Given an array interval of pairs of integers representing the starting and ending points of the interval of size N. The task is to find the smallest non-negative integer which is a non-overlapping number from the given set of intervals.
Input constraints:
Examples:
Input: interval = {{0, 4}, {6, 8}, {2, 3}, {9, 18}}
Output: 5
Explanation:
The smallest non-negative integer which is non-overlapping to all set of the intervals is 5.Input: interval = {{0, 14}, {86, 108}, {22, 30}, {5, 17}}
Output: 18
Naive Approach:
- Create a visited array of size MAX, and for every interval mark all value true from start to end.
- Finally, iterate from 1 to MAX and find the smallest value which is not visited.
However, this approach will not work if the interval coordinates are up to 10 9.
Time Complexity: O (N 2)
Auxiliary Space: O (MAX)
Efficient Approach:
- Instead of iterating from start to end just create a visited array and for each range, mark vis[start] = 1 and vis[end+1] = -1.
- Take the prefix sum of the array.
- Then iterate over the array to find the first integer with value 0.
Here is the implementation of the above approach:
// C++ program to find the // least non-overlapping number // from a given set intervals #include <bits/stdc++.h> using namespace std;
const int MAX = 1e5 + 5;
// function to find the smallest // non-overlapping number void find_missing(
vector<pair< int , int > > interval)
{ // create a visited array
vector< int > vis(MAX);
for ( int i = 0; i < interval.size(); ++i) {
int start = interval[i].first;
int end = interval[i].second;
vis[start]++;
vis[end + 1]--;
}
// find the first missing value
for ( int i = 1; i < MAX; i++) {
vis[i] += vis[i - 1];
if (!vis[i]) {
cout << i << endl;
return ;
}
}
} // Driver function int main()
{ vector<pair< int , int > > interval
= { { 0, 14 }, { 86, 108 },
{ 22, 30 }, { 5, 17 } };
find_missing(interval);
return 0;
} |
// Java program to find the // least non-overlapping number // from a given set intervals import java.io.*;
public class GFG{
static int MAX = ( int ) (1e5 + 5 );
static class pair
{ int first, second;
public pair( int first, int second)
{
this .first = first;
this .second = second;
}
} // function to find the smallest // non-overlapping number static void find_missing(
pair[] interval)
{ // create a visited array
int [] vis = new int [MAX];
for ( int i = 0 ; i < interval.length; ++i)
{
int start = interval[i].first;
int end = interval[i].second;
vis[start]++;
vis[end + 1 ]--;
}
// find the first missing value
for ( int i = 1 ; i < MAX; i++) {
vis[i] += vis[i - 1 ];
if (vis[i]== 0 ) {
System.out.print(i + "\n" );
return ;
}
}
} // Driver function public static void main(String[] args)
{ pair []interval = { new pair( 0 , 14 ),
new pair( 86 , 108 ),
new pair( 22 , 30 ),
new pair( 5 , 17 )};
find_missing(interval);
} } // This code is contributed by Rohit_ranjan |
# Python3 program to find the # least non-overlapping number # from a given set intervals MAX = int ( 1e5 + 5 )
# Function to find the smallest # non-overlapping number def find_missing(interval):
# Create a visited array
vis = [ 0 ] * ( MAX )
for i in range ( len (interval)):
start = interval[i][ 0 ]
end = interval[i][ 1 ]
vis[start] + = 1
vis[end + 1 ] - = 1
# Find the first missing value
for i in range ( 1 , MAX ):
vis[i] + = vis[i - 1 ]
if (vis[i] = = 0 ):
print (i)
return
# Driver code interval = [ [ 0 , 14 ], [ 86 , 108 ],
[ 22 , 30 ], [ 5 , 17 ] ]
find_missing(interval) # This code is contributed by divyeshrabadiya07 |
// C# program to find the // least non-overlapping number // from a given set intervals using System;
class GFG{
static int MAX = ( int )(1e5 + 5);
class pair
{ public int first, second;
public pair( int first, int second)
{
this .first = first;
this .second = second;
}
} // Function to find the smallest // non-overlapping number static void find_missing(pair[] interval)
{ // Create a visited array
int [] vis = new int [MAX];
for ( int i = 0; i < interval.Length; ++i)
{
int start = interval[i].first;
int end = interval[i].second;
vis[start]++;
vis[end + 1]--;
}
// Find the first missing value
for ( int i = 1; i < MAX; i++)
{
vis[i] += vis[i - 1];
if (vis[i] == 0)
{
Console.Write(i + "\n" );
return ;
}
}
} // Driver code public static void Main(String[] args)
{ pair []interval = { new pair(0, 14),
new pair(86, 108),
new pair(22, 30),
new pair(5, 17) };
find_missing(interval);
} } // This code is contributed by Amit Katiyar |
<script> // Javascript program to find the // least non-overlapping number // from a given set intervals var MAX = 100005;
// function to find the smallest // non-overlapping number function find_missing( interval)
{ // create a visited array
var vis = Array(MAX).fill(0);
for ( var i = 0; i < interval.length; ++i) {
var start = interval[i][0];
var end = interval[i][1];
vis[start]++;
vis[end + 1]--;
}
// find the first missing value
for ( var i = 1; i < MAX; i++) {
vis[i] += vis[i - 1];
if (!vis[i]) {
document.write( i + "<br>" );
return ;
}
}
} // Driver function var interval
= [ [ 0, 14 ], [ 86, 108 ],
[ 22, 30 ], [ 5, 17 ] ];
find_missing(interval); </script> |
Output
18
Time Complexity: O (N)
Auxiliary Space: O (MAX)
However, this approach will also not work if the interval coordinates are up to 10 9.
