Given an array arr[] consisting of N pairs [L, R], where L and R denotes the start and end indices of a segment, the task is to find the minimum number of segments that must be deleted from the array such that the remaining array contains at least one segment which intersects with all other segments present in the array.
Examples:
Input: arr[] = {{1, 2}, {5, 6}, {6, 7}, {7, 10}, {8, 9}}
Output: 2
Explanation: Delete the segments {1, 2} and {5, 6}. Therefore, the remaining array contains the segment {7, 10} which intersects with all other segments.Input: a[] = {{1, 2}, {2, 3}, {1, 5}, {4, 5}}
Output: 0
Explanation: The segment {1, 5} already intersects with all other remaining segments. Hence, no need to delete any segment.
Approach: The maximum possible answer is (N – 1), since after deleting (N – 1) segments from arr[], only one segment will be left. This segment intersects with itself. To achieve the minimum answer, the idea is to iterate through all the segments, and for each segment, check the number of segments which do not intersect with it.
Two segments [f1, s1] and [f2, s2] intersect only when max(f1, f2) ? min(s1, s2).
Therefore, if [f1, s1] does not intersect with [f2, s2], then there are only two possibilities:
- s1 < f2 i.e segment 1 ends before the start of segment 2
- f1 > s2 i.e segment 1 starts after the end of segment 2.
Follow the steps below to solve the problem:
- Traverse the array arr[] and store the starting point and ending point of each segment in startPoints[], and endPoints[] respectively.
- Sort both the arrays, startPoints[] and endPoints[] in increasing order.
- Initialize ans as (N – 1) to store the number of minimum deletions required.
- Again traverse the array, arr[] and for each segment:
- Store the number of segments satisfying the first and the second condition of non-intersection in leftDelete and rightDelete respectively.
- If leftDelete + rightDelete is less than ans, then set ans to leftDelete + rightDelete.
- After the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the minimum number // of segments required to be deleted void minSegments(pair< int , int > segments[],
int n)
{ // Stores the start and end points
int startPoints[n], endPoints[n];
// Traverse segments and fill the
// startPoints and endPoints
for ( int i = 0; i < n; i++) {
startPoints[i] = segments[i].first;
endPoints[i] = segments[i].second;
}
// Sort the startPoints
sort(startPoints, startPoints + n);
// Sort the startPoints
sort(endPoints, endPoints + n);
// Store the minimum number of
// deletions required and
// initialize with (N - 1)
int ans = n - 1;
// Traverse the array segments[]
for ( int i = 0; i < n; i++) {
// Store the starting point
int f = segments[i].first;
// Store the ending point
int s = segments[i].second;
// Store the number of segments
// satisfying the first condition
// of non-intersection
int leftDelete
= lower_bound(endPoints,
endPoints + n, f)
- endPoints;
// Store the number of segments
// satisfying the second condition
// of non-intersection
int rightDelete = max(
0, n - ( int )(upper_bound(startPoints,
startPoints + n, s)
- startPoints));
// Update answer
ans = min(ans,
leftDelete
+ rightDelete);
}
// Print the answer
cout << ans;
} // Driver Code int main()
{ pair< int , int > arr[] = {
{ 1, 2 }, { 5, 6 },
{ 6, 7 }, { 7, 10 }, { 8, 9 }
};
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
minSegments(arr, N);
return 0;
} |
// Java program for the above approach import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Pair class static class Pair
{ int first;
int second;
Pair( int first, int second)
{
this .first = first;
this .second = second;
}
} public static int lower_bound( int arr[], int key)
{ int l = - 1 , r = arr.length;
while (l + 1 < r)
{
int m = (l + r) >>> 1 ;
if (arr[m] >= key)
r = m;
else
l = m;
}
return r;
} public static int upper_bound( int arr[], int key)
{ int l = - 1 , r = arr.length;
while (l + 1 < r)
{
int m = (l + r) >>> 1 ;
if (arr[m] <= key)
l = m;
else
r = m;
}
return l + 1 ;
} // Function to find the minimum number // of segments required to be deleted static void minSegments(Pair segments[], int n)
{ // Stores the start and end points
int startPoints[] = new int [n];
int endPoints[] = new int [n];
// Traverse segments and fill the
// startPoints and endPoints
for ( int i = 0 ; i < n; i++)
{
startPoints[i] = segments[i].first;
endPoints[i] = segments[i].second;
}
// Sort the startPoints
Arrays.sort(startPoints);
// Sort the startPoints
Arrays.sort(endPoints);
// Store the minimum number of
// deletions required and
// initialize with (N - 1)
int ans = n - 1 ;
// Traverse the array segments[]
for ( int i = 0 ; i < n; i++)
{
// Store the starting point
int f = segments[i].first;
// Store the ending point
int s = segments[i].second;
// Store the number of segments
// satisfying the first condition
// of non-intersection
int leftDelete = lower_bound(endPoints, f);
// Store the number of segments
// satisfying the second condition
// of non-intersection
int rightDelete = Math.max(
0 , n - ( int )(upper_bound(startPoints, s)));
// Update answer
ans = Math.min(ans, leftDelete + rightDelete);
}
// Print the answer
System.out.println(ans);
} // Driver Code public static void main(String[] args)
{ Pair arr[] = { new Pair( 1 , 2 ), new Pair( 5 , 6 ),
new Pair( 6 , 7 ), new Pair( 7 , 10 ),
new Pair( 8 , 9 ) };
int N = arr.length;
// Function Call
minSegments(arr, N);
} } // This code is contributed by Kingash |
# Python3 program for the above approach from bisect import bisect_left,bisect_right
# Function to find the minimum number # of segments required to be deleted def minSegments(segments, n):
# Stores the start and end points
startPoints = [ 0 for i in range (n)]
endPoints = [ 0 for i in range (n)]
# Traverse segments and fill the
# startPoints and endPoints
for i in range (n):
startPoints[i] = segments[i][ 0 ]
endPoints[i] = segments[i][ 1 ]
# Sort the startPoints
startPoints.sort(reverse = False )
# Sort the startPoints
endPoints.sort(reverse = False )
# Store the minimum number of
# deletions required and
# initialize with (N - 1)
ans = n - 1
# Traverse the array segments[]
for i in range (n):
# Store the starting point
f = segments[i][ 0 ]
# Store the ending point
s = segments[i][ 1 ]
# Store the number of segments
# satisfying the first condition
# of non-intersection
leftDelete = bisect_left(endPoints, f)
# Store the number of segments
# satisfying the second condition
# of non-intersection
rightDelete = max ( 0 , n - bisect_right(startPoints,s))
# Update answer
ans = min (ans, leftDelete + rightDelete)
# Print the answer
print (ans)
# Driver Code if __name__ = = '__main__' :
arr = [[ 1 , 2 ],[ 5 , 6 ], [ 6 , 7 ],[ 7 , 10 ],[ 8 , 9 ]]
N = len (arr)
# Function Call
minSegments(arr, N)
