Discover the optimum variety of crossways lines

Provided an N horizontal line sectors are set up on the X-axis of a 2D aircraft. The start and end point of each line sector is given up an Nx2 matrix lines[ ][ ], the job is to discover the optimal number of crossways possible of any vertical line with the offered N line sectors.

Examples:

Input: N = 4, lines[][] = {{1, 3}, {2, 3}, {1, 2}, {4, 4}}
Output: 3
Description: A vertical line at X = 2 goes through {1, 3}, {2, 3}, {1, 2}, ie 3 of the offered horizontal lines.

Input: N = 3, lines[][] = {{1, 3}, {5, 6}, {3, 4}}
Output: 2
Description: A vertical line at X = 3 goes through 2 of the offered horizontal lines which are the optimum possible.

Technique: This can be resolved with the following concept:

Utilizing hashMap, we can identify the optimum worth amongst all the points and return it as a response.

Follow the actions pointed out listed below to execute the concept:

• Take the n, the variety of horizontal lines, and matrix lines[][]
• Produce a map to keep the variety of lines at each x-coordinate.
• Repeat through each line and upgrade the map appropriately.
• Initialize variables to track the optimum variety of crossways and an amount to track the variety of lines at each x-coordinate
• Repeat through the map and upgrade the amount and optimum appropriately.
• Return the optimum variety of crossways.

Below is the application of the above method:

// C++ code for the above method:

. #include < bits/stdc++. h>
> 
. utilizing namespace sexually transmitted disease;

.

.// Function to discover optimal crossways

. int maxIntersections( vector<< vector < int > > lines, int N) 
.

{
. 
.// Produce a map to keep the number 
.// of lines at each x-coordinate

<. map < int>, int > mp;


. 
.// Repeat through each line and 
.// upgrade the map appropriately

. for (int i < =0; i < lines.size();
i++) {
.
int a= lines[i][0];

.

int b =lines[i][1]; 
. mp[a]++;

.
mp[b + 1]--;

.

} 
. 
.// Initialize with minimum worth 
.
// of integer 
. int res= INT_MIN; 
.
int amount= 0; 
. 

.// Repeat through the map and upgrade 
.
// the amount and optimum appropriately 
.

for(
automobile it: mp) {
.
amount += it.second; 
. res= max( res, amount);

.
} 

.

.
// Return the optimum number 
.
// of crossways 
.

return res;

.} 
. 
./ >/ Motorist code 
.
int primary( )
. {
. 
. int n = 4; 
. vector < vector < int > > mat 
. = {{1, 3}, {2
,

3
}
, {1, << 2}, {4 <<, 4}}; 
. 
.// Function call 
. cout < < maxIntersections( mat, n) < < endl; 
.} 

Time Intricacy: O( N * log( N))
Auxiliary Area: O (N)