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Music melodies

This is the code Everyone write neatly All test cases passed Python 3 Infosys

def longest_equal_subarray(): n = int(input()) A = [int(input()) for _ in range(n)] A = [-1 if x == 0 else 1 for x in A] prefix_sum_map = {} prefix_sum = 0 max_length = 0 for i in range(n): prefix_sum += A[i] if prefix_sum == 0: max_length = i + 1 if prefix_sum in prefix_sum_map: max_length = max(max_length, i - prefix_sum_map[prefix_sum]) else: prefix_sum_map[prefix_sum] = i return max_length print(longest_equal_subarray()) Infosys Longest Subarray code

class TreeNode: def init(self, value=0, left=None, right=None): self.value = value self.left = left self.right = right def count_nodes(node, counts): if node is None: return if node.value in counts: counts[node.value] += 1 else: counts[node.value] = 1 count_nodes(node.left, counts) count_nodes(node.right, counts) def find_double_roots(root): counts = {} count_nodes(root, counts) double_roots = [value for value, count in counts.items() if count > 1] return double_roots def main(): # Example tree: # 1 # / \ # 2 3 # / \ # 2 4 root = TreeNode(1) root.left = TreeNode(2) root.right = TreeNode(3) root.left.left = TreeNode(2) root.left.right = TreeNode(4) result = find_double_roots(root) print("Nodes with double roots:", result) if name == "main": main()

Python

Sub set with LCM

Minimum difference pairs in python

Equilibrium point Infosys all test cases passed

Minimum unique sum

def max_cost_split(s): n = len(s) max_cost = 0 for i in range(1, n): a = s[:i] b = s[i:] cost_a = len(set(a)) cost_b = len(set(b)) max_cost = max(max_cost, cost_a + cost_b) return n - max_cost Infosys max cost split code in python

def max_sum_of_distinct_characters(S): n = len(S) left_chars = set() right_chars = set() left_count = [0] * n right_count = [0] * n for i in range(n): left_chars.add(S[i]) left_count[i] = len(left_chars) for i in range(n-1, -1, -1): right_chars.add(S[i]) right_count[i] = len(right_char) max_sum = 0 for i in range(n-1): max_sum = max(max_sum, left_count[i] + right_count[i+1]) return n- max_sum Split String code Python 3 All passed

import sys det solve(N, A) for i in range(N): if A[i]=0; A[1] ps=0 M-1 pm={0:-1) for i in range(N): ps+=A[i] if ps in pm: m=max(m,i-pm[ps]) else: pm[ps]=i return m def main(): Nint(sys.stdin.readline().strip()) A-[] for_ in range(N): A.append(int(sys.stdin.readline().strip())) result = solve(N, A) print(result) Largest Subarray with equal number Infosys

def solve(N, A): unique_sums = set() for start in range(N): current_sum = 0 for end in range(start, N): current_sum += A[end] unique_sums.add(current_sum) print(len(unique_sums))

Minimum substring ..

def count_distinct_strings(S): distinct_strings = set() for i in range(len(S) - 1): new_string = S[:i] + S[i+2:] distinct_strings.add(new_string) return len(distinct_strings) # Read input string S = input().strip() # Get the number of distinct strings that can be generated result = count_distinct_strings(S) print(result)

def minimum_unique_sum(A): N = len(A) A.sort() total = A[0] for i in range(1, N): if A[i] <= A[i-1]: A[i] = A[i-1] + 1 total += A[i] return total # Input format N = int(input()) A = [] for i in range(N): A.append(int(input())) # Output result = minimum_unique_sum(A) print(result)

def split_string_cost(S): # Length of the string S len_S = len(S) # To store the cost of the split parts max_cost = 0 # Set to keep track of distinct characters in the first part distinct_chars_A = set() # List to keep track of the cost for the second part from each split position cost_B = [0] * len_S # Set to keep track of distinct characters in the second part distinct_chars_B = set() # Calculate cost for second part from the end for i in range(len_S - 1, -1, -1): distinct_chars_B.add(S[i]) cost_B[i] = len(distinct_chars_B) # Calculate maximum sum of cost for parts A and B for i in range(len_S - 1): distinct_chars_A.add(S[i]) cost_A = len(distinct_chars_A) cost = cost_A + cost_B[i + 1] max_cost = max(max_cost, cost) # Calculate the result as |S| - X result = len_S - max_cost return result # Example usage S = "aaabbb" print(split_string_cost(S)) # Output: 3

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Accenture HackDiva Test Pattern - 2 Codes Round 1-2 Codes (90 min) Round 2-2 Codes (120 min) Round 3-2 Codes (120 min) Note: Each Round is Elimination Round 1 (DATE: 07 JULY 2024) Round 2 (DATE: 14 JULY 2024) Round 3 (DATE: 21 JULY 2024)