An interval is represented as a combination of start time and end time. Given a set of intervals, check if any two intervals overlap.
Examples:
Input: arr[] = {{1, 3}, {5, 7}, {2, 4}, {6, 8}}
Output: true
The intervals {1, 3} and {2, 4} overlap
Input: arr[] = {{1, 3}, {7, 9}, {4, 6}, {10, 13}}
Output: false
No pair of intervals overlap.
Expected time complexity is O(nLogn) where n is number of intervals.
We strongly recommend to minimize your browser and try this yourself first.
A Simple Solution is to consider every pair of intervals and check if the pair overlaps or not. The time complexity of this solution is O(n2)
Method 1
A better solution is to Use Sorting. Following is complete algorithm.
1) Sort all intervals in increasing order of start time. This step takes O(nLogn) time.
2) In the sorted array, if start time of an interval is less than end of previous interval, then there is an overlap. This step takes O(n) time.
So overall time complexity of the algorithm is O(nLogn) + O(n) which is O(nLogn).
Below is C++ implementation of above idea.
// A C++ program to check if any two intervals overlap #include <algorithm> #include <iostream> using namespace std; // An interval has start time and end time struct Interval { int start; int end; }; // Compares two intervals according to their staring time. // This is needed for sorting the intervals using library // function std::sort(). See http:// goo.gl/iGspV bool compareInterval(Interval i1, Interval i2) { return (i1.start < i2.start) ? true : false; } // Function to check if any two intervals overlap bool isOverlap(Interval arr[], int n) { // Sort intervals in increasing order of start time sort(arr, arr + n , compareInterval); // In the sorted array, if start time of an interval // is less than end of previous interval, then there // is an overlap for (int i = 1; i < n; i++) if (arr[i - 1].end > arr[i].start) return true; // If we reach here, then no overlap return false; } // Driver program int main() { Interval arr1[] = { { 1, 3 }, { 7, 9 }, { 4, 6 }, { 10, 13 } }; int n1 = sizeof(arr1) / sizeof(arr1[0]); isOverlap(arr1, n1) ? cout << "Yes\n" : cout << "No\n"; Interval arr2[] = { { 6, 8 }, { 1, 3 }, { 2, 4 }, { 4, 7 } }; int n2 = sizeof(arr2) / sizeof(arr2[0]); isOverlap(arr2, n2) ? cout << "Yes\n" : cout << "No\n"; return 0; } |
Output:
No Yes
Method 2: This approach is suggested by Anjali Agarwal. Following are the steps:
1. Find the overall maximum element. Let it be max_ele
2. Initialize an array of size max_ele with 0.
3. For every interval [start, end], increment the value at index start, i.e. arr[start]++ and decrement the value at index (end + 1), i.e. arr[end + 1]- -.
4. Compute the prefix sum of this array (arr[]).
5. Every index, i of this prefix sum array will tell how many times i has occurred in all the intervals taken together. If this value is greater than 1, then it occurs in 2 or more intervals.
6. So, simply initialize the result variable as false and while traversing the prefix sum array, change the result variable to true whenever the value at that index is greater than 1.
Below is the implementation of this (Method 2) approach.
