# 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
Expalination:
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 co-ordinates are up to 10 9.

Time Complexity: O (N 2)
Auxiliary Space: O (MAX)

Efficient Approach:

1. Istead of iterating from start to end just create a visited array and for each range, mark vis[start] = 1 and vis[end+1] = -1.
2. Take the prefix sum of the array.
3. 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 ` `using` `namespace` `std; ` `const` `int` `MAX = 1e5 + 5; ` ` `  `// function to find the smallest ` `// non-overlapping number ` `void` `find_missing( ` `    ``vector > 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 > interval ` `        ``= { { 0, 14 }, { 86, 108 }, ` `            ``{ 22, 30 }, { 5, 17 } }; ` `    ``find_missing(interval); ` `    ``return` `0; ` `} `

Output:

```18
```

Time Complexity: O (N)
Auxiliary Space: O (MAX)

However, this approach will also not work if the interval co-ordinates are up to 10 9.

Efficient Approach:

1. Sort the range by their start-coordinate and for each next range.
2. 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 ` `using` `namespace` `std; ` ` `  `// function to find the smallest ` `// non-overlapping number ` `void` `find_missing( ` `    ``vector > 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 vaue 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 > interval ` `        ``= { { 0, 14 }, { 86, 108 }, ` `            ``{ 22, 30 }, { 5, 17 } }; ` `    ``find_missing(interval); ` `    ``return` `0; ` `} `

Output:

```18
```

Time Complexity: O (N * logN)
Auxiliary Space: O (1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.