[p] = trie.size();
            trie.push_back(null);
        }
        act = trie[act].h[p]; 
        trie[act].n++;
    }
}

int findPartial(string &prefix) {
    int act = 0; 
    for (char c : prefix) {
        int p = c - 'a'; 
        if (!trie[act].h[p]) return 0; 
        act = trie[act].h[p]; 
    }
    return trie[act].n; 
}

int main() {
    int n;
    cin >> n; 
    string operation, contact;

    null.n = 0;
    for (int i = 0; i < 26; i++) null.h[i] = 0;
    trie.push_back(null);

    vector<int> results; 

    for (int i = 0; i < n; i++) {
        cin >> operation >> contact;
        if (operation == "add") {
            addContact(contact);
        } else if (operation == "find") {
            results.push_back(findPartial(contact)); 
        }
    }

    for (int res : results) {
        cout << res << "\n";
    }

    return 0;
}

Jaguar ✅

O(log n) labs hiring Frontend Developer ( SDE l )

3 - 6 LPA

Apply Here : https://wellfound.com/jobs/3076985-frontend-developer-sde-1

Frontend required for a specific task
To develop a single landing page for product based company
Dm him with the best piece of work

@kalihaxor

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

vector<int> frequencyOfMaxValue(vector<int>& numbers, vector<int>& queries) {
    int n = numbers.size();
    vector<int> answer(n, -1);   
    int maxvalue = -1;           
    int count = 1;              
    for (int i = n - 1; i >= 0; --i) {
        if (numbers[i] == maxvalue) {
            count++;
        } else if (numbers[i] > maxvalue) {
            maxvalue = numbers[i];
            count = 1;
        }
        answer[i] = count;       
    }

    vector<int> result;
    for (int index : queries) {
        result.push_back(answer[index - 1]);  
    }

    return result;
}

 

Frequency ofMaxValue
Wells fargo✅

#include <bits/stdc++.h>
using namespace std;
vector<int> getPopularityOrder(vector<vector<int>> song_preferences) {
    int n = song_preferences[0].size();
    vector<vector<int>> boxScore(n, vector<int>(n, 0));
    for (const vector<int>& pref : song_preferences) {
        for (int i = 0; i < pref.size() - 1; i++) {
            for (int j = i + 1; j < pref.size(); j++) {
                boxScore[pref[i]][pref[j]] += 1;
            }
        }
    }

    vector<array<int, 2>> wins(n, {0, 0});
    for (int i = 0; i < n; i++) {
        wins[i][0] = i; 
        for (int j = i + 1; j < n; j++) {
            int score = boxScore[i][j] - boxScore[j][i];
            if (score >= 0) {
                wins[i][1]++;
            } else {
                wins[j][1]++;
            }
        }
    }

    sort(wins.begin(), wins.end(), [](const array<int, 2>& a, const array<int, 2>& b) {
        if (a[1] == b[1]) {
            return a[0] < b[0];
        } else {
            return b[1] < a[1];
        }
    });

    vector<int> result;
    for (const auto& win : wins) {
        result.emplace_back(win[0]);
    }

    return result;
}

Songs Popularity
Wells Fargo ✅