**Wolf of Wall Street 2 - Amazon Top Interview Questions**

### Problem Statement :

Given a list of integers prices representing the stock prices of a company in chronological order, return the maximum profit you could have made from buying and selling that stock once. You must buy before you can sell it. Note: You are not required to buy or sell the stock. Constraints n ≤ 100,000 where n is the length of prices Example 1 Input prices = [9, 11, 8, 5, 7, 10] Output 5 Explanation You can buy at 5 and sell at 10. Example 2 Input prices = [1, 2, 3, 4, 5, 6, 7, 8, 9] Output 8 Explanation You can buy at 1 and sell at 9. Example 3 Input prices = [9, 8, 7, 6, 5, 4, 3, 2, 1] Output 0 Explanation The stock's only going down so we don't buy at all.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& prices) {
int profit = 0;
int low = INT_MAX;
for (auto x : prices) {
low = min(low, x);
profit = max(profit, x - low);
}
return profit;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] prices) {
int len = prices.length;
if (len == 0 || len == 1)
return 0;
int min = prices[0], diff = 0;
for (int i = 1; i < len; i++) {
if (prices[i] - min > diff) {
diff = prices[i] - min;
}
if (prices[i] < min) {
min = prices[i];
}
}
return diff;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, prices):
if len(prices) < 2:
return 0
maxProfit = 0
min = prices[0]
for i in prices[1:]:
if i - min > maxProfit:
maxProfit = i - min
if i < min:
min = i
return maxProfit
```

## View More Similar Problems

## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

View Solution →## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

View Solution →## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

View Solution →## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

View Solution →## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =

View Solution →