Efficient Approach:
- Sort the range by their start-coordinate and for each next range.
- Check if the starting point is greater than the maximum end-coordinate encountered so far, then a missing number can be found, and it will be previous_max + 1.
Illustration:
Consider the following example:
interval[][] = { { 0, 14 }, { 86, 108 }, { 22, 30 }, { 5, 17 } };
After sorting, interval[][] = { { 0, 14 }, { 5, 17 }, { 22, 30 }, { 86, 108 }};
Initial mx = 0 and after considering first interval mx = max(0, 15) = 15
Since mx = 15 and 15 > 5 so after considering second interval mx = max(15, 18) = 18
now 18 < 22 so 18 is least non-overlapping number.
Here is the implementation of the above approach:
// C++ program to find the // least non-overlapping number // from a given set intervals #include <bits/stdc++.h> using namespace std;
// function to find the smallest // non-overlapping number void find_missing(
vector<pair< int , int > > interval)
{ // Sort the intervals based on their
// starting value
sort(interval.begin(), interval.end());
int mx = 0;
for ( int i = 0; i < ( int )interval.size(); ++i) {
// check if any missing value exist
if (interval[i].first > mx) {
cout << mx;
return ;
}
else
mx = max(mx, interval[i].second + 1);
}
// finally print the missing value
cout << mx;
} // Driver function int main()
{ vector<pair< int , int > > interval
= { { 0, 14 }, { 86, 108 },
{ 22, 30 }, { 5, 17 } };
find_missing(interval);
return 0;
} |
// Java program to find the // least non-overlapping number // from a given set intervals import java.util.*;
import java.io.*;
class GFG{
static class Pair implements Comparable<Pair>
{ int start,end;
Pair( int s, int e)
{
start = s;
end = e;
}
public int compareTo(Pair p)
{
return this .start - p.start;
}
} // Function to find the smallest // non-overlapping number static void findMissing(ArrayList<Pair> interval)
{ // Sort the intervals based on their
// starting value
Collections.sort(interval);
int mx = 0 ;
for ( int i = 0 ; i < interval.size(); ++i)
{
// Check if any missing value exist
if (interval.get(i).start > mx)
{
System.out.println(mx);
return ;
}
else
mx = Math.max(mx, interval.get(i).end + 1 );
}
// Finally print the missing value
System.out.println(mx);
} // Driver code public static void main(String []args)
{ ArrayList<Pair> interval = new ArrayList<>();
interval.add( new Pair( 0 , 14 ));
interval.add( new Pair( 86 , 108 ));
interval.add( new Pair( 22 , 30 ));
interval.add( new Pair( 5 , 17 ));
findMissing(interval);
} } // This code is contributed by Ganeshchowdharysadanala |
# Python3 program to find the # least non-overlapping number # from a given set intervals # function to find the smallest # non-overlapping number def find_missing(interval):
# Sort the intervals based
# on their starting value
interval.sort()
mx = 0
for i in range ( len (interval)):
# Check if any missing
# value exist
if (interval[i][ 0 ] > mx):
print (mx)
return
else :
mx = max (mx,
interval[i][ 1 ] + 1 )
# Finally print the missing value
print (mx)
# Driver code if __name__ = = "__main__" :
interval = [[ 0 , 14 ], [ 86 , 108 ],
[ 22 , 30 ], [ 5 , 17 ]]
find_missing(interval);
# This code is contributed by Chitranayal |
// C# program to find the // least non-overlapping number // from a given set intervals using System;
using System.Collections.Generic;
class GFG{
class Pair : IComparable<Pair>
{ public int start,end;
public Pair( int s, int e)
{
start = s;
end = e;
}
public int CompareTo(Pair p)
{
return this .start - p.start;
}
} // Function to find the smallest // non-overlapping number static void findMissing(List<Pair> interval)
{ // Sort the intervals based on their
// starting value
interval.Sort();
int mx = 0;
for ( int i = 0; i < interval.Count; ++i)
{
// Check if any missing value exist
if (interval[i].start > mx)
{
Console.WriteLine(mx);
return ;
}
else
mx = Math.Max(mx, interval[i].end + 1);
}
// Finally print the missing value
Console.WriteLine(mx);
} // Driver code public static void Main(String []args)
{ List<Pair> interval = new List<Pair>();
interval.Add( new Pair(0, 14));
interval.Add( new Pair(86, 108));
interval.Add( new Pair(22, 30));
interval.Add( new Pair(5, 17));
findMissing(interval);
} } // This code is contributed by shikhasingrajput |
<script> // Javascript program to find the // least non-overlapping number // from a given set intervals // Function to find the smallest // non-overlapping number function findMissing(interval)
{ // Sort the intervals based on their
// starting value
