# Storage Optimization Solution Amazon OA 2021

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## Storage Optimization Solution

You are given a rectangular cake of size `h x w` and two arrays of integers `horizontalCuts` and `verticalCuts` where:

• `horizontalCuts[i]` is the distance from the top of the rectangular cake to the `ith` horizontal cut and similarly, and
• `verticalCuts[j]` is the distance from the left of the rectangular cake to the `jth` vertical cut.

Return the maximum area of a piece of cake after you cut at each horizontal and vertical position provided in the arrays `horizontalCuts` and `verticalCuts`. Since the answer can be a large number, return this modulo `109 + 7`.

Also See: Amazon OA Online Assessment 2021 Questions and Answers

Example 1:

```Input: h = 5, w = 4, horizontalCuts = [1,2,4], verticalCuts = [1,3]
Output: 4
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green piece of cake has the maximum area.
```

Example 2:

```Input: h = 5, w = 4, horizontalCuts = [3,1], verticalCuts = 
Output: 6
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green and yellow pieces of cake have the maximum area.
```

Example 3:

```Input: h = 5, w = 4, horizontalCuts = , verticalCuts = 
Output: 9
```

Constraints:

• `2 <= h, w <= 109`
• `1 <= horizontalCuts.length <= min(h - 1, 105)`
• `1 <= verticalCuts.length <= min(w - 1, 105)`
• `1 <= horizontalCuts[i] < h`
• `1 <= verticalCuts[i] < w`
• All the elements in `horizontalCuts` are distinct.
• All the elements in `verticalCuts` are distinct.

## Solution

Program C++

```class Solution {
public:
int maxArea(int h, int w, vector<int>& hc, vector<int>& vc) {
int mod = pow(10, 9) + 7;
sort(hc.begin(), hc.end());
sort(vc.begin(), vc.end());
int maxl = INT_MIN, maxw = INT_MIN;
long long res;
for(int i = 0; i < hc.size(); i++) {
if(i == 0) maxl = max(maxl, hc[i] - 0);
if(i == hc.size() - 1) maxl = max(maxl,h - hc[i]);
if(i > 0)  maxl = max(maxl, hc[i] - hc[i - 1]);
}
for(int i = 0; i < vc.size(); i++) {
if(i == 0) maxw = max(maxw, vc[i] - 0);
if(i == vc.size() - 1) maxw = max(maxw, w - vc[i]);
if(i > 0)  maxw = max(maxw, vc[i] - vc[i - 1]);
}
res = (long(maxl)* long(maxw)) % mod;
return res;
}
};```

Program Java

```class Solution {
public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {
Arrays.sort(horizontalCuts);
Arrays.sort(verticalCuts);

long maxHorizontalGap = horizontalCuts;
int index = 0;
for(index = 1; index < horizontalCuts.length; index++){
maxHorizontalGap = (long)Math.max(maxHorizontalGap, horizontalCuts[index] - horizontalCuts[index - 1]);
}
maxHorizontalGap = (long)Math.max(maxHorizontalGap, h - horizontalCuts[index - 1]);

long maxVerticalGap = verticalCuts;
index = 0;
for(index = 1; index < verticalCuts.length; index++){
maxVerticalGap = (long)Math.max(maxVerticalGap, verticalCuts[index] - verticalCuts[index - 1]);
}
maxVerticalGap = (long)Math.max(maxVerticalGap, w - verticalCuts[index - 1]);

return (int)((maxHorizontalGap * maxVerticalGap) % 1000000007 );
}
}```

Program Python

```class Solution:
def maxArea(self, h: int, w: int, horizontalCuts: List[int], verticalCuts: List[int]) -> int:
#SORT BOTH INPUT LISTS
horizontalCuts.sort()
verticalCuts.sort()

#INSERT '0' AT BEGINNING AND 'h' AT END OF 'horizontalCuts' LIST
horizontalCuts.insert(0,0)
horizontalCuts.append(h)

#CREATE A NEW LIST OF DIFFERENCES BETWEEN  ALL 'horizontalCuts' ELEMENTS
horDifs = []
for x in range(1,len(horizontalCuts)):
horDifs.append(horizontalCuts[x] - horizontalCuts[x-1])

#DO THE SAME WITH 'verticalCuts' LIST
verticalCuts.insert(0,0)
verticalCuts.append(w)
verDifs = []
for x in range(1,len(verticalCuts)):
verDifs.append(verticalCuts[x] - verticalCuts[x-1])

#TAKE THE MAX VALUES FROM BOTH LISTS OF DIFFERENCES AND MULTIPLY
#TAKE MODULO AT THE END
return max(horDifs) * max(verDifs) % (10**9 + 7)```

Also See: AMCAT Study Materials, Preparation Guide

Also See: Microsoft Online Assessment Questions and Solution

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