Find Intersection of all Intervals
Last Updated :
15 Nov, 2022
Given N intervals of the form of [l, r], the task is to find the intersection of all the intervals. An intersection is an interval that lies within all of the given intervals. If no such intersection exists then print -1.
Examples:
Input: arr[] = {{1, 6}, {2, 8}, {3, 10}, {5, 8}}
Output: [5, 6]
[5, 6] is the common interval that lies in all the given intervals.
Input: arr[] = {{1, 6}, {8, 18}}
Output: -1
No intersection exists between the two given ranges.
Approach:
- Start by considering first interval as the required answer.
- Now, starting from the second interval, try searching for the intersection. Two cases can arise:
- There exists no intersection between [l1, r1] and [l2, r2]. Possible only when r1 < l2 or r2 < l1. In such a case answer will be 0 i.e. no intersection exists.
- There exists an intersection between [l1, r1] and [l2, r2]. Then the required intersection will be [max(l1, l2), min(r1, r2)].
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void findIntersection( int intervals[][2], int N)
{
int l = intervals[0][0];
int r = intervals[0][1];
for ( int i = 1; i < N; i++) {
if (intervals[i][0] > r || intervals[i][1] < l) {
cout << -1;
return ;
}
else {
l = max(l, intervals[i][0]);
r = min(r, intervals[i][1]);
}
}
cout << "[" << l << ", " << r << "]" ;
}
int main()
{
int intervals[][2] = {
{ 1, 6 },
{ 2, 8 },
{ 3, 10 },
{ 5, 8 }
};
int N = sizeof (intervals) / sizeof (intervals[0]);
findIntersection(intervals, N);
}
|
Java
import java.io.*;
class GFG
{
static void findIntersection( int intervals[][], int N)
{
int l = intervals[ 0 ][ 0 ];
int r = intervals[ 0 ][ 1 ];
for ( int i = 1 ; i < N; i++)
{
if (intervals[i][ 0 ] > r ||
intervals[i][ 1 ] < l)
{
System.out.println(- 1 );
return ;
}
else
{
l = Math.max(l, intervals[i][ 0 ]);
r = Math.min(r, intervals[i][ 1 ]);
}
}
System.out.println ( "[" + l + ", " + r + "]" );
}
public static void main (String[] args)
{
int intervals[][] = {{ 1 , 6 },
{ 2 , 8 },
{ 3 , 10 },
{ 5 , 8 }};
int N = intervals.length;
findIntersection(intervals, N);
}
}
|
Python
def findIntersection(intervals,N):
l = intervals[ 0 ][ 0 ]
r = intervals[ 0 ][ 1 ]
for i in range ( 1 ,N):
if (intervals[i][ 0 ] > r or intervals[i][ 1 ] < l):
print ( - 1 )
else :
l = max (l, intervals[i][ 0 ])
r = min (r, intervals[i][ 1 ])
print ( "[" ,l, ", " ,r, "]" )
intervals = [
[ 1 , 6 ],
[ 2 , 8 ],
[ 3 , 10 ],
[ 5 , 8 ]
]
N = len (intervals)
findIntersection(intervals, N)
|
C#
using System;
class GFG
{
static void findIntersection( int [,]intervals, int N)
{
int l = intervals[0, 0];
int r = intervals[0, 1];
for ( int i = 1; i < N; i++)
{
if (intervals[i, 0] > r ||
intervals[i, 1] < l)
{
Console.WriteLine(-1);
return ;
}
else
{
l = Math.Max(l, intervals[i, 0]);
r = Math.Min(r, intervals[i, 1]);
}
}
Console.WriteLine( "[" + l + ", " + r + "]" );
}
public static void Main()
{
int [,]intervals = {{ 1, 6 }, { 2, 8 },
{ 3, 10 }, { 5, 8 }};
int N = intervals.GetLength(0);
findIntersection(intervals, N);
}
}
|
PHP
<?php
function findIntersection( $intervals , $N )
{
$l = $intervals [0][0];
$r = $intervals [0][1];
for ( $i = 1; $i < $N ; $i ++)
{
if ( $intervals [ $i ][0] > $r ||
$intervals [ $i ][1] < $l )
{
echo -1;
return ;
}
else
{
$l = max( $l , $intervals [ $i ][0]);
$r = min( $r , $intervals [ $i ][1]);
}
}
echo "[" . $l . ", " . $r . "]" ;
}
$intervals = array ( array (1, 6), array (2, 8),
array (3, 10), array (5, 8));
$N = sizeof( $intervals );
findIntersection( $intervals , $N );
?>
|
Javascript
<script>
function findIntersection(intervals, N)
{
let l = intervals[0][0];
let r = intervals[0][1];
for (let i = 1; i < N; i++)
{
if (intervals[i][0] > r ||
intervals[i][1] < l)
{
document.write(-1 + "</br>" );
return ;
}
else
{
l = Math.max(l, intervals[i][0]);
r = Math.min(r, intervals[i][1]);
}
}
document.write( "[" + l + ", " + r + "]" + "</br>" );
}
let intervals = [[ 1, 6 ],
[ 2, 8 ],
[ 3, 10],
[ 5, 8 ]];
let N = intervals.length;
findIntersection(intervals, N);
</script>
|
Time Complexity: O(N), where N is the size of the given 2D-array
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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