Kth Smallest Element of a Matrix of given dimensions filled with product of indices

Given an integer K and a matrix of size N x M, where each matrix element is equal to the product of its indices(i * j), the task is to find the K^th Smallest element in the given Matrix.
Examples:

Input: N = 2, M = 3, K = 5 
Output: 4 
Explanation:
The matrix possible for given dimensions is {{1, 2, 3}, {2, 4, 6}} 
Sorted order of the matrix elements: {1, 2, 2, 3, 4, 6} 
Therefore, the 5^th smallest element is 4.
Input: N = 1, M = 10, K = 8 
Output: 8 
Explanation:
The matrix possible for given dimensions is {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}} 
Therefore, the 8^th smallest element is 8. 

Naive Approach: The simplest approach is to store all the elements of the matrix in an array and then find the K^th smallest element by sorting the array. 
Time Complexity: O(N×M×log(N×M)) 
Auxiliary Space: O(N×M)
Efficient Approach:
To optimize the naive approach the idea is to use the Binary Search algorithm. Follow the steps below to solve the problem:

1. Initialize low as 1 and high as N×M, as the K^th smallest element lies between 1 and N×M. 
2. Find the mid element between the low and high elements.
3. If the number of elements less than mid is greater than or equal to K, then update high to mid-1 as the Kth smallest element lies between low and mid.
4. If the number of elements less than mid is less than K, then update low to mid+1 as the Kth smallest element lies between mid and high.
5. As the elements in the ith row are the multiple of i, the number of elements less than mid in the ith row can be calculated easily by min(mid/i, M). So, the time complexity to find the number of elements less than mid can be done in only O(N).
6. Perform binary search till low is less than or equal to high and return high + 1 as the Kth smallest element of the matrix N×M.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

#define LL long long

// Function that returns true if the
// count of elements is less than mid
bool countLessThanMid(LL mid, LL N,
                    LL M, LL K)
{
    // To store count of elements
    // less than mid
    LL count = 0;

    // Loop through each row
    for (int i = 1;
        i <= min((LL)N, mid); ++i) {

        // Count elements less than
        // mid in the ith row
        count = count + min(mid / i, M);
    }

    if (count >= K)
        return false;
    else
        return true;
}

// Function that returns the Kth
// smallest element in the NxM
// Matrix after sorting in an array
LL findKthElement(LL N, LL M, LL K)
{
    // Initialize low and high
    LL low = 1, high = N * M;

    // Perform binary search
    while (low <= high) {

        // Find the mid
        LL mid = low + (high - low) / 2;

        // Check if the count of
        // elements is less than mid
        if (countLessThanMid(mid, N, M, K))
            low = mid + 1;
        else
            high = mid - 1;
    }

    // Return Kth smallest element
    // of the matrix
    return high + 1;
}

// Driver Code
int main()
{
    LL N = 2, M = 3;

    LL int K = 5;

    cout << findKthElement(N, M, K) << endl;

    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{

// Function that returns true if the
// count of elements is less than mid
public static boolean countLessThanMid(int mid, int N,
                                       int M, int K)
{

    // To store count of elements
    // less than mid
    int count = 0;

    // Loop through each row
    for(int i = 1;
            i <= Math.min(N, mid); ++i)
    {

        // Count elements less than
        // mid in the ith row
        count = count + Math.min(mid / i, M);
    }

    if (count >= K)
        return false;
    else
        return true;
}

// Function that returns the Kth
// smallest element in the NxM
// Matrix after sorting in an array
public static int findKthElement(int N, int M, int K)
{

    // Initialize low and high
    int low = 1, high = N * M;

    // Perform binary search
    while (low <= high)
    {

        // Find the mid
        int mid = low + (high - low) / 2;

        // Check if the count of
        // elements is less than mid
        if (countLessThanMid(mid, N, M, K))
            low = mid + 1;
        else
            high = mid - 1;
    }

    // Return Kth smallest element
    // of the matrix
    return high + 1;
}

// Driver code
public static void main(String[] args)
{
    int N = 2, M = 3;
    int K = 5;

    System.out.println(findKthElement(N, M, K));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
# Function that returns true if the
# count of elements is less than mid
def countLessThanMid(mid, N, M, K):

    # To store count of elements
    # less than mid
    count = 0

    # Loop through each row
    for i in range (1, min(N, mid) + 1):

        # Count elements less than
        # mid in the ith row
        count = count + min(mid // i, M)

    if (count >= K):
        return False
    else:
        return True

# Function that returns the Kth
# smallest element in the NxM
# Matrix after sorting in an array
def findKthElement(N, M, K):

    # Initialize low and high
    low = 1
    high = N * M

    # Perform binary search
    while (low <= high):

        # Find the mid
        mid = low + (high - low) // 2

        # Check if the count of
        # elements is less than mid
        if (countLessThanMid(mid, N, M, K)):
            low = mid + 1
        else:
            high = mid - 1

    # Return Kth smallest element
    # of the matrix
    return high + 1

# Driver Code
if __name__ == "__main__":
    N = 2
    M = 3
    K = 5
    print(findKthElement(N, M, K))

# This code is contributed by Chitranayal

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function that returns true if the
// count of elements is less than mid
public static bool countLessThanMid(int mid, int N,
                                    int M, int K)
{

    // To store count of elements
    // less than mid
    int count = 0;

    // Loop through each row
    for(int i = 1;
            i <= Math.Min(N, mid); ++i)
    {

        // Count elements less than
        // mid in the ith row
        count = count + Math.Min(mid / i, M);
    }

    if (count >= K)
        return false;
    else
        return true;
}

// Function that returns the Kth
// smallest element in the NxM
// Matrix after sorting in an array
public static int findKthElement(int N, int M,
                                        int K)
{

    // Initialize low and high
    int low = 1, high = N * M;

    // Perform binary search
    while (low <= high)
    {

        // Find the mid
        int mid = low + (high - low) / 2;

        // Check if the count of
        // elements is less than mid
        if (countLessThanMid(mid, N, M, K))
            low = mid + 1;
        else
            high = mid - 1;
    }

    // Return Kth smallest element
    // of the matrix
    return high + 1;
}

// Driver code
public static void Main(String[] args)
{
    int N = 2, M = 3;
    int K = 5;

    Console.WriteLine(findKthElement(N, M, K));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript program for the above approach

// Function that returns true if the
// count of elements is less than mid
function countLessThanMid(mid, N, M, K)
{
    // To store count of elements
    // less than mid
    let count = 0;

    // Loop through each row
    for (let i = 1;
        i <= Math.min(N, mid); ++i) {

        // Count elements less than
        // mid in the ith row
        count = count + Math.min(parseInt(mid / i), M);
    }

    if (count >= K)
        return false;
    else
        return true;
}

// Function that returns the Kth
// smallest element in the NxM
// Matrix after sorting in an array
function findKthElement(N, M, K)
{
    // Initialize low and high
    let low = 1, high = N * M;

    // Perform binary search
    while (low <= high) {

        // Find the mid
        let mid = low + parseInt((high - low) / 2);

        // Check if the count of
        // elements is less than mid
        if (countLessThanMid(mid, N, M, K))
            low = mid + 1;
        else
            high = mid - 1;
    }

    // Return Kth smallest element
    // of the matrix
    return high + 1;
}

// Driver Code
    let N = 2, M = 3;

    let K = 5;

    document.write(findKthElement(N, M, K));

</script>

Output: 

4

Time Complexity: O(N×log(N×M)) 
Auxiliary Space: O(1)