# Max Profit Amazon OA Solution

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## Max Profit Amazon OA Solution

An Amazon seller is deciding which of their products to invest in for the next quarter to maximize their profits. They have each of their products listed as segments of a circle. Due to varying market conditions, the products do not sell consistently. The seller wants to achieve maximum profit using limited resources for investment. The product list is segmented into a number of equal segments, and a projected profit is calculated for each segment. The projected profit is the cost to invest versus the sale price of the product. The seller has chosen to invest in a number of contiguous segments along with those opposite. Determine the maximum profit the seller can achieve using this approach.

For example, the product list Is divided into n = 6 sections and will select k = 2 contiguous sections and those opposite to invest in. The profit estimates are `profit = [1, 5, 1, 3, 7, -3]` respectively. The diagrams below show the possible choices with `profits` at the 9 o’clock position and filling counterclockwise.

The profit levels, from left to right, are `1+5+7+3=16``5+1+7+-3=10`, and `1+3+-3+1=2`. The maximum profit is 16.

### Input

• `k`: an integer that denotes half of the needed number of products within the list
• `profit`: an array of integers that denote the profit from investing in each of the products

### Output

the maximum profit possible

### Examples

#### Example 1:

Input:

1k = 2
2profit = [1, 5, 1, 3, 7 -3]

Output`16`

Explanation:

The profit levels, from left to right, are `1+5+7+3=16``5+1+7+-3=10`, and `1+3+-3+1=2`. The maximum profit is 16.

#### Example 2:

Input:

1k = 1
2profit = 3 -5

Output`-2`

Explanation:

Here `k=1` and `n=2`. The seller will choose `2*k=2` products. In this case, it is the entire list, so overall profit is `3+-5=-2`.

Constraints

• `1 <= k <= n/2`
• `2 <= n <= 10^5`
• n is even
• `0 <= |profit[i]| <= 10^9`

## SOLUTION

Program: Max Profit Amazon OA Solution in Python

Python Approach – the question gives you a lot of unnecessary information. The main thing we need to be concerned about is the edge case of getting the array to be “circular”. There are some cases my code might not pass (i.e if k is larger than 2 times the size of the array), but this is the general idea of it.
Time: O(n^2) – basically doing a double for loop for a window of the array
Space: O(1)

```def contprofit(arr, k):
profits = 0
for i in range(len(arr) - k):
window = sum(arr[i : i + k])
for j in range(i + k, len(arr)):
nextwindow = sum(arr[j : j + k])
if j + k > len(arr) - 1:
diff = (j + k) - len(arr)
nextwindow += sum(arr[0:diff])
profits = max(profits, window + nextwindow)
return profits```

Program: Max Profit Amazon OA Solution in C++

```#include<bits/stdc++.h>
using namespace std;

void solve() {

int n, k;
cin >> n >> k;
vector<int>a(n);
for (int i = 0; i < n; i++) cin >> a[i];
int i = 0, j = 0, sum = 0, p = INT_MIN;
while (j < n) {
sum += a[j] + a[(n / 2 +  j) % n];
while (j - i + 1 == k) {
p = max(p, sum);
sum -= (a[i] + a[(i + n / 2 ) % n]);
i++;
}
j++;
}
cout << p << endl;
}

int main() {

#ifndef ONLINE_JUDGE
//for getting input from input.txt
freopen("input1.txt", "r", stdin);
// for writing output to output.txt
freopen("output1.txt", "w", stdout);
#endif

int t ;
cin >> t;

while (t--) {

solve();
}
}```

Program: Max Profit Amazon OA Solution in C++

Time Complexity – `O(n)`
Space Complexity
– `O(n)`

Let me know your thoughts on this

```// efficient harvest
int getQuerySum(vector<int> &prefixSum, int a, int b) // function to get interval sum in constant time
{
if(a <= b)
return prefixSum[b] - (a >= 1 ? prefixSum[a-1] : 0);

return getQuerySum(prefixSum, a, prefixSum.size()-1) + getQuerySum(prefixSum, 0, b);
}

int findMaxProfit(vector<int> profit, int n, int k)
{
vector<int> prefixSum;

int sum = 0, maxProfit = INT_MIN, l = (n >> 1);

for(auto &x : profit)
{
sum += x;
prefixSum.push_back(sum);
}

for(int i = 0;i < (n >> 1); i++)
{
int temp = getQuerySum(prefixSum, i, i + k - 1) + getQuerySum(prefixSum, (i + l) % n, (i + l + k - 1) % n);

maxProfit = max(maxProfit, temp);
}

return maxProfit;
}

int main()
{
cout<<findMaxProfit({1, 5, 1, 3, 7, -3}, 6, 2)<<"\n"; // 16
cout<<findMaxProfit({-3, -6, 3, 6}, 4, 1)<<"\n"; // 0
cout<<findMaxProfit({3, -5}, 2, 1)<<"\n"; // -2
return 0;
}```

Program: Max Profit Amazon OA Solution in JavaScript

TC: O(n + k)

```(function main() {
var k = 2;
var profit = [7, 3, 1, 5, 1, -3];
console.log(maxProfit(k, profit));

k = 1;
profit = [6, 3, -6, -3];
console.log(maxProfit(k, profit));

k = 1;
profit = [-5, 3];
console.log(maxProfit(k, profit));
}());

function maxProfit(k, profit) {
var length = profit.length;
var half_length = length / 2;
var max = 0;
//get the profit for index 0;
for (var i = 0; i < k; i ++) {
max += profit[i];
max += profit[(i+half_length) % length];
}

var current_profit = max;
for (var i = 1; i < length/2; i ++) {
current_profit -= profit[i - 1];
current_profit -= profit[(i - 1 + half_length) % length]

current_profit += profit[i + k - 1]
current_profit += profit[(i + k - 1 + half_length) % length]
max = Math.max(max, current_profit);
}

return max;
}```

Program: Max Profit Amazon OA Solution in Golang
Solution 1: O(N*K)
The key is how to get the oppsite index:
for circle 6: 0->3; 1->4;2->5;3->0; 5->2;

opIndex:=(curIndex+len(profits)/2)%len(profits).

```       size, res := len(profits), math.MinInt32
for i := 0; i < size/2; i++ {   // try each!
var sum int
for j := i; j < i+k; j++ {sum += profits[j] + profits[(j+size/2)%size]}
if sum > res {	res = sum}
}```

Solution 2: O(N) with preSum

```	size := len(profits)
preSum := make([]int, 2*size+1)
for i := 1; i < len(preSum); i++ {preSum[i] = preSum[i-1] + profits[i%size]}
res := math.MinInt32
for i := 0; i < size/2; i++ {
thisSum := preSum[i+k] - preSum[i]
oppSum := preSum[i+size/2+k] - preSum[i+size/2]
if thisSum+oppSum > res { res = thisSum + oppSum}
}```

Program: Max Profit Amazon OA Solution in Java

```public class EffecientHarvest {

public static void main(String[] args) {
// TODO Auto-generated method stub
int[] table = {3,-5};
int k = 1;
int n = table.length;

System.out.println(effecient(table, k, n));

}

private static int effecient(int[] table, int k, int n) {
// TODO Auto-generated method stub
n=n/2;
if(k>n) {
return 0;
}

int i=0;
int max = Integer.MIN_VALUE;
while(i<n) {
int sum=0;
for(int j=i; j<i+k; j++) {
int opp;
if(j<n) {
opp = table[j+n];
}
else {
opp = table[j+n-(n*2)];
}
sum+=opp+table[j];
}
if(sum>max) {
max=sum;
}
i++;
}
return max;
}

}```