# Find intersection of intervals given by two lists

Given two 2-D arrays which represent intervals. Each 2-D array represents a list of intervals. Each list of intervals is disjoint and sorted in increasing order. Find the intersection or set of ranges that are common to both the lists.

Disjoint means no element is common in a list. Example: {1, 4} and {5, 6} are disjoint while {1, 4} and {2, 5} are not as 2, 3 and 4 are common to both intervals.

Examples:

Input: arr1[][] = {{0, 4}, {5, 10}, {13, 20}, {24, 25}}
arr2[][] = {{1, 5}, {8, 12}, {15, 24}, {25, 26}}
Output: {{1, 4}, {5, 5}, {8, 10}, {15, 20}, {24, 24}, {25, 25}}

Explanation:
{1, 4} lies completely within range {0, 4} and {1, 5}. Hence, {1, 4} is the desired intersection. Similarly, {24, 24} lies completely within two intervals {24, 25} and {15, 24}.

Input: arr1[][] = {{0, 2}, {5, 10}, {12, 22}, {24, 25}}
arr2[][] = {{1, 4}, {9, 12}, {15, 24}, {25, 26}}
Output: {{1, 2}, {9, 10}, {12, 12}, {15, 22}, {24, 24}, {25, 25}}
Explanation:
{1, 2} lies completely within range {0, 2} and {1, 4}. Hence, {1, 2} is the desired intersection. Similarly, {12, 12} lies completely within two intervals {12, 22} and {9, 12}.

Approach:

To solve the problem mentioned above, two pointer technique can be used, as per the steps given below:

• Maintain two pointers i and j to traverse the two interval lists, arr1 and arr2 respectively.
• Now, if arr1[i] has smallest endpoint, it can only intersect with arr2[j]. Similarly, if arr2[j] has smallest endpoint, it can only intersect with arr1[i]. If intersection occurs, find the intersecting segment.
• [l, r] will be the intersecting segment iff l <= r, where l = max(arr1[i], arr2[j]) and r = min(arr1[i], arr2[j]).
• Increment the i and j pointers accordingly to move ahead.

Below is the implementation of the approach:

 `// C++ implementation to find the ` `// intersection of the two intervals ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to print intersecting intervals ` `void` `printIntervals(vector > arr1, ` `                    ``vector > arr2) ` `{ ` ` `  `    ``// i and j pointers for ` `    ``// arr1 and arr2 respectively ` `    ``int` `i = 0, j = 0; ` ` `  `    ``// Size of the two lists ` `    ``int` `n = arr1.size(), m = arr2.size(); ` ` `  `    ``// Loop through all intervals unless ` `    ``// one of the interval gets exhausted ` `    ``while` `(i < n && j < m) { ` `        ``// Left bound for intersecting segment ` `        ``int` `l = max(arr1[i], arr2[j]); ` ` `  `        ``// Right bound for intersecting segment ` `        ``int` `r = min(arr1[i], arr2[j]); ` ` `  `        ``// If segment is valid print it ` `        ``if` `(l <= r) ` `            ``cout << ``"{"` `<< l << ``", "` `                 ``<< r << ``"}\n"``; ` ` `  `        ``// If i-th interval's right ` `        ``// bound is smaller ` `        ``// increment i else ` `        ``// increment j ` `        ``if` `(arr1[i] < arr2[j]) ` `            ``i++; ` `        ``else` `            ``j++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``vector > arr1 ` `        ``= { { 0, 4 }, { 5, 10 }, ` `            ``{ 13, 20 }, { 24, 25 } }; ` ` `  `    ``vector > arr2 ` `        ``= { { 1, 5 }, { 8, 12 }, ` `            ``{ 15, 24 }, { 25, 26 } }; ` ` `  `    ``printIntervals(arr1, arr2); ` ` `  `    ``return` `0; ` `} `

Output:

```{1, 4}
{5, 5}
{8, 10}
{15, 20}
{24, 24}
{25, 25}
```

Time Complexity: O(N + M), where N and M are lengths of the 2-D arrays

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

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.