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Nvidia code Stone weight Python3 ✅
Nvidia code Stone weight Python3 ✅

Nvidia code
Nvidia code

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import bisect import functools def jobScheduling(pickUp, drop, tip): jobs = sorted(zip(pickup, drop, tip), key=lambda v: v[1]) print(jobs) dp = [[0, 0]] for s, e, p in jobs:      i = bisect.bisect(dp, [s + 1]) - 1      if dp[i][1] + e-s+p > dp[-1][1]:          dp.append([e, dp[i][1] + e-s+p]) return dp[-1][1] Maximum Earnings ✅

from functools import cmp_to_key def romanToInt(s):     rtoi = {'I': 1, 'V': 5, 'X': 10, 'L': 50}     length = len(s)     value = 0         for i in range(length-1):         if (rtoi[s[i]] >= rtoi[s[i+1]]):             value += rtoi[s[i]]         else:             value -= rtoi[s[i]]     value += rtoi[s[length-1]]     return value def compare(name1, name2):     namelst1 = name1.split()     namelst2 = name2.split()     if (namelst1[0] < namelst2[0]):         return -1     elif (namelst1[0] > namelst2[0]):         return 1     else:         roman1 = romanToInt(namelst1[1])         roman2 = romanToInt(namelst2[1])         if (roman1 < roman2):             return -1         elif (roman1 > roman2):             return 1     return 0 def func(names):     return sorted(names, key=cmp_to_key(compare)) Sort Roman numeral✅

long long solution(const vector &no_adjacent, const vector &one_adjacent, const vector &both_adjacent) {     const int n = no_adjacent.size();     vector dp = {no_adjacent[0], one_adjacent[0]};     for (int i = 2; i < n; ++i) {         dp = {             max(dp[0] + one_adjacent[i - 1], dp[1] + no_adjacent[i - 1]),             max(dp[0] + both_adjacent[i - 1], dp[1] + one_adjacent[i - 1])         };     }     return max(dp[0] + one_adjacent[n - 1], dp[1] + no_adjacent[n - 1]); } Efficient Deployments✅

long long solution(const vector &amp;no_adjacent, const vector &amp;one_adjacent, const vector &amp;both_adjacent) { &nbsp;&n
long long solution(const vector &no_adjacent, const vector &one_adjacent, const vector &both_adjacent) {     const int n = no_adjacent.size();     vector dp = {no_adjacent[0], one_adjacent[0]};     for (int i = 2; i < n; ++i) {         dp = {             max(dp[0] + one_adjacent[i - 1], dp[1] + no_adjacent[i - 1]),             max(dp[0] + both_adjacent[i - 1], dp[1] + one_adjacent[i - 1])         };     }     return max(dp[0] + one_adjacent[n - 1], dp[1] + no_adjacent[n - 1]); } Efficient Deployments✅

ibm ✅ share @whitehatcoding ❤️
ibm ✅ share @whitehatcoding ❤️

🚩Jai Shree Ram 🕉❤
  🚩Jai Shree Ram 🕉❤

#include <iostream> #include <vector> #include <deque> #include <unordered_map> using namespace std; int smartTaxiDriver(int N, int K, vector<int>& T, vector<int>& P, vector<int>& C) {     unordered_map<int, vector<pair<int, int>>> g;     for (int i = 0; i < N - 1; ++i) {         g[P[i] - 1].push_back({i + 2, C[i]});     }     int mpc = 0;     for (int sn = 1; sn <= N; ++sn) {         vector<int> d(N + 1, -1);         d[sn] = 0;         deque<pair<int, int>> q = {{sn, 0}};         while (!q.empty()) {             auto [cn, cd] = q.front();             q.pop_front();             for (auto& [ne, rd] : g[cn]) {                 if (d[ne] == -1 || d[ne] > cd + rd) {                     d[ne] = cd + rd;                     q.push_back({ne, d[ne]});                 }             }         }         int pc = 0;         for (int des : T) {             if (d[des] <= K) {                 pc++;             }         }         mpc = max(mpc, pc);     }     return mpc - 1; } int main() {     int N, K;     cin >> N >> K;     vector<int> T(N - 1), P(N - 1), C(N - 1);     for (int i = 0; i < N - 1; ++i) {         cin >> T[i];     }     for (int i = 0; i < N - 1; ++i) {         cin >> P[i];     }     for (int i = 0; i < N - 1; ++i) {         cin >> C[i];     }     int res = smartTaxiDriver(N, K, T, P, C);     cout << res << endl;     return 0; } //Smart Taxi Driver

#include <bits/stdc++.h> using namespace std; #define ll long long ll dp[10005][2]; const int mod = (1e9 + 7); ll solve(ll i, ll o, ll n, ll k, vector<ll> &v) {     if (i > n) {         return 0;     }     if (dp[i][o] != -1)         return dp[i][o];     ll ans = 0;     if (o == 0) {         for (int j = i; j < min(n + 1, i + k); j++) {             ans = max(ans, solve(j + 1, 1 - o, n, k, v)) % mod;         }     } else {         ll ta = 0;         ll mx = 0;         for (int j = i; j < min(n + 1, i + k); j++) {             if (ta + v[j] >= 0) {                 ta += v[j];                 mx = max(mx, ta);             } else {                 ta = 0;             }             ll c = solve(j + 1, 1 - o, n, k, v) % mod;             ans = max(ans, ((j - i + 1) * mx) % mod + c % mod);         }     }     return dp[i][o] = ans; } int main() {     ll n, k;     cin >> n >> k;     vector<ll> v;     ll neg = 0;     for (int i = 0; i < n; i++) {         ll x;         cin >> x;         v.push_back(x);         if (x <= 0)             neg++;     }     if (neg == v.size()) {         cout << 0 << endl;         return 1;     }     memset(dp, -1, sizeof(dp));       cout << max(solve(0, 0, v.size() - 1, k, v), solve(0, 1, v.size() - 1, k, v)) << endl; } // Array Segments

