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#include<bits/stdc++.h>    using namespace std;    int closestToTarget(vector<int> oxygenLevels, int target)  {      int current_AND_value = oxygenLevels[0];      int best_diff = abs(current_AND_value - target);        unordered_map<int, int> window_bit_counts;                       int num = oxygenLevels[0];          for (int j = 0; j < 32; j++)          {              if (num & (1 << j))              {                  window_bit_counts[j]++;              }          }              int left_idx = 0, right_idx = 0;      while (right_idx < oxygenLevels.size())      {          if (current_AND_value > target || left_idx > right_idx)          {              right_idx++;              if (right_idx >= oxygenLevels.size())              {                  break;              }              int num = oxygenLevels[right_idx];              for (int j = 0; j < 32; j++)              {                  if (num & (1 << j))                  {                      window_bit_counts[j]++;                  }              }                // cout<<"right_idx: "<<right_idx<<endl;          }          else          {                 int num = oxygenLevels[left_idx];              for (int j = 0; j < 32; j++)              {                  if (num & (1 << j))                  {                      window_bit_counts[j]--;                  }              }              left_idx++;                        }            current_AND_value = 0;          for (auto it = window_bit_counts.begin(); it != window_bit_counts.end(); it++)          {                 if (it->second == right_idx - left_idx + 1)              {                  current_AND_value += (1 << it->first);              }          }              if (left_idx <= right_idx)          {              best_diff = min(best_diff, abs(current_AND_value - target));          }                  }        return best_diff;  } Optimum oxygen code Salesforce ✅

int closestToTarget(vector oxygenLevels, int target) {     int n = oxygenLevels.size();     int minDiff = abs(oxygenLevels[0] - target); // Initialize the minimum difference with the first oxygen level     // Iterate through the oxygen levels to find the minimum difference     for (int i = 1; i < n; i++) {         int diff = abs(oxygenLevels[i] - target);         minDiff = min(minDiff, diff);     }     return minDiff; } Optimum oxygen Salesforce ✅

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#include<bits/stdc++.h> typedef double C; typedef complex<C> P; #define X real() #define Y imag() P a[205]; int n, x, y, k; bool check(int r){ int c=0; for(int i = 0; i < n; i++){ for(int j = i+1; j < n; j++){ P g = a[j] - a[i]; double d = abs(g); if(d > 2*r) continue; P mid = (a[i] + a[j]) * (1.0/2); double h = sqrt(r*1.0*r - d*d/4); P per = P(-g.Y,g.X) * (h/d); int c1 = 2, c2 = 2; for(int l = 0; l < n; l++){ if(l == i or l == j) continue; if(abs(a[l] - (mid-per)) <= r) c1++; if(abs(a[l] - (mid+per)) <= r) c2++; } c=max({c,c1,c2}); } } return c>=k; } int Solution::solve(vector<vector<int> > &A, int B ) { n = A.size(); for(int i = 0; i < n;i++) a[i] = {1.0*A[i][0], 1.0*A[i][1]}; k = B; int l = 1, h = 2000000000, ans; while(h >= l){ int m = (l+h)/2; if(check(m)) { ans = m; h = m-1; } else l = m+1; } return ans; } Red Zone Media. Net ✅

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Median Path (Google) ✅ #include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; template <class T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; struct Median {     priority_queue<int> max_heap;     priority_queue<int, vector<int>, greater<int>> min_heap;     void add(int x)     {         if (max_heap.empty() || x <= max_heap.top())             max_heap.push(x);         else             min_heap.push(x);         if (max_heap.size() > min_heap.size() + 1)         {             min_heap.push(max_heap.top());             max_heap.pop();         }         else if (min_heap.size() > max_heap.size())         {             max_heap.push(min_heap.top());             min_heap.pop();         }     }     int get()     {         return max_heap.top();     } }; void solve() {     int n;     cin >> n;     vector<int> a(n);     for (auto &x : a)         cin >> x;     vector<vector<int>> g(n);     for (int i = 0; i < n - 1; i++)     {         int u, v;         cin >> u >> v, u--, v--;         g[u].push_back(v);         g[v].push_back(u);     }     oset<pair<int, int>> s;     long ans = 0;     function<void(int, int, int)> dfs = [&](int u, int p, int d)     {         s.insert({a[u], u});         if (d & 1)             ans += s.find_by_order((d - 1) / 2)->first;         for (auto v : g[u])             if (v != p)                 dfs(v, u, d + 1);         s.erase({a[u], u});     };     dfs(0, -1, 1);     cout << ans << '\n'; } int main() {     ios_base::sync_with_stdio(false);     cin.tie(nullptr);     cout.tie(nullptr);     int t;     cin >> t;     while (t--)         solve(); }

def solve(i, num, n, k, arr, cost, dp, pf): if i == n: return 0 if dp[i][num] != -1: return dp[i][num] ans = float('inf') for j in range(1, n): sum_val = arr[i] + k * num sum_val += pf[j+1] - pf[i] ans = min(ans, sum_val + solve(i+1, num+1, n, k, arr, cost, dp, pf)) dp[i][num] = ans return ans def getMinCost(arr, cost, k): n = len(arr) for i in range(1, n): arr[i] += arr[i - 1] pf = [0] * (n + 1) for i in range(1, n): pf[i] = pf[i-1] + cost[i - 1] dp = [[-1] * (n + 1) for _ in range(n)] return solve(0, 0, n, k, arr, cost, dp, pf) De Shaw ✅