Minimum points to be selected such that removal of line segments passing through them empties given array
Given a 2D array arr[][], where each row is of the form {start, end} representing the start and endpoints of each line segment on the X-axis. In one step, select a point on the X-axis and delete all the line segments passing through that point. The task is to find the minimum number of such points that need to be selected to delete all the line segments of the array.
Examples:
Input: arr[][]= { {9, 15}, {3, 8}, {1, 6}, {7, 12}, {5,10} }
Output : 2
Explanation:
Select the point arr[2][1](= (6, 0) on the X-axis and delete the second(= arr[1]), third(= arr[2]), and fifth(= arr[4]) line segments.
Select the point arr[3][1](= (12, 0)) on the X-axis and delete the first(=arr[0]) and the fourth(=arr[3]) line segments.
Therefore, the required count is 2.Input: arr[][]={ {1, 4}, {5, 7}, {9, 13} }
Output: 3
Approach: The problem can be solved using the Greedy technique. Follow the steps below to solve the problem:
- Initialize a variable, say cntSteps to count total number of steps required to delete all the line segments.
- Sort the array arr[][] based on the end points of the line segments.
- Initialize a variable, say Points = arr[0][1] to store the points of the X-axis.
- Traverse the array and check if the value of arr[i][0] greater than Points or not. If found to be true then increment the value cntSteps by 1 and update the value of Points = arr[i][1].
- Finally, print the value of cntSteps.
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Comparator function bool comp(vector< int > &x, vector< int > y) { return x[1] < y[1]; } // Function to count the minimum number of // steps required to delete all the segments int cntMinSteps(vector<vector< int > > &arr, int N) { // Stores count of steps required // to delete all the line segments int cntSteps = 1; // Sort the array based on end points // of line segments sort(arr.begin(), arr.end(), comp); // Stores point on X-axis int Points = arr[0][1]; // Traverse the array for ( int i = 0; i < N; i++) { // If arr[1][0] is // greater than Points if (arr[i][0] > Points) { // Update cntSteps cntSteps++; // Update Points Points = arr[i][1]; } } return cntSteps; } // Driver Code int main() { vector<vector< int > > arr = { { 9, 15 }, { 3, 8 }, { 1, 6 }, { 7, 12 }, { 5, 10 } }; int N = arr.size(); cout<< cntMinSteps(arr, N); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ // Function to sort by column public static void sortbyColumn( int arr[][], int col) { // Using built-in sort function Arrays.sort Arrays.sort(arr, new Comparator< int []>() { @Override // Compare values according to columns public int compare( final int [] entry1, final int [] entry2) { // To sort in descending order revert // the '>' Operator if (entry1[col] > entry2[col]) return 1 ; else return - 1 ; } }); // End of function call sort(). } // Function to count the minimum number of // steps required to delete all the segments static int cntMinSteps( int [][] arr, int N) { // Stores count of steps required // to delete all the line segments int cntSteps = 1 ; // Sort the array based on end points // of line segments sortbyColumn(arr, 1 ); // Stores point on X-axis int Points = arr[ 0 ][ 1 ]; // Traverse the array for ( int i = 0 ; i < N; i++) { // If arr[1][0] is // greater than Points if (arr[i][ 0 ] > Points) { // Update cntSteps cntSteps++; // Update Points Points = arr[i][ 1 ]; } } return cntSteps; } // Driver Code public static void main(String[] args) { int [][] arr = { { 9 , 15 }, { 3 , 8 }, { 1 , 6 }, { 7 , 12 }, { 5 , 10 } }; int N = arr.length; System.out.print(cntMinSteps(arr, N)); } } // This code is contributed by shikhasingrajput |
C#
