Tip 1 : work on communication skills
Tip 2 : Data Structures and Algorithums
Tip 1 : Include al teast two good project
Tip 2 : good to mention your skills
How to Design BookMyShow, Fandango (or similar movie ticketing application)
A movie ticket booking application such as BookMyShow facilitates users to browse through movies and live events, watch trailers, check seat availability and purchase a ticket. Let’s discuss how to design a similar online ticketing system
Two strings are said to be a permutation of each other when either of the string's characters can be rearranged so that it becomes identical to the other one.
Example:
str1= "sinrtg"
str2 = "string"
The character of the first string(str1) can be rearranged to form str2 and hence we can say that the given strings are a permutation of each other.
bool isAnagram(string s, string t) {
int a[256]={0};
for(int i=0;i {
a[s[i]]++;
}
for(int i=0;i {
a[t[i]]--;
}
for(int i=0;i<256;i++)
if(a[i]!=0)
return false;
return true;
}
1. There are no 2 adjacent elements having same value (as mentioned in the constraints).
2. Do not print anything, just return the index of the peak element (0 - indexed).
3. 'True'/'False' will be printed depending on whether your answer is correct or not.
Input: 'arr' = [1, 8, 1, 5, 3]
Output: 3
Explanation: There are two possible answers. Both 8 and 5 are peak elements, so the correct answers are their positions, 1 and 3.
int findPeakElement(vector& nums) {
int max=-2147483648;
int pos=0;
for(int i=0;imax)
{
max=nums[i];
pos=i;
}
}
return pos;
}
The diameter of a binary tree is the length of the longest path between any two end nodes in a tree.
The number of edges between two nodes represents the length of the path between them.
Input: Consider the given binary tree:
Output: 6
Explanation:
Nodes in the diameter are highlighted. The length of the diameter, i.e., the path length, is 6.
int height(TreeNode* root,int *x)
{
int lh=0;
int rh=0;
int ld=0;
int rd=0;
if(root==NULL)
return 0;
ld=height(root->left,&lh);
rd=height(root->right,&rh);
*x=max(lh,rh)+1;
return max(lh+rh+1,max(ld,rd));
}
int diameterOfBinaryTree(TreeNode* root) {
int x=0;
return height(root,&x)-1;
}
Make a fair coin from a biased coin
You are given a function foo() that represents a biased coin. When foo() is called, it returns 0 with 60% probability, and 1 with 40% probability. Write a new function that returns 0 and 1 with a 50% probability each. Your function should use only foo(), no other library method.
We know foo() returns 0 with 60% probability. How can we ensure that 0 and 1 are returned with a 50% probability?
The solution is similar to this post. If we can somehow get two cases with equal probability, then we are done. We call foo() two times. Both calls will return 0 with a 60% probability. So the two pairs (0, 1) and (1, 0) will be generated with equal probability from two calls of foo(). Let us see how.
(0, 1): The probability to get 0 followed by 1 from two calls of foo() = 0.6 * 0.4 = 0.24
(1, 0): The probability to get 1 followed by 0 from two calls of foo() = 0.4 * 0.6 = 0.24
So the two cases appear with equal probability. The idea is to return consider only the above two cases, return 0 in one case, return 1 in other case. For other cases [(0, 0) and (1, 1)], recur until you end up in any of the above two cases.
Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
What does ROLLBACK do in DBMS?