# Number of intersections between two ranges

Given N ranges of type1 ranges and M ranges of type2.The task is to find the total number of intersections between all possible type1 and type2 range pairs. All start and end points of type1 and type2 ranges are given.

Examples:

Input : N = 2, M = 2
type1[ ] = { { 1, 3 }, { 5, 9 } }
type2[ ] = { { 2, 8 }, { 9, 12 } }
Output : 3
Range {2, 8} intersects with type1 ranges {1, 3} and {5, 9}
Range {9, 12} intersects with {5, 9} only.
So the total number of intersections is 3.

Input : N = 3, M = 1
type1[ ] = { { 1, 8 }, { 5, 10 }, { 14, 28 }
type2[ ] = { { 2, 8 } }
Output : 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Idea is to use inclusion-exclusion method to determine the total number of intersections.
• Total possible number of intersections are n * m. Now subtract those count of type1 ranges which do not intersect with ith type2 range.
• Those type1 ranges will not intersect with ith type2 range which ends before starts of ith type2 range and starts after the end of ith type2 range.
• This count can be determined by using binary search . The C++ inbuilt function upper_bound can be used directly.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return total ` `// number of intersections ` `int` `FindIntersection(pair<``int``, ``int``> type1[], ``int` `n, ` `                     ``pair<``int``, ``int``> type2[], ``int` `m) ` `{ ` ` `  `    ``// Maximum possible number ` `    ``// of intersections ` `    ``int` `ans = n * m; ` ` `  `    ``vector<``int``> start, end; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Store all starting ` `        ``// points of type1 ranges ` `        ``start.push_back(type1[i].first); ` ` `  `        ``// Store all endpoints ` `        ``// of type1 ranges ` `        ``end.push_back(type1[i].second); ` `    ``} ` ` `  `    ``sort(start.begin(), start.end()); ` `    ``sort(end.begin(), end.end()); ` ` `  `    ``for` `(``int` `i = 0; i < m; i++) { ` ` `  `        ``// Starting point of type2 ranges ` `        ``int` `L = type2[i].first; ` ` `  `        ``// Ending point of type2 ranges ` `        ``int` `R = type2[i].second; ` ` `  `        ``// Subtract those ranges which ` `        ``// are starting after R ` `        ``ans -= (start.end() -  ` `        ``upper_bound(start.begin(), start.end(), R)); ` ` `  `        ``// Subtract those ranges which ` `        ``// are ending before L ` `        ``ans -=  ` `        ``(upper_bound(end.begin(), end.end(), L - 1)  ` `        ``- end.begin()); ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``pair<``int``, ``int``> type1[] = ` `    ``{ { 1, 2 }, { 2, 3 }, { 4, 5 }, { 6, 7 } }; ` ` `  `    ``pair<``int``, ``int``> type2[] =  ` `    ``{ { 1, 5 }, { 2, 3 }, { 4, 7 }, { 5, 7 } }; ` ` `  `    ``int` `n = ``sizeof``(type1) / (``sizeof``(type1)); ` `    ``int` `m = ``sizeof``(type2) / ``sizeof``(type2); ` ` `  `    ``cout << FindIntersection(type1, n, type2, m); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of above approach ` `from` `bisect ``import` `bisect as upper_bound ` ` `  `# Function to return total ` `# number of intersections ` `def` `FindIntersection(type1, n, type2, m): ` ` `  `    ``# Maximum possible number ` `    ``# of intersections ` `    ``ans ``=` `n ``*` `m ` ` `  `    ``start ``=` `[] ` `    ``end ``=` `[] ` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Store all starting ` `        ``# points of type1 ranges ` `        ``start.append(type1[i][``0``]) ` ` `  `        ``# Store all endpoints ` `        ``# of type1 ranges ` `        ``end.append(type1[i][``1``]) ` ` `  `    ``start ``=` `sorted``(start) ` `    ``start ``=` `sorted``(end) ` ` `  `    ``for` `i ``in` `range``(m): ` ` `  `        ``# Starting poof type2 ranges ` `        ``L ``=` `type2[i][``0``] ` ` `  `        ``# Ending poof type2 ranges ` `        ``R ``=` `type2[i][``1``] ` ` `  `        ``# Subtract those ranges which ` `        ``# are starting after R ` `        ``ans ``-``=` `(``len``(start)``-` `upper_bound(start, R)) ` ` `  `        ``# Subtract those ranges which ` `        ``# are ending before L ` `        ``ans ``-``=` `(upper_bound(end, L ``-` `1``)) ` ` `  `    ``return` `ans ` ` `  `# Driver Code ` `type1 ``=` `[ [ ``1``, ``2` `], [ ``2``, ``3` `],  ` `          ``[ ``4``, ``5` `], [ ``6``, ``7` `] ] ` ` `  `type2 ``=` `[ [ ``1``, ``5` `], [ ``2``, ``3` `],  ` `          ``[ ``4``, ``7` `], [ ``5``, ``7` `] ] ` ` `  `n ``=` `len``(type1) ` `m ``=` `len``(type2) ` ` `  `print``(FindIntersection(type1, n, type2, m)) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```9
```

Time Complexity: O(M*log(N))

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Improved By : mohit kumar 29