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Find m-th smallest value in k sorted arrays in C++
In this problem, we are given k different arrays of different sizes. Our task is to find m-th smallest value in k sorted arrays.
Problem Description: Here, we need to find m-th smallest element of the merged array of all k arrays.
Let’s take an example to understand the problem,
Input: m = 4
arr[][] = { {4 , 7},
{2, 5, 6},
{3, 9, 12, 15, 19}}
Output: 5
Explanation:
Merged Sorted array : 2, 3, 4, 5, 6, 7, 9, 12, 15, 19
The 4th element is 5.
Solution approach:
A simple solution to find the m-th smallest elements is by creating a merged array and then sort the array in ascending order which will have the m-th smallest element of the array at index (m-1). Return this output value.
A more effective solution would be using the min heap data structure.
For this, we will create a min heap and insert elements from all arrays. And then m times remove the min element element from the heap and store the output to array. Then insert the next element to the heap.
Print the m-th removed item.
Program to illustrate the working of our solution,
Example
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, pair<int, int> > ppi;
int findMSmallestElement(vector<vector<int> > sortedArr, int m) {
priority_queue<ppi, vector<ppi>, greater<ppi> > priorQueue;
for (int i = 0; i < sortedArr.size(); i++)
priorQueue.push({ sortedArr[i][0], { i, 0 } });
int count = 0;
int i = 0, j = 0;
while (count < m && priorQueue.empty() == false) {
ppi curr = priorQueue.top();
priorQueue.pop();
i = curr.second.first;
j = curr.second.second;
if (j + 1 < sortedArr[i].size())
priorQueue.push( { sortedArr[i][j + 1], { i, (j + 1) } });
count++;
}
return sortedArr[i][j];
}
int main() {
vector<vector<int> > arr{ {4 , 7},
{2, 5, 6},
{3, 9, 12, 15, 19}};
int m = 6;
cout<<m<<"th smallest value in k sorted arrays is "<<findMSmallestElement(arr, m);
return 0;
}Output
6th smallest value in k sorted arrays is 7
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