Check if N given lines can be intersected by K vertical lines
Given N horizontal lines represented by an array position[][] of size N, where position[i] represents the ith horizontal line which has x-coordinates from position[i][0] to position[i][1] and an integer K, which represents the maximum number of vertical lines that can be drawn, the task is to check if N given lines can be intersected by at most K vertical lines.
Examples:
Input: N = 4, K = 2, position[][] = [[1, 4], [6, 8], [7, 9], [10, 15]]
Output: NO
Explanation: In the given example, draw lines at [1, 7, 10]. The line drawn at 1 will intersect the first line, the line at position 7 will intersect both the second and third lines, and the line drawn at 10 will intersect the fourth line. Hence, a minimum of 3 rods is required. Hence, it is not possible
Input: N = 5, K = 3, position[][] = [[1, 6], [3, 5], [2, 4], [8, 12], [10, 24]]
Output : YES
Approach : The idea is to use greedy approach to solve this problem by sorting the positions[][] array and, with 1 line, intersect as many lines as possible and so on. Follow the steps below to solve the problem:
- Sort the array position[][] in ascending order.
- Initialize the variables ans as 1 to store the answer and r as position[0][1] to store the end point till a particular point other horizontal lines can be for intersection with the given considered vertical line.
- Iterate over the range [1, N] using the variable, say I, and perform the following steps:
- If position[i][0] is less than r, then set the value of r to the minimum of r or position[i][1].
- Otherwise, add the value of ans by 1 and set the value of r as position[i][1].
- If k is greater than equal to ans, then print “YES” and print “NO” otherwise.
Below is the implementation of the above approach:
C++14
#include <bits/stdc++.h>
using namespace std;
void findIfPossible( int n, int k,
vector<vector< int > > position)
{
sort(position.begin(), position.end());
int ans = 1;
int r = position[0][1];
for ( int i = 1; i < n; i++) {
if (position[i][0] <= r) {
r = min(r, position[i][1]);
}
else {
ans++;
r = position[i][1];
}
}
if (k >= ans)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
int main()
{
int n = 5;
int k = 2;
vector<vector< int > > position = {
{ 2, 5 }, { 4, 6 }, { 7, 16 }, { 9, 10 }, { 10, 17 }
};
findIfPossible(n, k, position);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void findIfPossible( int n, int k,
int [][] position)
{
Arrays.sort(position, Comparator.comparingDouble(o -> o[ 0 ]));
int ans = 1 ;
int r = position[ 0 ][ 1 ];
for ( int i = 1 ; i < n; i++) {
if (position[i][ 0 ] <= r) {
r = Math.min(r, position[i][ 1 ]);
}
else {
ans++;
r = position[i][ 1 ];
}
}
if (k >= ans)
System.out.println( "YES" );
else
System.out.println( "NO" );
}
public static void main(String[] args)
{
int n = 5 ;
int k = 2 ;
int [][] position = {
{ 2 , 5 }, { 4 , 6 }, { 7 , 16 }, { 9 , 10 }, { 10 , 17 }
};
findIfPossible(n, k, position);
}
}
|
Python3
def findIfPossible(n, k, position):
position.sort()
ans = 1
r = position[ 0 ][ 1 ]
for i in range ( 1 , n):
if (position[i][ 0 ] < = r):
r = min (r, position[i][ 1 ])
else :
ans + = 1
r = position[i][ 1 ]
if (k > = ans):
print ( "YES" )
else :
print ( "NO" )
n = 5
k = 2
position = [[ 2 , 5 ], [ 4 , 6 ], [ 7 , 16 ], [ 9 , 10 ], [ 10 , 17 ]]
findIfPossible(n, k, position)
|
C#
using System;
class GFG{
static void Sort( int [,] arr)
{
for ( int i = 0; i < arr.GetLength(0); i++)
{
for ( int j = arr.GetLength(1) - 1; j > 0; j--)
{
for ( int k = 0; k < j; k++)
{
if (arr[i, k] > arr[i, k + 1])
{
int myTemp = arr[i, k];
arr[i, k] = arr[i, k + 1];
arr[i, k + 1] = myTemp;
}
}
}
}
}
static void findIfPossible( int n, int k,
int [,] position)
{
Sort(position);
int ans = 1;
int r = position[0, 1];
for ( int i = 1; i < n; i++)
{
if (position[i, 0] <= r)
{
r = Math.Min(r, position[i, 1]);
}
else
{
ans++;
r = position[i, 1];
}
}
if (k >= ans)
Console.Write( "YES" );
else
Console.Write( "NO" );
}
public static void Main( string [] args)
{
int n = 5;
int k = 2;
int [,] position = { { 2, 5 }, { 4, 6 },
{ 7, 16 }, { 9, 10 },
{ 10, 17 } };
findIfPossible(n, k, position);
}
}
|
Javascript
<script>
function findIfPossible(n, k, position)
{
position = position.sort( function (a, b)
{
return a[0]-b[0]
});
var i;
var ans = 1;
var r = position[0][1];
for (i = 1; i < n; i++) {
if (position[i][0] <= r) {
r = Math.min(r, position[i][1]);
}
else {
ans++;
r = position[i][1];
}
}
if (k >= ans)
document.write( "YES" );
else
document.write( "NO" );
}
var n = 5;
var k = 2;
var position = [[2, 5],[4, 6], [7, 16],[9, 10],[10, 17]];
findIfPossible(n, k, position);
</script>
|
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
Last Updated :
27 Jul, 2021
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