Problem of the day
Given array ‘PAR’ of size ‘N’, representing a binary tree, where ‘PAR[i]’ denotes the parent of node ‘i’. And a binary array ‘LOCK’ of ‘N’ integers, ‘LOCK[i] = 1’ means the ‘ith’ node is locked. Find out whether a target node ‘K’ can be locked or not.
A node will be locked only when some or all of the ancestors or the node itself is locked.
EXAMPLE:Input:
'N' = 5, ‘K’ = 3
‘ARR’ = [-1, 0, 0, 1, 2]
‘LOCK’ = [0, 0, 1, 0, 0]
Output: ‘1’
In the above tree in the simple path from node ‘4’ to root ‘1,’ the nodes encountered are [0, 1, 3], and no node from the set is locked. Hence node ‘3’ can be locked.
The first line will contain the integer 'T', the number of test cases. For each test case
The first line of each test case contains two integers ‘N’, and ‘K’.
The second line of each test case contains ‘N’ integers denoting the parent of node ‘i’.
The third line of each test case contains ‘N’ integers denoting elements of array ‘LOCK’.
Output format :
For each test case, print ‘1’ or ’0’, denoting whether the node can be locked.
Note :
You don't need to print anything. It has already been taken care of. Just implement the given function.
1 <= 'T' <= 10
1 <= 'N' <= 10^5
0 <= ‘K’ <= ‘N-1’
0 <= ‘PAR[i]’ <= ‘N-1’
0 <= ‘LOCK[i]’ <= 1
Time Limit: 1 sec
2
5 0
-1 0 3 0 3
1 1 1 0 0
4 1
3 2 -1 1
1 0 0 1
0
1
In the first test case,
In the above tree the target node ‘0’ is itself locked, so it cannot be locked.
Hence the answer is ‘false’.
In the second test case,
In the above tree in the simple path from node ‘1’ to root ‘2’ the nodes encountered are [1, 2], and no node from the set is locked.
Hence the answer is ‘true’.
2
4 2
-1 0 1 1
0 1 0 0
4 3
-1 0 1 2
1 0 0 0
0
0
There is only one path in between the root and the target node.
Approach:
Algorithm :
O(H), Where ‘H’ is the height of the tree.
We are only traversing up the tree in the simple path from target to node, which takes the height of tree time, the time complexity will be O(H).
O(1).
The ‘RES’ variable will only take ~1 space, Hence the space complexity will be O(1).