leetcode2975. The maximum area of a square field obtained by removing the fence

 topic

There is a large  (m - 1) x (n - 1) rectangular field whose two opposite corners are  (1, 1) and  (m, n) , and inside the field there are some horizontal and vertical fences, given by the arrays  hFences and ,  respectively vFences .

The horizontal fence is the coordinate  (hFences[i], 1) to  (hFences[i], n), and the vertical fence is the coordinate  (1, vFences[i]) to  (m, vFences[i]) .

Returns the area of ​​the largest  square  field that can be formed by  removing  some ( possibly no ) fences, or returns the area if a square field cannot be formed  .-1

Since the answer may be large, return the remainder of the result  109 + 7 pair  .

NOTE: The field is surrounded by two horizontal fences (coordinates  (1, 1) to  (1, n) and coordinates  (m, 1) to  (m, n) ) and two vertical fences (coordinates  (1, 1) to  (m, 1) and coordinates  (1, n) to  (m, n) ). These fences  cannot  be removed.

Example 1:

Input: m = 4, n = 3, hFences = [2,3], vFences = [2]
 Output: 4
 Explanation: Removing the horizontal fence at 2 and the vertical fence at 2 will result in a square field with area 4 .

Example 2:

Input: m = 6, n = 7, hFences = [2], vFences = [4]
 Output: -1
 Explanation: It can be shown that a square field cannot be formed by removing fences.

hint:

• 3 <= m, n <= 109
• 1 <= hFences.length, vFences.length <= 600
• 1 < hFences[i] < m
• 1 < vFences[i] < n
• hFences The elements in and  vFences are unique.

Problem: https://leetcode.cn/problems/maximum-squar100169 e-area-by-removing-fences-from-a-field/description/

[TOC]

Ideas

Traverse all the horizontal railings and vertical railings, calculate the difference between the horizontal railings and the difference between the longitudinal railings and store the two sets. The final answer is the square of the intersection of the two sets.

the complexity

time complexity:

O(m^2+n^2)

Space complexity:

O(m+n)

Code

class Solution:
    def maximizeSquareArea(self, m: int, n: int, hFences: List[int], vFences: List[int]) -> int:
        
        ans = -1
        hFences.append(m)
        hFences.insert(0,1)

        row = set()
        for i in range(len(hFences)):
            for j in range(0,i):
                row.add(abs( hFences[i] - hFences[j] ))
                    
        vFences.append(n)
        vFences.insert(0,1)

        col = set()
        for i in range(len(vFences)):
            for j in range(0,i):
                col.add(abs(vFences[i] - vFences[j]))
                
                if abs(vFences[i] - vFences[j]) in row :
                    ans = max(ans,pow(abs(vFences[i] - vFences[j]),2))

        for val in row:
            if val in col:
                ans = max(ans,pow(val,2))
       
        return ans % 1_000_000_007 if