# This code is contributed by ipg2016107.
|
// C# program for the above approach using System;
using System.Collections.Generic;
public class GFG {
// Pair class
class Pair {
public int first;
public int second;
public Pair( int first, int second)
{
this .first = first;
this .second = second;
}
}
public static int lower_bound( int [] arr, int key)
{
int l = -1, r = arr.Length;
while (l + 1 < r) {
int m = (l + r) >> 1;
if (arr[m] >= key)
r = m;
else
l = m;
}
return r;
}
public static int upper_bound( int [] arr, int key)
{
int l = -1, r = arr.Length;
while (l + 1 < r) {
int m = (l + r) >> 1;
if (arr[m] <= key)
l = m;
else
r = m;
}
return l + 1;
}
static void minSegments(Pair[] segments, int n)
{
// Stores the start and end points
int [] startPoints = new int [n];
int [] endPoints = new int [n];
// Traverse segments and fill the
// startPoints and endPoints
for ( int i = 0; i < n; i++) {
startPoints[i] = segments[i].first;
endPoints[i] = segments[i].second;
}
// Sort the startPoints
Array.Sort(startPoints);
// Sort the startPoints
Array.Sort(endPoints);
// Store the minimum number of
// deletions required and
// initialize with (N - 1)
int ans = n - 1;
// Traverse the array segments[]
for ( int i = 0; i < n; i++) {
// Store the starting point
int f = segments[i].first;
// Store the ending point
int s = segments[i].second;
// Store the number of segments
// satisfying the first condition
// of non-intersection
int leftDelete = lower_bound(endPoints, f);
// Store the number of segments
// satisfying the second condition
// of non-intersection
int rightDelete = Math.Max(
0, n - ( int )(upper_bound(startPoints, s)));
// Update answer
ans = Math.Min(ans, leftDelete + rightDelete);
}
// Print the answer
Console.WriteLine(ans);
}
static public void Main()
{
// Code
Pair[] arr = { new Pair(1, 2), new Pair(5, 6),
new Pair(6, 7), new Pair(7, 10),
new Pair(8, 9) };
int N = arr.Length;
minSegments(arr, N);
}
} // This code is contributed by lokesh (lokeshmvs21). |
// JS program for the above approach function lowerBound(arr, key) {
let l = -1, r = arr.length;
while (l + 1 < r) {
let m = (l + r) >>> 1;
if (arr[m] >= key)
r = m;
else
l = m;
}
return r;
} function upperBound(arr, key) {
let l = -1, r = arr.length;
while (l + 1 < r) {
let m = (l + r) >>> 1;
if (arr[m] <= key)
l = m;
else
r = m;
}
return l + 1;
} // Function to find the minimum number // of segments required to be deleted function minSegments(segments, n) {
// Stores the start and end points
let startPoints = [];
let endPoints = [];
// Traverse segments and fill the
// startPoints and endPoints
for (let i = 0; i < n; i++) {
startPoints.push(segments[i].first);
endPoints.push(segments[i].second);
}
// Sort the startPoints
startPoints.sort();
// Sort the startPoints
endPoints.sort( function (a, b) { return a - b });
// Store the minimum number of
// deletions required and
// initialize with (N - 1)
let ans = n - 1;
// Traverse the array segments[]
for (let i = 0; i < n; i++) {
// Store the starting point
let f = segments[i].first;
// Store the ending point
let s = segments[i].second;
// Store the number of segments
// satisfying the first condition
// of non-intersection
let leftDelete
= lowerBound(endPoints, f);
// Store the number of segments
// satisfying the second condition
// of non-intersection
let rightDelete = Math.max(0, n - Math.floor(upperBound(startPoints, s)));
// Update answer
ans = Math.min(ans, leftDelete + rightDelete);
}
// Print the answer
console.log(ans);
} // Driver Code arr = [{ "first" : 1, "second" : 2 },
{ "first" : 5, "second" : 6 },
{ "first" : 6, "second" : 7 },
{ "first" : 7, "second" : 10 },
{ "first" : 8, "second" : 9 }];
let N = arr.length; // Function Call minSegments(arr, N); // This code is contributed by akashish__. |
2
Time Complexity: O(N*(log N2))
Auxiliary Space: O(N)