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DDeloitte | COGNIZANT | IBM | CAPGEMINI

Deloitte | COGNIZANT | IBM | CAPGEMINI

@python_java_Coding · canal · Tecnologia · indexada desde 2026-07-19
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7posts em 30 dias
Deloitte | COGNIZANT | IBM | CAPGEMINI
import math def is_possible(K, houses, H): total_time = 0 for house in houses: total_time += math.ceil(house / K) return total_time <= H def find_min_K(n, H, houses): low = 1 high = max(houses) answer = high while low <= high: mid = (low + high) // 2 if is_possible(mid, houses, H): answer = mid high = mid - 1 else: low = mid + 1 return answer n, H = map(int, input().split()) houses = list(map(int, input().split())) print(find_min_K(n, H, houses)) q1 //quantiphi✅
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def max_points(points): n = len(points) if n == 0: return 0 if n == 1: return points[0] dp = [0] * n dp[0] = points[0] dp[1] = max(points[0], points[1]) for i in range(2, n): dp[i] = max(dp[i-1], points[i] + dp[i-2]) return dp[-1] n = int(input()) points = list(map(int, input().split())) print(max_points(points)) q2 //quantiphi ✅
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import heapq from collections import defaultdict def solve(n, m, k, treasures, bridges, queries): graph = defaultdict(list) for u, v, w in bridges: graph[u - 1].append((v - 1, w)) results = [] for start, end in queries: start -= 1 end -= 1 pq = [(-treasures[start], 0, start)] visited = {} while pq: neg_tres, time, node = heapq.heappop(pq) total_treasure = -neg_tres if node == end: results.append((total_treasure, time)) break if node in visited and visited[node] >= total_treasure: continue visited[node] = total_treasure for nei, wt in graph[node]: if nei not in visited or visited[nei] < total_treasure + treasures[nei]: heapq.heappush(pq, (-(total_treasure + treasures[nei]), time + wt, nei)) return results n, m, k = map(int, input().split()) treasures = list(map(int, input().split())) bridges = [tuple(map(int, input().split())) for _ in range(m)] queries = [tuple(map(int, input().split())) for _ in range(k)] result = solve(n, m, k, treasures, bridges, queries) for treasure, time in result: print(treasure, time) q3 //quantiphi ✅
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Deloitte | COGNIZANT | IBM | CAPGEMINI
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Capgemini offer letter for SAP role ✅ Contact : @MLCODER2
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Deloitte | COGNIZANT | IBM | CAPGEMINI
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MOSCHIP ON-CAMPUS EXAM ✅✅✅ Contact : @MLCODER2
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def ABpair(N): res=1 while True: if N+res==2*(N^res): return res res+=1 Arithmetic bitwise pairs✅ //GalaxE solutions✅
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def solve(N,Q,commands): slots=[0]*N for idx in commands: i=idx[0] slots[i-1]^=1 return sum(slots) Memory slot Flip✅
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def solve(N, M, grid, K): def is_star(r, c, grid): val = grid[r][c] if val == 0: return False for dr, dc in [(-1,0), (1,0), (0,-1), (0,1)]: nr, nc = r + dr, c + dc if not (0 <= nr < N and 0 <= nc < M): return False if grid[nr][nc] != val: return False return True def check_win(grid): for r in range(1, N-1): for c in range(1, M-1): if is_star(r, c, grid): return True return False def serialize(grid): return tuple(tuple(row) for row in grid) def dfs(state, turn): grid = [list(row) for row in state] if check_win(grid): return (0, 1, 0) if turn == 1 else (0, 0, 1) moves = [(r, c) for r in range(N) for c in range(M) if grid[r][c] == 0] if not moves: return (1, 0, 0) total = [0, 0, 0] for r, c in moves: for fruit in range(1, K+1): grid[r][c] = fruit res = dfs(serialize(grid), 1 - turn) for i in range(3): total[i] += res[i] grid[r][c] = 0 # backtrack return tuple(total) result = dfs(serialize(grid), 0) print(*result) The game of grid✅✅
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def solve(N, K, Count, A, Q, XY): results = [] for x, y in XY: p = A[x - 1] c = 0 for i in range(N): if A[i] > p: c += Count[i] e = (c // K) + 1 z = c + 1 if e <= y <= z + Count[x - 1] - 1: results.append("YES") else: results.append("NO") return results Favourite Candy✅
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Wipro 9AM✅✅
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IBM EXAM ✅✅ Contact : @MLCODER2
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