Storage Optimization Amazon OA 2023 Solution

Storage Optimization Amazon OA 2023 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 i^th horizontal cut and similarly, and
• verticalCuts[j] is the distance from the left of the rectangular cake to the j^th 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.

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 = [1]

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 = [3], verticalCuts = [3]

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: Storage Optimization Amazon OA Solution in 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: Storage Optimization Amazon OA Solution in Java

class Solution {
    public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {
        Arrays.sort(horizontalCuts);
        Arrays.sort(verticalCuts);
        
        long maxHorizontalGap = horizontalCuts[0];
        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[0];
        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: Storage Optimization Amazon OA Solution in 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)

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