Company: GOLDMANSACHS

There are 'N' coders standing in a line, where i denotes the ith position of a coder with a rating of Ri. All ratings are distinct.

You have to form a team of 3 from amongst them with the condition:
- Any three coders with positions (i, j, k) and ratings (Ri, Rj, Rk) can form a team when ("Ri < Rj < Rk" or "Ri > Rj > Rk") and (1 <= i < j < k <= N).

You have to find out how many such teams e xist (1 coder can be part of multiple teams).

Input:
N - 5
Ratings - 5 23 14

Output:
3
(5,2,1), (5,3,1), (2,3,4) Show More

A birthday party was attended by N number of kids, and each kid was given a unique ID ranging from 1 to N. As a return gift, there are T toys that must be distributed to the kids.

The host decided the best way to do this is by sitting the kids down in a circle (ordered by ascending ID), and then, starting with a random number D (between 1 and N), distribute one toy at a time to each sequentiall y numbered kid until all toys are distributed.

For example, if the host picks a random number D = 2, the gift distribution order would be: (2, 3, 4, 5, ... N-1, N, 1, 2, 3, 4, ...) until all T toys are distributed. The very last toy is damaged.

Your task is to determine which kid will receive the last (damaged) toy, so they can be informed and asked to exchange it at the shop.

Input:
- N (number of kids)
- T (number of toys)
- D (starting ID where distribution begins)

Output:
Print the ID of the last kid who will receive the damaged toy.

Constraints:
- 1 ≤ N ≤ 1000
- 1 ≤ T ≤ 1000
- 1 ≤ D ≤ N

Sample Input:
5
2
1

Sample Output:
2

Explanation:
There are N=5 kids and T=2 toys. Distribution starts at kid ID 1 (D=1). The distribution follows this pattern:
- Kid 1 receives the first toy.
- Kid 2 receives the second (and last) toy.

Thus, we must inform Kid 2 about the damaged toy, so we print `2` on a new line. Show More

Given a matrix of size m * n, where m denotes the number of rows (starting with index 0) and n denotes the number of columns (starting with index 0), the matrix will hold distinct integers as elements.

We need to find the distinct number of positional elements that are either the minimum or maximum in their corresponding row or column. If any row or any column has multiple minimum or maximum el ements, return -1.

Example:

Matrix of size 3 * 3:
1 3 5
8 2 7
4 9 6

The expected output is 7.

In this example, we identified the output as 7 based on the following:

- 1: Minimum in both its row and column.
- 4: Maximum in its row.
- 2: Minimum in both its row and column.
- 9: Maximum in both its row and column.
- 8: Maximum in both its row and column.
- 7: Maximum in its column.
- 6: Minimum in its row.

Thus, the total number of distinct elements that are either the minimum or maximum in their respective row or column is 7.

Input:
- m: Integer representing the number of rows.
- n: Integer representing the number of columns.
- matrix: m x n matrix of integers.

Output:
- r: Integer representing the total number of distinct elements which are either the minimum or maximum in their corresponding row or column.

Constraints:
- 0 < m, n < 100
- Elements in the matrix are positive integers. Show More

A traveler is traveling from the city of Zeta to Omega. He starts with X amount of money. Every day he spends some money and may also work on certain days to earn money. On some days, he may earn more than he spends, and on other days, he may spend more than he earns.

You are given an array of integers, which represents his net savings (earnings - expenses) on any day. Your task is to find out the minimum amount the traveler should begin with to ensure that he always has some money (> 0) at the end of any day.

Constraints:
- -200 <= a <= 200, where `a` are the elements of the array.
- 0 < i <= 100, where `i` is the array length.
- X >= 0

Example:
Input:
3 // Array length
4 // Array elements start
2
-3

Output:
0

Explanation:
Traveler saves $4 on the first day, $2 on the second day, and $-3 on the third day (expenses are more on the third day than earnings).

- End of the first day, he has X + $4.
- End of the second day, he has X + $(4 + 2).
- End of the third day, he has X + $(4 + 2 - 3).

So, effectively, the traveler can start with X = $0 to always ensure he has some money at the end of each day. Show More