// C# program to implement // the above approach class GFG { // A C# Function to sort the array by specified column. static public int [,] Sort_By_Column( int [,] array, int [,] sort_directive) { // number of rows inside array int array_rows = array.GetLength(0); // number of columns inside array int array_columns = array.Length/array_rows; // number of columns to be sorted int sort_directive_columns = sort_directive.GetLength(0); // for ( int i=0;i<array_rows-1;i++) { for ( int j=i+1;j<array_rows;j++) { for ( int c=0;c<sort_directive_columns;c++) { // // sort array values in descending sort order // if (sort_directive[c,1]==-1 && array[i,sort_directive[c,0]].CompareTo(array[j,sort_directive[c,0]])<0) { // // if values are in ascending sort order // swap values // for ( int d=0;d<array_columns;d++) { int h = array[j,d]; array[j,d]=array[i,d]; array[i,d]=h; } break ; } // // if values are in correct sort order break // else if (sort_directive[c,1]==-1 && array[i,sort_directive[c,0]].CompareTo(array[j,sort_directive[c,0]])>0) break ; // // sort array values in ascending sort order // else if (sort_directive[c,1]==1 && array[i,sort_directive[c,0]].CompareTo(array[j,sort_directive[c,0]])>0) { // // if values are in descending sort order // swap values // for ( int d=0;d<array_columns;d++) { int h = array[j,d]; array[j,d]=array[i,d]; array[i,d]=h; } break ; } // // if values are in correct sort order break // else if (sort_directive[c,1]==1 && array[i,sort_directive[c,0]].CompareTo(array[j,sort_directive[c,0]])<0) break ; // // if values are equal // select next sort directive // } } } return array; } // Function to count the minimum number of // steps required to delete all the segments static public int cntMinSteps( int [,]arr, int N) { // Stores count of steps required // to delete all the line segments int cntSteps = 1; // Sort the array based on end points // of line segments int [,] SORT_DIRECTIVE= new int [1,2]{ {1, 1} }; Sort_By_Column(arr, SORT_DIRECTIVE); // Stores point on X-axis int Points = arr[0,1]; // Traverse the array for ( int i = 0; i < N; i++) { // If arr[1][0] is // greater than Points if (arr[i,0] > Points) { // Update cntSteps cntSteps = cntSteps + 1; // Update Points Points = arr[i,1]; } } return cntSteps; } // Driver code static void Main() { int [,] arr = new int [,]{ { 9, 15 }, { 3, 8 }, { 1, 6 }, { 7, 12 }, { 5, 10 } }; int N = arr.GetLength(0); System.Console.WriteLine(cntMinSteps(arr, N)); } } // The code is contributed by Gautam goel (gautamgoel962) |
Python3
# Python3 program to implement # the above approach # Comparator function def comp(x, y): return x[ 1 ] < y[ 1 ] # Function to count the # minimum number of steps # required to delete all # the segments def cntMinSteps(arr, N): # Stores count of steps # required to delete all # the line segments cntSteps = 1 # Sort the array based # on end points of line # segments arr.sort(reverse = False ) # Stores point on X-axis Points = arr[ 0 ][ 1 ] # Traverse the array for i in range (N): # If arr[1][0] is # greater than Points if (arr[i][ 0 ] > Points): # Update cntSteps cntSteps + = 1 # Update Points Points = arr[i][ 1 ] return cntSteps # Driver Code if __name__ = = '__main__' : arr = [[ 9 , 15 ], [ 3 , 8 ], [ 1 , 6 ], [ 7 , 12 ], [ 5 , 10 ]] N = len (arr) print (cntMinSteps(arr, N)) # This code is contributed by bgangwar59 |
Javascript
<script> // JavaScript program to implement // the above approach // Comparator function function comp(x, y){ return x[1] - y[1] } // Function to count the // minimum number of steps // required to delete all // the segments function cntMinSteps(arr, N){ // Stores count of steps // required to delete all // the line segments let cntSteps = 1 // Sort the array based // on end points of line // segments arr.sort(comp) // Stores point on X-axis let Points = arr[0][1] // Traverse the array for (let i=0;i<N;i++){ // If arr[1][0] is // greater than Points if (arr[i][0] > Points){ // Update cntSteps cntSteps += 1 // Update Points Points = arr[i][1] } } return cntSteps } // Driver Code let arr = [[9, 15], [3, 8], [1, 6], [7, 12], [5, 10]] let N = arr.length document.write(cntMinSteps(arr, N)) // This code is contributed by shinjanpatra </script> |
2
Time Complexity: O(N * log(N)), for using inbuilt sort() function.
Auxiliary Space: O(1), constant extra space is required.