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int min_operations(string s) {     int n = s.length();     vector<vector<int>> dp(n, vector<int>(n, INT_MAX));     for (int i = 0; i < n; i++) {         dp[i][i] = 0;     }     for (int len = 2; len <= n; len++) {         for (int i = 0; i <= n - len; i++) {             int j = i + len - 1;             if (s[i] == s[j]) {                 dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);             } else {                 for (int k = i; k < j; k++) {                     dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + 1);                 }             }         }     }     return dp[0][n - 1]; } // String dot

#include <bits/stdc++.h> using namespace std; void solve(vector<vector<int>> vv, int operation, int xx, int yy, int &res) {     for (int i = 1; i < 3; i++)     {         int sum1 = 0;         for (int j = 0; j < vv.size(); j++)         {             sum1 += vv[j][i - 1];         }         int sum2 = 0;         for (int j = 0; j < vv.size(); j++)         {             sum2 += vv[j][i];         }         if (sum1 == sum2)         {             res = min(res, operation);         }         return;     }     for (int i = 0; i < vv.size(); i++)     {         solve(vv, operation, xx, yy, res);         vector<int> p1 = vv[i];         reverse(p1.begin(), p1.end());         solve(vv, operation + yy, xx, yy, res);         vector<int> p2 = vv[i];         int temp1 = p2[0];         int temp2 = p2[1];         int temp3 = p2[2];         p2[2] = temp1;         p2[1] = temp3;         p2[0] = temp2;         solve(vv, operation + xx, xx, yy, res);         vector<int> p3 = vv[i];         temp1 = p2[0];         temp2 = p2[1];         temp3 = p2[2];         p2[2] = temp2;         p2[1] = temp1;         p2[0] = temp3;         solve(vv, operation + xx, xx, yy, res);     }     return; } int main() {     ios_base::sync_with_stdio(false);     cin.tie(NULL);     int t = 1;     while (t--)     {         int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, sum = 0, diff = 0, maxN = 0, minN = 0, count = 0, temp = 0;         bool flag = false;         cin >> n;         cin >> m;         int xx;         cin >> xx;         int yy;         cin >> yy;         vector<vector<int>> vv(n, vector<int>(m));         for (int i = 0; i < n; i++)         {             for (int j = 0; j < m; j++)             {                 cin >> vv[i][j];             }         }         if (n == 1)         {             cout << -1 << endl;             continue;         }         int res = INT_MAX;         solve(vv, 0, xx, yy, res);         cout << res << endl;     }     return 0; } // Pay for a gift

#include <iostream> #include <vector> #include <queue> #include <algorithm> using namespace std; const int INF = 1e9; struct Node {     int value, dist;     Node(int v, int d) : value(v), dist(d) {} }; int main() {     int n, m;     cin >> n >> m;     vector<int> A(n);     vector<vector<int>> graph(n, vector<int>());     vector<vector<int>> dist(n, vector<int>(n, INF));     for (int i = 0; i < n; ++i) {         cin >> A[i];     }     for (int i = 0; i < m; ++i) {         int x, y;         cin >> x >> y;         --x; --y; // Convert to 0-based indexing         graph[x].push_back(y);         graph[y].push_back(x);         dist[x][y] = dist[y][x] = 1;     }     // Floyd-Warshall algorithm to calculate shortest distances     for (int k = 0; k < n; ++k) {         for (int i = 0; i < n; ++i) {             for (int j = 0; j < n; ++j) {                 dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);             }         }     }     priority_queue<Node, vector<Node>, greater<Node>> pq;     for (int i = 0; i < n; ++i) {         if (A[i] > 0) {             pq.push(Node(i, 0));         }     }     long long cost = 0;     while (!pq.empty()) {         Node node = pq.top();         pq.pop();         int u = node.value;         int d = node.dist;         if (A[u] > 0) {             cost += d * A[u];             A[u] = 0;             for (int v : graph[u]) {                 if (A[v] > 0) {                     pq.push(Node(v, d + 1));                 }             }         }     }     for (int i = 0; i < n; ++i) {         if (A[i] > 0) {             cout << -n << endl;             return 0;         }     }     cout << cost << endl;     return 0; } Tree=0✅ All pass✅ Share @whitehatcoding❤️ More Share More Solutions❤️👨‍💻

The Infosys online test for the SP role has been scheduled for Sunday,  21st January, 2024 from 2:30 PM to 5:30 PM.  The test window is stuck and the team is working from backend and it will be sorted in next 10 minutes  Don't refresh and close the window. Stay there. The test will be starting soon and the time will be extended.

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