Question

Sc Python 1 Task 2 3 Consider a binary tree of N vertices 4 such that children of node K are 2* K + 1. Vertex 1 is the root Kand 2 of the tree and each node has an integer value associated with it. Such a tree may be represented as an array of N integers by writing down values from consecutive nodes For example, the tree below 8 Test might be represented as an array o A node is said to be at level X if the length of the shortest path 2 Python 1 Task 2 A node is said to be at level X if the length of the shortest path between that node and root is X - 1. So, the root is at level 1, the children of the root are at level 2, and so on. Your task is to find the smallest level number X such that the sum of all the nodes at level X is maximal. Write a function: def solution(A) that, given an array A consisting of N integers denoting a tree of N vertices, returns the level with the maximal sum of values of its vertices. If there is more than one of such levels, your function should return the smallest level. Note that A[K denotes value associated with node K+1 For example, given array A such that: A[2] = e A[3] 7 A[4]-8 the function should return 2. Explanation: Nodes at level 1: [1], sum equals-1 Nodes at level 2: [2, 3], sum equals 7 +0 Nodes at level 3: [4, 51, sum equals 7+ The sum of associated values of nodes at level 2 equals 7, and this is the maximum sum among all levels. Test O Assume that: . N is an integer within the range [1..1,000] . each element of 2 The sum of associated values of nodes at level 2 equals 7, and this is the maximum sum among all levels. Assume that: . N is an integer within the range [1..1,000]; . each element of array A is an integer within the range I-100,000..100,000]. In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment. 2

Answer 1

EXPLANATION IN CODE COMMENTS:

import queue
q = queue.Queue(20)

def solution(A):
#q to keep track of levels and nodes at that level
#First push the root node level
q.put(1)
#s to track the global maximum sum so far
s = -10000
#lvl to track the global maximum sum level
lvl = 0
#to iterate all levels
k=1
#exit condition variable for loop
flag = True
while flag:
#lsum keeps track of sum at each level
lsum = 0
#Gets no. of nodes at current level
siz = q.qsize()
#loop at all nodes at current level
while siz>0:
v = q.get()
#if that node does not exist in the array break the loop
if v > len(A):
#all nodes are traversed , exit all loops
flag = False
break
#add it to lsum
lsum+=A[v-1]
#push its child nodes into queue
q.put(2*v)
q.put(2*v+1)
siz-=1
#if globalSum < localSum then update globalSum and lvl of that sum
if s<lsum:
s=lsum
lvl = k
#increase the level
k+=1
#return level
return lvl
  
a = [-1,7,0,7,-8]
print(solution(a))

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In [32]: runfile('C:/Users/Rogue/Desktop/sumLevelB.py', wdir-'C:/Users/Rogue/ Desktop')