interval.sort( function (a,b){ return a[0]-b[0];});
let mx = 0;
for (let i = 0; i < interval.length; ++i)
{
// Check if any missing value exist
if (interval[i][0] > mx)
{
document.write(mx+ "<br>" );
return ;
}
else
mx = Math.max(mx, interval[i][1] + 1);
}
// Finally print the missing value
document.write(mx);
} // Driver code let interval = [[0, 14], [86, 108], [22, 30], [5, 17]];
findMissing(interval); // This code is contributed by avanitrachhadiya2155. </script> |
Output
18
Time Complexity: O (N * logN)
Auxiliary Space: O (1)
Another Approach:
Approach:
Sort the intervals in ascending order based on their end points. This can be done using any efficient sorting algorithm such as merge sort or quicksort.
Initialize a variable “last” to the smallest possible integer value. This variable will keep track of the end point of the last interval that was added to the output.
Iterate through each interval in the sorted list. If the start point of the current interval is greater than “last”, add the end point of the current interval to the output and update “last” to be the end point of the current interval.
Return the value of “last” as the least non-overlapping number.
#include <bits/stdc++.h> using namespace std;
// Interval structure struct Interval {
int start;
int end;
}; // Comparison function to sort intervals based on their end points bool compare( const Interval& a, const Interval& b) {
return a.end < b.end;
} // Function to find the least non-overlapping number from a given set of intervals int findLeastNonOverlappingNumber(Interval intervals[], int n) {
// Sort intervals in ascending order based on their end points
sort(intervals, intervals + n, compare);
// Initialize last to the smallest possible integer value
int last = -2147483648;
// Iterate through each interval
for ( int i = 0; i < n; i++) {
// If the start point of the current interval is greater than last,
// add the end point of the current interval to the output and update last
if (intervals[i].start > last) {
last = intervals[i].end;
}
}
// Return the value of last as the least non-overlapping number
return last;
} // Driver code int main() {
// Sample input
Interval intervals[] = {{1, 3}, {2, 4}, {3, 6}, {5, 7}, {7, 8}};
int n = sizeof (intervals) / sizeof (intervals[0]);
// Find the least non-overlapping number from the input set of intervals
int leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals, n);
// Print the result
cout << "The least non-overlapping number is " << leastNonOverlappingNumber << endl;
return 0;
} |
#include <stdio.h> #include <stdlib.h> // Interval structure struct Interval {
int start;
int end;
}; // Comparison function to sort intervals based on their end points int compare( const void * a, const void * b) {
return (( struct Interval*)a)->end - (( struct Interval*)b)->end;
} // Function to find the least non-overlapping number from a given set of intervals int findLeastNonOverlappingNumber( struct Interval intervals[], int n) {
// Sort intervals in ascending order based on their end points
qsort (intervals, n, sizeof ( struct Interval), compare);
// Initialize last to the smallest possible integer value
int last = -2147483648;
// Iterate through each interval
for ( int i = 0; i < n; i++) {
// If the start point of the current interval is greater than last,
// add the end point of the current interval to the output and update last
if (intervals[i].start > last) {
last = intervals[i].end;
}
}
// Return the value of last as the least non-overlapping number
return last;
} // Driver code int main() {
// Sample input
struct Interval intervals[] = {{1, 3}, {2, 4}, {3, 6}, {5, 7}, {7, 8}};
int n = sizeof (intervals) / sizeof (intervals[0]);
// Find the least non-overlapping number from the input set of intervals
int leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals, n);
// Print the result
printf ( "The least non-overlapping number is %d\n" , leastNonOverlappingNumber);
return 0;
} |
import java.util.Arrays;
import java.util.Comparator;
// Interval class class Interval {
int start;
int end;
public Interval( int start, int end) {
this .start = start;
this .end = end;
}
} public class Main {
// Function to find the least non-overlapping number from a given set of intervals
public static int findLeastNonOverlappingNumber(Interval[] intervals) {
// Sort intervals in ascending order based on their end points
Arrays.sort(intervals, new Comparator<Interval>() {
@Override
public int compare(Interval a, Interval b) {
return a.end - b.end;
}
});
