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from collections import deque, defaultdict def Solve(n, r, t, k, s, e): graph = defaultdict(list) for u, v in r: graph[u].append(v) graph[v].append(u) visited_t = set(t) queue_t = deque([(th, 0) for th in t]) while queue_t: current_t, distance_t = queue_t.popleft() if distance_t < k: for neighbor in graph[current_t]: if neighbor not in visited_t: visited_t.add(neighbor) queue_t.append((neighbor, distance_t + 1)) visited_s = set() queue_s = deque([(s, 0)]) while queue_s: current_s, distance_s = queue_s.popleft() if current_s in visited_s or current_s in visited_t: continue if current_s == e: return distance_s visited_s.add(current_s) for neighbor in graph[current_s]: if neighbor not in visited_s and neighbor not in visited_t: queue_s.append((neighbor, distance_s + 1)) return -1 Uber ✅

typedef long long ll; ll solution(int n, vector a1, vector a2) { vector left(n + 1, 0), right(n + 2, 0), prefixSum(n + 1, 0), maxL(n + 1, 0), maxM(n + 1, LLONG_MIN); for (int i = 1; i <= n; i++) { left[i] = i == 1 ? a1[0] : max(left[i - 1] + (ll)a1[i - 1], (ll)a1[i - 1]); } ll res = LLONG_MIN; for (int i = 1; i <= n; i++) { res = max(res, left[i]); } for (int i = n; i >= 1; i--) { right[i] = i == n ? (ll)a1[n - 1] : max(right[i + 1] + (ll)a1[i - 1], (ll)a1[i - 1]); } for (int i = 1; i <= n; i++) { prefixSum[i] = prefixSum[i - 1] + (ll)a2[i - 1]; } maxL[1] = left[0] - prefixSum[0]; maxM[1] = maxL[1]; for (int i = 2; i <= n; i++) { maxL[i] = left[i - 1] - prefixSum[i - 1]; maxM[i] = max(maxM[i - 1], maxL[i]); } for (int j = 1; j <= n; j++) { res = max(res, maxM[j] + prefixSum[j] + right[j + 1]); } return res; } Tichnas sir and bonds mam Uber ✅

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#include <iostream> #include <string> #include <unordered_map> #include <queue> using namespace std; string reorganizeString(string s) {     unordered_map<char, int> freq_map;     for (char c : s) {         freq_map[c]++;     }     priority_queue<pair<int, char>> max_heap;     for (auto &[ch, freq] : freq_map) {         max_heap.push({freq, ch});     }     string res;     while (max_heap.size() >= 2) {         auto [freq1, char1] = max_heap.top(); max_heap.pop();         auto [freq2, char2] = max_heap.top(); max_heap.pop();         res += char1;         res += char2;         if (--freq1 > 0) max_heap.push({freq1, char1});         if (--freq2 > 0) max_heap.push({freq2, char2});     }     if (!max_heap.empty()) {         auto [freq, ch] = max_heap.top();         if (freq > 1) return "";         res += ch;     }     return res; } int main() {     string input;     cout << "Enter a string: ";     cin >> input;     string result = reorganizeString(input);     if (result.empty()) {         cout << "The string cannot be reorganized to avoid adjacent repeating characters." << endl;     } else {         cout << "Reorganized string: " << result << endl;     }     return 0; } IBM ✅

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