Leetcode 2943: Maximize Area of Square Hole in Grid
You are given a grid with n + 2 horizontal bars and m + 2 vertical bars, creating 1x1 unit cells. You can remove some bars from the given arrays hBars (horizontal bars) and vBars (vertical bars). The remaining bars are fixed and cannot be removed. Your task is to determine the maximum area of a square-shaped hole that can be formed in the grid after removing some bars.
Problem
Approach
Steps
Complexity
Input: You are given the integers n and m, and two arrays of integers: hBars and vBars. The arrays represent the positions of horizontal and vertical bars in the grid.
Example: n = 2, m = 1, hBars = [2, 3], vBars = [2]
Constraints:
• 1 <= n <= 10^9
• 1 <= m <= 10^9
• 1 <= hBars.length <= 100
• 1 <= vBars.length <= 100
• 2 <= hBars[i] <= n + 1
• 2 <= vBars[i] <= m + 1
• All values in hBars are distinct.
• All values in vBars are distinct.
Output: Return an integer representing the maximum area of a square-shaped hole that can be formed in the grid after removing some bars.
Example: For n = 2, m = 1, hBars = [2, 3], vBars = [2], the output is 4.
Constraints:
Goal: The goal is to determine the maximum possible area of a square-shaped hole that can be created in the grid by removing some horizontal and vertical bars.
Steps:
• Sort the hBars and vBars arrays.
• Calculate the maximum distance between consecutive bars in both horizontal and vertical directions.
• The maximum square area is the square of the minimum distance between the bars in the horizontal and vertical directions.
Goal: The values of n and m can be very large (up to 10^9), but the lengths of hBars and vBars are relatively small (up to 100).
Steps:
• 1 <= n <= 10^9
• 1 <= m <= 10^9
• hBars.length <= 100
• vBars.length <= 100
Assumptions:
• The input arrays hBars and vBars contain distinct values and represent valid positions of the bars in the grid.
• Input: n = 2, m = 1, hBars = [2, 3], vBars = [2]
• Explanation: The maximum square-shaped hole can be obtained by removing horizontal bar 2 and vertical bar 2, creating a square of area 4.
• Input: n = 1, m = 1, hBars = [2], vBars = [2]
• Explanation: By removing horizontal bar 2 and vertical bar 2, a square-shaped hole of area 4 is formed.
• Input: n = 2, m = 3, hBars = [2, 3], vBars = [2, 4]
• Explanation: By removing horizontal bar 3 and vertical bar 4, a square-shaped hole with an area of 4 is created.
Approach: The approach involves sorting the bars and calculating the maximum square area by finding the largest gap between consecutive bars in both the horizontal and vertical directions.
Observations:
• The problem boils down to finding the largest gap between consecutive bars in both horizontal and vertical directions.
• Sorting the bars and calculating the maximum gap between consecutive bars will help us find the largest square that can be formed.
Steps:
• Sort the arrays hBars and vBars.
• Find the maximum gap between consecutive horizontal bars and the maximum gap between consecutive vertical bars.
• The largest square hole will have a side length equal to the minimum of these two maximum gaps.
• Return the area of the square, which is the side length squared.
Empty Inputs:
• The input arrays hBars and vBars should never be empty, as per the problem's constraints.
Large Inputs:
• The problem allows very large values for n and m (up to 10^9), but the lengths of hBars and vBars are small (up to 100), making the problem feasible for the given constraints.
Special Values:
• If hBars and vBars contain bars placed very close to each other, the maximum square hole area will be smaller.
Constraints:
• The arrays hBars and vBars are guaranteed to have distinct values within the given range.
int maximizeSquareHoleArea(int n, int m, vector<int>& hBars, vector<int>& vBars) {
vector<int>h;
vector<int>v;
sort(hBars.begin(),hBars.end());
sort(vBars.begin(),vBars.end());
for(auto i:hBars)
h.push_back(i);
for(auto i:vBars)
v.push_back(i);
int mv=1,mh=1;
int c=1;
for(int i=1;i<h.size();i++)
{
if(h[i]==h[i-1]+1)
c++;
else
c=1;
mh=max(mh,c);
}
c=1;
for(int i=1;i<v.size();i++)
{
if(v[i]==v[i-1]+1)
c++;
else
c=1;
mv=max(mv,c);
}
int x=min(mh+1,mv+1);
return x*x;
}
1 : Function Definition
int maximizeSquareHoleArea(int n, int m, vector<int>& hBars, vector<int>& vBars) {
Defines the function 'maximizeSquareHoleArea' that takes the dimensions of the grid and vectors of horizontal and vertical bars as input, and returns the area of the largest square that can be formed.