// Initialize last to the smallest possible integer value
int last = Integer.MIN_VALUE;
// Iterate through each interval
for ( int i = 0 ; i < intervals.length; i++) {
// If the start point of the current interval is greater than last,
// add the end point of the current interval to the output and update last
if (intervals[i].start > last) {
last = intervals[i].end;
}
}
// Return the value of last as the least non-overlapping number
return last;
}
// Driver code
public static void main(String[] args) {
// Sample input
Interval[] intervals = { new Interval( 1 , 3 ), new Interval( 2 , 4 ), new Interval( 3 , 6 ), new Interval( 5 , 7 ), new Interval( 7 , 8 )};
// Find the least non-overlapping number from the input set of intervals
int leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals);
// Print the result
System.out.println( "The least non-overlapping number is " + leastNonOverlappingNumber);
}
} |
from typing import List , Tuple
# Interval structure class Interval:
def __init__( self , start: int , end: int ):
self .start = start
self .end = end
# Function to find the least non-overlapping number from a given set of intervals def findLeastNonOverlappingNumber(intervals: List [Interval]) - > int :
# Sort intervals in ascending order based on their end points
intervals.sort(key = lambda x: x.end)
# Initialize last to the smallest possible integer value
last = float ( '-inf' )
# Iterate through each interval
for interval in intervals:
# If the start point of the current interval is greater than last,
# add the end point of the current interval to the output and update last
if interval.start > last:
last = interval.end
# Return the value of last as the least non-overlapping number
return last
# Driver code if __name__ = = '__main__' :
# Sample input
intervals = [Interval( 1 , 3 ), Interval( 2 , 4 ), Interval( 3 , 6 ), Interval( 5 , 7 ), Interval( 7 , 8 )]
# Find the least non-overlapping number from the input set of intervals
leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals)
# Print the result
print (f 'The least non-overlapping number is {leastNonOverlappingNumber}' )
|
using System;
using System.Collections.Generic;
// Make the Interval class public public class Interval {
public int start;
public int end;
public Interval( int start, int end)
{
this .start = start;
this .end = end;
}
} public class Program {
// Make the FindLeastNonOverlappingNumber method public
// and static
public static int
findLeastNonOverlappingNumber(Interval[] intervals)
{
Array.Sort(intervals, (a, b) = > a.end - b.end);
int last = int .MinValue;
foreach (Interval interval in intervals)
{
if (interval.start > last) {
last = interval.end;
}
}
return last;
}
static void Main( string [] args)
{
Interval[] intervals
= { new Interval(1, 3), new Interval(2, 4),
new Interval(3, 6), new Interval(5, 7),
new Interval(7, 8) };
int leastNonOverlappingNumber
= findLeastNonOverlappingNumber(intervals);
Console.WriteLine(
"The least non-overlapping number is "
+ leastNonOverlappingNumber);
}
} |
// Interval structure class Interval { constructor(start, end) {
this .start = start;
this .end = end;
}
} // Comparison function to sort intervals based on their end points function compare(a, b) {
return a.end - b.end;
} // Function to find the least non-overlapping number from a given set of intervals function findLeastNonOverlappingNumber(intervals) {
const n = intervals.length;
// Sort intervals in ascending order based on their end points
intervals.sort(compare);
// Initialize last to the smallest possible integer value
let last = -2147483648;
// Iterate through each interval
for (let i = 0; i < n; i++) {
// If the start point of the current interval is greater than last,
// add the end point of the current interval to the output and update last
if (intervals[i].start > last) {
last = intervals[i].end;
}
}
// Return the value of last as the least non-overlapping number
return last;
} // Driver code ( function ()
{ // Sample input
const intervals = [ new Interval(1, 3), new Interval(2, 4), new Interval(3, 6), new Interval(5, 7), new Interval(7, 8)];
// Find the least non-overlapping number from the input set of intervals
const leastNonOverlappingNumber = findLeastNonOverlappingNumber(intervals);
// Print the result
console.log(`The least non-overlapping number is ${leastNonOverlappingNumber}`);
})(); |
Output
The least non-overlapping number is 7
time complexity of O(nlogn), Where n is the size of the intervals
space complexity of O(n), Where n is the size of the intervals