Find k-th smallest element in given n ranges

Given n and q, i.e, the number of ranges and number of queries, find the kth smallest element for each query (assume k>1).Print the value of kth smallest element if it exists, else print -1.

Examples :

Input : arr[] = {{1, 4}, {6, 8}}
queries[] = {2, 6, 10};
Output : 2
7
-1
After combining the given ranges, the numbers
become 1 2 3 4 6 7 8. As here 2nd element is 2,
so we print 2. As 6th element is 7, so we print
7 and as 10th element doesn't exist, so we
print -1.

Input : arr[] = {{2, 6}, {5, 7}}
queries[] = {5, 8};
Output : 6
-1
After combining the given ranges, the numbers
become 2 3 4 5 6 7. As here 5th element is 6,
so we print 6 and as 8th element doesn't exist,
so we print -1.

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

The idea is to first Prerequisite : Merge Overlapping Intervals and keep all intervals sorted in ascending order of start time. After merging in an array merged[], we use linear search to find kth smallest element. Below is the implementation of the above approach :

 // C++ implementation to solve k queries // for given n ranges #include using namespace std;    // Structure to store the // start and end point struct Interval {     int s;     int e; };    // Comparison function for sorting bool comp(Interval a, Interval b) {     return a.s < b.s; }    // Function to find Kth smallest number in a vector // of merged intervals int kthSmallestNum(vector merged, int k) {     int n = merged.size();        // Traverse merged[] to find     // Kth smallest element using Linear search.     for (int j = 0; j < n; j++)     {         if (k <= abs(merged[j].e -                      merged[j].s + 1))             return (merged[j].s + k - 1);            k = k - abs(merged[j].e -                      merged[j].s + 1);     }        if (k)         return -1; }    // To combined both type of ranges, // overlapping as well as non-overlapping. void mergeIntervals(vector &merged,                  Interval arr[], int n) {     // Sorting intervals according to start     // time     sort(arr, arr + n, comp);        // Merging all intervals into merged     merged.push_back(arr);     for (int i = 1; i < n; i++)     {         // To check if starting point of next         // range is lying between the previous         // range and ending point of next range         // is greater than the Ending point         // of previous range then update ending         // point of previous range by ending         // point of next range.         Interval prev = merged.back();         Interval curr = arr[i];         if ((curr.s >= prev.s &&              curr.s <= prev.e) &&             (curr.e > prev.e))                merged.back().e = curr.e;            else         {             // If starting point of next range             // is greater than the ending point             // of previous range then store next range             // in merged[].             if (curr.s > prev.e)                 merged.push_back(curr);         }     } }    // Driver\'s Function int main() {     Interval arr[] = {{2, 6}, {4, 7}};     int n = sizeof(arr)/sizeof(arr);     int query[] = {5, 8};     int q = sizeof(query)/sizeof(query);        // Merge all intervals into merged[]     vectormerged;     mergeIntervals(merged, arr, n);        // Processing all queries on merged     // intervals     for (int i = 0; i < q; i++)         cout << kthSmallestNum(merged, query[i])              << endl;        return 0; }

Output:

6
-1

Time Complexity : O(nlog(n))

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