2 : Vector Initialization
vector<int>h;
Initializes an empty vector 'h' to store the horizontal bar positions.
3 : Vector Initialization
vector<int>v;
Initializes an empty vector 'v' to store the vertical bar positions.
4 : Sort Horizontal Bars
sort(hBars.begin(),hBars.end());
Sorts the 'hBars' vector in ascending order to process the bars in the correct order.
5 : Sort Vertical Bars
sort(vBars.begin(),vBars.end());
Sorts the 'vBars' vector in ascending order to process the bars in the correct order.
6 : Push Horizontal Bars to Vector
for(auto i:hBars)
Iterates over the sorted 'hBars' and pushes each element into the 'h' vector.
7 : Push Horizontal Bar to Vector
h.push_back(i);
Adds the current horizontal bar position to the 'h' vector.
8 : Push Vertical Bars to Vector
for(auto i:vBars)
Iterates over the sorted 'vBars' and pushes each element into the 'v' vector.
9 : Push Vertical Bar to Vector
v.push_back(i);
Adds the current vertical bar position to the 'v' vector.
10 : Initialize Variables
int mv=1,mh=1;
Initializes two variables 'mv' and 'mh' to store the maximum consecutive vertical and horizontal distances, respectively.
11 : Initialize Counter
int c=1;
Initializes a counter 'c' to track the number of consecutive bars (either horizontal or vertical).
12 : Loop Through Horizontal Bars
for(int i=1;i<h.size();i++)
Starts a loop to iterate through the 'h' vector (horizontal bars) starting from the second element.
13 : Consecutive Horizontal Bars Check
if(h[i]==h[i-1]+1)
Checks if the current horizontal bar is consecutive to the previous one (i.e., if the difference between the current and previous bar is 1).
14 : Increment Counter
c++;
If the current horizontal bar is consecutive to the previous one, increments the counter 'c'.
15 : Reset Counter
else
If the bars are not consecutive, resets the counter 'c' to 1.
16 : Update Maximum Horizontal Distance
c=1;
Resets the counter to 1, as the sequence of consecutive bars was broken.
17 : Update Maximum Horizontal Distance
mh=max(mh,c);
Updates the maximum horizontal distance 'mh' to the maximum of the current 'mh' and the counter 'c'.
18 : Reset Counter for Vertical Bars
c=1;
Resets the counter 'c' to 1 before checking the vertical bars.
19 : Loop Through Vertical Bars
for(int i=1;i<v.size();i++)
Starts a loop to iterate through the 'v' vector (vertical bars) starting from the second element.
20 : Consecutive Vertical Bars Check
if(v[i]==v[i-1]+1)
Checks if the current vertical bar is consecutive to the previous one (i.e., if the difference between the current and previous bar is 1).
21 : Increment Counter
c++;
If the current vertical bar is consecutive to the previous one, increments the counter 'c'.
22 : Reset Counter
else
If the bars are not consecutive, resets the counter 'c' to 1.
23 : Update Maximum Vertical Distance
c=1;
Resets the counter to 1, as the sequence of consecutive bars was broken.
24 : Update Maximum Vertical Distance
mv=max(mv,c);
Updates the maximum vertical distance 'mv' to the maximum of the current 'mv' and the counter 'c'.
25 : Calculate Square Size
int x=min(mh+1,mv+1);
Calculates the size of the largest square possible using the minimum of the maximum horizontal and vertical distances, incremented by 1.
26 : Return Result
return x*x;
Returns the area of the square, which is the square of the calculated size 'x'.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The time complexity is dominated by the sorting of the hBars and vBars arrays.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the input arrays.
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