Minimum unique sum

Minimum difference pairs in python

Sub set with LCM

Python

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()

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

This is the code


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All test cases passed

Python 3 Infosys

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import sys

def get_ans(N, K, A):



def main():

N = int(sys.stdin.readline().strip())

K = int(sys.stdin.readline().strip())

A = []

for _ in range(N):

A.append(int(sys.stdin.readline().strip()))

result = get_ans(N, K, A)

print(result)


if name == "main":

main()


Subset with LCM code in python

Tree lis Query

Unique subarray sum solution

Minimum unique sum