×
Detail Form
We will send your result on your email id and phone no. please fill detail
1.
Your Task:
You don't need to read input or print anything. Complete the function AllParenthesis() which takes N as input parameter and returns the list of balanced parenthesis.
Expected Time Complexity: O(2N).
Expected Auxiliary Space: O(2*N*X), X = Number of valid Parenthesis.
Constraints:
1 ≤ N ≤ 12
Given an integer N representing the number of pairs of parentheses, the task is to generate all combinations of well-formed(balanced) parentheses.
Example 1:
Input:
N = 3
Output:
((()))
(()())
(())()
()(())
()()()
Example 2:
N = 3
Output:
((()))
(()())
(())()
()(())
()()()
Input:
N = 1
Output:
()
N = 1
Output:
()
Your Task:
You don't need to read input or print anything. Complete the function AllParenthesis() which takes N as input parameter and returns the list of balanced parenthesis.
Expected Time Complexity: O(2N).
Expected Auxiliary Space: O(2*N*X), X = Number of valid Parenthesis.
Constraints:
1 ≤ N ≤ 12
2.
Given a number N, generate bit patterns from 0 to 2^N-1 such that successive patterns differ by one bit.
Example 1:
Example 2:
You don't need to read input or print anything. Your task is to complete the function graycode() which takes an integer N as input and returns a la list of patterns.
Expected Time Complexity: O(2n)
Expected Auxiliary Space: O(2n)
Constraints :
1<=N<=16
Example 1:
Input:
N = 2
Output:
00 01 11 10
Explanation:
00 and 01 differ by one bit.
01 and 11 differ by one bit.
11 and 10 also differ by one bit.
N = 2
Output:
00 01 11 10
Explanation:
00 and 01 differ by one bit.
01 and 11 differ by one bit.
11 and 10 also differ by one bit.
Example 2:
Input:
N=3
Output:
000 001 011 010 110 111 101 100
Explanation:
000 and 001 differ by one bit.
001 and 011 differ by one bit.
011 and 010 differ by one bit.
Similarly, every successive pattern
differs by one bit.
Your task:N=3
Output:
000 001 011 010 110 111 101 100
Explanation:
000 and 001 differ by one bit.
001 and 011 differ by one bit.
011 and 010 differ by one bit.
Similarly, every successive pattern
differs by one bit.
You don't need to read input or print anything. Your task is to complete the function graycode() which takes an integer N as input and returns a la list of patterns.
Expected Time Complexity: O(2n)
Expected Auxiliary Space: O(2n)
Constraints :
1<=N<=16
3.
Given a String S, Find all possible Palindromic partitions of the given String.
Example 1:
Input:
S = "madam"
Output:
m a d a m
m ada m
madam
S = "madam"
Output:
m a d a m
m ada m
madam
Your Task:
You don't need to read input or print anything. Your task is to complete the function allPalindromicPerms() which takes a String S as input parameter and returns a list of lists denoting all the possible palindromic partitions.
Expected Time Complexity: O(N*2N)
Expected Auxiliary Space: O(N2), where N is the length of the String
Constraints:
1 <= |S| <= 20
4.
Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle in a fixed direction.
The task is to choose the safe place in the circle so that when you perform these operations starting from 1st place in the circle, you are the last one remaining and survive.
Example 1:
Input:
n = 3 k = 2
Output: 3
Explanation: There are 3 persons so
skipping 1 person i.e 1st person 2nd
person will be killed. Thus the safe
position is 3.
n = 3 k = 2
Output: 3
Explanation: There are 3 persons so
skipping 1 person i.e 1st person 2nd
person will be killed. Thus the safe
position is 3.
Example 2:
Input:
n = 5 k = 3
Output: 4
Explanation: There are 5 persons so
skipping 2 person i.e 3rd person will
be killed. Thus the safe position is 4.
n = 5 k = 3
Output: 4
Explanation: There are 5 persons so
skipping 2 person i.e 3rd person will
be killed. Thus the safe position is 4.
Your Task:
You don't need to read input or print anything. You are required to complete the function josephus () that takes two parameters n and k and returns an integer denoting safe position.
Expected Time Complexity: O(N).
Expected Auxiliary Space: O(N).
Constraints:
1 <= k, n <= 20
5.
Given a natural number N, the task is to find the sum of the divisors of all the divisors of N.
Example 1:
Input:
N = 54
Output:
232
Explanation:
Divisors of 54 = 1, 2, 3, 6, 9, 18, 27, 54.
Sum of divisors of 1, 2, 3, 6, 9, 18, 27, 54
are 1, 3, 4, 12, 13, 39, 40, 120 respectively.
Sum of divisors of all the divisors of 54
= 1 + 3 + 4 + 12 + 13 + 39 + 40 + 120 = 232.
N = 54
Output:
232
Explanation:
Divisors of 54 = 1, 2, 3, 6, 9, 18, 27, 54.
Sum of divisors of 1, 2, 3, 6, 9, 18, 27, 54
are 1, 3, 4, 12, 13, 39, 40, 120 respectively.
Sum of divisors of all the divisors of 54
= 1 + 3 + 4 + 12 + 13 + 39 + 40 + 120 = 232.
Example 2:
Input:
N = 10
Output:
28
Explanation:
Divisors of 10 are 1, 2, 5, 10
Sums of divisors of all the divisors are
1, 3, 6, 18.
Overall sum = 1 + 3 + 6 + 18 = 28
N = 10
Output:
28
Explanation:
Divisors of 10 are 1, 2, 5, 10
Sums of divisors of all the divisors are
1, 3, 6, 18.
Overall sum = 1 + 3 + 6 + 18 = 28
Your Task:
You don't need to read input or print anything. Your task is to complete the function sumOfDivisors() which takes an integer N as an input parameter and return the sum of the divisors of all the divisors of N.
Expected Time Complexity: O(NloglogN)
Expected Auxiliary Space: O(N)
Constraints:
1 <= N <= 104