Find m-th smallest value in k sorted arrays

• Difficulty Level : Hard
• Last Updated : 25 May, 2021

Given k sorted arrays of possibly different sizes, find m-th smallest value in the merged array.
Examples:

Input: m = 5
arr[][] = { {1, 3},
{2, 4, 6},
{0, 9, 10, 11}} ;
Output: 4
Explanation The merged array would
be {0 1 2 3 4 6 9 10 11}.  The 5-th
smallest element in this merged
array is 4.

Input: m = 2
arr[][] = { {1, 3, 20},
{2, 4, 6}} ;
Explanation The merged array would
be {1 2 3 4 6 20}. The 2nd smallest element would be 2.
Output: 2

A simple solution is to create an output array and and one by one copy all arrays to it. Finally, sort the output array using. This approach takes O(N Logn N) time where N is count of all elements.
An efficient solution is to use heap data structure. The time complexity of heap based solution is O(m Log k).
1. Create a min heap of size k and insert 1st element in all the arrays into the heap
2. Repeat following steps m times
…..a) Remove minimum element from heap (minimum is always at root) and store it in output array.
…..b) Insert next element from the array from which the element is extracted. If the array doesn’t have any more elements, then do nothing.
3. Print the last removed item.

CPP

 // C++ program to find m-th smallest element // in the merged arrays. #include using namespace std;   // A pair of pairs, first element is going to // store value, second element index of array // and third element index in the array. typedef pair > ppi;   // This function takes an array of arrays as an // argument and all arrays are assumed to be // sorted. It returns m-th smallest element in // the array obtained after merging the given // arrays. int mThLargest(vector > arr, int m) {     // Create a min heap with k heap nodes. Every     // heap node has first element of an array     priority_queue, greater > pq;       for (int i = 0; i < arr.size(); i++)         pq.push({ arr[i], { i, 0 } });       // Now one by one get the minimum element     // from min heap and replace it with next     // element of its array     int count = 0;     int i = 0, j = 0;     while (count < m && pq.empty() == false) {         ppi curr = pq.top();         pq.pop();           // i ==> Array Number         // j ==> Index in the array number         i = curr.second.first;         j = curr.second.second;           // The next element belongs to same array as         // current.         if (j + 1 < arr[i].size())             pq.push({ arr[i][j + 1], { i, j + 1 } });           count++;     }       return arr[i][j]; }   // Driver program to test above functions int main() {     vector > arr{ { 2, 6, 12 },                               { 1, 9 },                               { 23, 34, 90, 2000 } };       int m = 4;     cout << mThLargest(arr, m);       return 0; }

Output:

9

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