C++
// A C++ program to check if any two intervals overlap #include <algorithm> #include <iostream> using namespace std; // An interval has start time and end time struct Interval { int start; int end; }; // Function to check if any two intervals overlap bool isOverlap(Interval arr[], int n) { int max_ele = 0; // Find the overall maximum element for (int i = 0; i < n; i++) { if (max_ele < arr[i].end) max_ele = arr[i].end; } // Initialize an array of size max_ele int aux[max_ele + 1] = { 0 }; for (int i = 0; i < n; i++) { // starting point of the interval int x = arr[i].start; // end point of the interval int y = arr[i].end; aux[x]++, aux[y + 1]--; } for (int i = 1; i <= max_ele; i++) { // Calculating the prefix Sum aux[i] += aux[i - 1]; // Overlap if (aux[i] > 1) return true; } // If we reach here, then no Overlap return false; } // Driver program int main() { Interval arr1[] = { { 1, 3 }, { 7, 9 }, { 4, 6 }, { 10, 13 } }; int n1 = sizeof(arr1) / sizeof(arr1[0]); isOverlap(arr1, n1) ? cout << "Yes\n" : cout << "No\n"; Interval arr2[] = { { 6, 8 }, { 1, 3 }, { 2, 4 }, { 4, 7 } }; int n2 = sizeof(arr2) / sizeof(arr2[0]); isOverlap(arr2, n2) ? cout << "Yes\n" : cout << "No\n"; return 0; } // This Code is written by Anjali Agarwal |
Java
// A Java program to check if any two intervals overlap class GFG { // An interval has start time and end time static class Interval { int start; int end; public Interval(int start, int end) { super(); this.start = start; this.end = end; } }; // Function to check if any two intervals overlap static boolean isOverlap(Interval arr[], int n) { int max_ele = 0; // Find the overall maximum element for (int i = 0; i < n; i++) { if (max_ele < arr[i].end) max_ele = arr[i].end; } // Initialize an array of size max_ele int []aux = new int[max_ele + 1]; for (int i = 0; i < n; i++) { // starting point of the interval int x = arr[i].start; // end point of the interval int y = arr[i].end; aux[x]++; aux[y ]--; } for (int i = 1; i <= max_ele; i++) { // Calculating the prefix Sum aux[i] += aux[i - 1]; // Overlap if (aux[i] > 1) return true; } // If we reach here, then no Overlap return false; } // Driver program public static void main(String[] args) { Interval arr1[] = { new Interval(1, 3), new Interval(7, 9), new Interval(4, 6), new Interval(10, 13) }; int n1 = arr1.length; if(isOverlap(arr1, n1)) System.out.print("Yes\n"); else System.out.print("No\n"); Interval arr2[] = { new Interval(6, 8), new Interval(1, 3), new Interval(2, 4), new Interval(4, 7) }; int n2 = arr2.length; if(isOverlap(arr2, n2)) System.out.print("Yes\n"); else System.out.print("No\n"); } } // This code is contributed by 29AjayKumar |
C#
// C# program to check if // any two intervals overlap using System; class GFG { // An interval has start time and end time class Interval { public int start; public int end; public Interval(int start, int end) { this.start = start; this.end = end; } }; // Function to check if // any two intervals overlap static bool isOverlap(Interval []arr, int n) { int max_ele = 0; // Find the overall maximum element for (int i = 0; i < n; i++) { if (max_ele < arr[i].end) max_ele = arr[i].end; } // Initialize an array of size max_ele int []aux = new int[max_ele + 1]; for (int i = 0; i < n; i++) { // starting point of the interval int x = arr[i].start; // end point of the interval int y = arr[i].end; aux[x]++; aux[y ]--; } for (int i = 1; i <= max_ele; i++) { // Calculating the prefix Sum aux[i] += aux[i - 1]; // Overlap if (aux[i] > 1) return true; } // If we reach here, then no Overlap return false; } // Driver Code public static void Main(String[] args) { Interval []arr1 = { new Interval(1, 3), new Interval(7, 9), new Interval(4, 6), new Interval(10, 13) }; int n1 = arr1.Length; if(isOverlap(arr1, n1)) Console.Write("Yes\n"); else Console.Write("No\n"); Interval []arr2 = { new Interval(6, 8), new Interval(1, 3), new Interval(2, 4), new Interval(4, 7) }; int n2 = arr2.Length; if(isOverlap(arr2, n2)) Console.Write("Yes\n"); else Console.Write("No\n"); } } // This code is contributed by Rajput-Ji |
Output:
No Yes
Time Complexity : O(max_ele + n)
Note: This method is more efficient than Method 1 if there are more number of intervals and at the same time maximum value among all intervals should be low, since time complexity is directly proportional to O(max_ele).
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