Find Intersection of all Intervals
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++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to print the intersection void findIntersection( int intervals[][2], int N) { // First interval int l = intervals[0][0]; int r = intervals[0][1]; // Check rest of the intervals and find the intersection for ( int i = 1; i < N; i++) { // If no intersection exists if (intervals[i][0] > r || intervals[i][1] < l) { cout << -1; return ; } // Else update the intersection else { l = max(l, intervals[i][0]); r = min(r, intervals[i][1]); } } cout << "[" << l << ", " << r << "]" ; } // Driver code int main() { int intervals[][2] = { { 1, 6 }, { 2, 8 }, { 3, 10 }, { 5, 8 } }; int N = sizeof (intervals) / sizeof (intervals[0]); findIntersection(intervals, N); } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Function to print the intersection static void findIntersection( int intervals[][], int N) { // First interval int l = intervals[ 0 ][ 0 ]; int r = intervals[ 0 ][ 1 ]; // Check rest of the intervals // and find the intersection for ( int i = 1 ; i < N; i++) { // If no intersection exists if (intervals[i][ 0 ] > r || intervals[i][ 1 ] < l) { System.out.println(- 1 ); return ; } // Else update the intersection else { l = Math.max(l, intervals[i][ 0 ]); r = Math.min(r, intervals[i][ 1 ]); } } System.out.println ( "[" + l + ", " + r + "]" ); } // Driver code public static void main (String[] args) { int intervals[][] = {{ 1 , 6 }, { 2 , 8 }, { 3 , 10 }, { 5 , 8 }}; int N = intervals.length; findIntersection(intervals, N); } } // This Code is contributed by ajit.. |
Python
# Python3 implementation of the approach # Function to print the intersection def findIntersection(intervals,N): # First interval l = intervals[ 0 ][ 0 ] r = intervals[ 0 ][ 1 ] # Check rest of the intervals # and find the intersection for i in range ( 1 ,N): # If no intersection exists if (intervals[i][ 0 ] > r or intervals[i][ 1 ] < l): print ( - 1 ) # Else update the intersection else : l = max (l, intervals[i][ 0 ]) r = min (r, intervals[i][ 1 ]) print ( "[" ,l, ", " ,r, "]" ) # Driver code intervals = [ [ 1 , 6 ], [ 2 , 8 ], [ 3 , 10 ], [ 5 , 8 ] ] N = len (intervals) findIntersection(intervals, N) # this code is contributed by mohit kumar |
C#
// C# implementation of the approach using System; class GFG { // Function to print the intersection static void findIntersection( int [,]intervals, int N) { // First interval int l = intervals[0, 0]; int r = intervals[0, 1]; // Check rest of the intervals // and find the intersection for ( int i = 1; i < N; i++) { // If no intersection exists if (intervals[i, 0] > r || intervals[i, 1] < l) { Console.WriteLine(-1); return ; } // Else update the intersection else { l = Math.Max(l, intervals[i, 0]); r = Math.Min(r, intervals[i, 1]); } } Console.WriteLine( "[" + l + ", " + r + "]" ); } // Driver code public static void Main() { int [,]intervals = {{ 1, 6 }, { 2, 8 }, { 3, 10 }, { 5, 8 }}; int N = intervals.GetLength(0); findIntersection(intervals, N); } } // This code is contributed by Ryuga |
PHP
<?php // PHP implementation of the approach // Function to print the intersection function findIntersection( $intervals , $N ) { // First interval $l = $intervals [0][0]; $r = $intervals [0][1]; // Check rest of the intervals and // find the intersection for ( $i = 1; $i < $N ; $i ++) { // If no intersection exists if ( $intervals [ $i ][0] > $r || $intervals [ $i ][1] < $l ) { echo -1; return ; } // Else update the intersection else { $l = max( $l , $intervals [ $i ][0]); $r = min( $r , $intervals [ $i ][1]); } } echo "[" . $l . ", " . $r . "]" ; } // Driver code $intervals = array ( array (1, 6), array (2, 8), array (3, 10), array (5, 8)); $N = sizeof( $intervals ); findIntersection( $intervals , $N ); // This code is contributed // by Akanksha Rai ?> |
Javascript
<script> // Javascript implementation of the approach // Function to print the intersection function findIntersection(intervals, N) { // First interval let l = intervals[0][0]; let r = intervals[0][1]; // Check rest of the intervals // and find the intersection for (let i = 1; i < N; i++) { // If no intersection exists if (intervals[i][0] > r || intervals[i][1] < l) { document.write(-1 + "</br>" ); return ; } // Else update the intersection 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> |
Output
[5, 6]
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.