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Check if any interval completely overlaps the other

  • Difficulty Level : Easy
  • Last Updated : 12 Jan, 2022

An interval is represented as a combination of start time and end time. Given a set of intervals, we need to write a program to check if any interval completely overlaps the other. 

Examples: 

Input:  arr[] = {{1, 3}, {1, 7}, {4, 8}, {2, 5}}
Output: true
The intervals {1, 3} completely overlaps in {1, 7}. 

Input:  arr[] = {{1, 3}, {7, 9}, {4, 6}, {10, 13}}
Output: false
No pair of intervals overlap. 

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). 

A better solution is to use Sorting. Following is the complete algorithm. 
1) Sort all intervals in increasing order of start time. This step takes O(n Logn) time. 
2) In the sorted array, if the end time of an interval is not more than the end of the previous interval, then there is a complete overlap. This step takes O(n) time. 

Given below is an implementation of the above approach:  

C++




// A C++ program to check if any two intervals
// completely overlap
#include <bits/stdc++.h>
using namespace std;
 
// An interval has start time and end time
struct Interval {
    int start;
    int end;
};
 
// Compares two intervals according to their starting
// time. This is needed for sorting the intervals
// using library function std::sort().
bool compareInterval(Interval i1, Interval i2)
{
    return (i1.start < i2.start) ? true : false;
}
 
// Function to check if any two intervals
// completely overlap
bool isOverlap(Interval arr[], int n)
{
    // Sort intervals in increasing order of
    // start time
    sort(arr, arr + n - 1, compareInterval);
 
    // In the sorted array, if end time of an
    // interval is not more than that of
    // end of previous interval, then there
    // is an overlap
    for (int i = 1; i < n; i++)
        if (arr[i].end <= arr[i - 1].end)
            return true;
 
    // If we reach here, then no overlap
    return false;
}
 
// Driver code
int main()
{
    // 1st example
    Interval arr1[] = { { 1, 3 }, { 1, 7 }, { 4, 8 },
                                          { 2, 5 } };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    if (isOverlap(arr1, n1))
        cout << "Yes\n";
    else
        cout << "No\n";
 
    // 2nd example
    Interval arr2[] = { { 1, 3 }, { 7, 9 }, { 4, 6 },
                                         { 10, 13 } };
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    if (isOverlap(arr2, n2))
        cout << "Yes\n";
    else
        cout << "No\n";
 
    return 0;
}

Javascript




<script>
    // A JavaScript program to check if any two intervals
    // completely overlap
 
    // An interval has start time and end time
 
    // Compares two intervals according to their starting
    // time. This is needed for sorting the intervals
    // using library function std::sort().
    const compareInterval = (i1, i2) => i1.start - i2.start;
 
    // Function to check if any two intervals
    // completely overlap
    const isOverlap = (arr, n) => {
     
        // Sort intervals in increasing order of
        // start time
        arr.sort(compareInterval);
 
        // In the sorted array, if end time of an
        // interval is not more than that of
        // end of previous interval, then there
        // is an overlap
        for (let i = 1; i < n; i++)
            if (arr[i].end <= arr[i - 1].end)
                return true;
 
        // If we reach here, then no overlap
        return false;
    }
 
    // Driver code
 
    // 1st example
    let arr1 = [
        { start: 1, end: 3 },
        { start: 1, end: 7 },
        { start: 4, end: 8 },
        { start: 2, end: 5 }
    ];
    let n1 = arr1.length;
    if (isOverlap(arr1, n1))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");
 
    // 2nd example
    let arr2 = [
        { start: 1, end: 3 },
        { start: 7, end: 9 },
        { start: 4, end: 6 },
        { start: 10, end: 13 }
    ];
    let n2 = arr2.length;
    if (isOverlap(arr2, n2))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");
 
    // This code is contributed by rakeshsahni
 
</script>

Output: 

Yes
No

 


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