Find least non-overlapping number from a given set of intervals
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++
// 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
// 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
# 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#
// 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 |
Javascript
<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> |
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++
// 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
// 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
# 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#
// 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 |
Javascript
<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> |
18
Time Complexity: O (N * logN)
Auxiliary Space: O (1)
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