Find the element at R’th row and C’th column in given a 2D pattern

Given two integers R and C, the task is to find the element at the R^th row and C^th column.
Pattern:

• First Element of i^th row =
• Every element is a Arithmetic progression increasing difference where common difference is 1.
• Initial Difference Term = 

Examples:

Input: R = 4, C = 4 
Output: 25 
Explanation:
Pattern of size 4 * 4 is – 
1 3 6 10 
2 5 9 14 
4 8 13 19 
7 12 18 25 
Therefore, Element at Pat[4][4] = 25
Input: R = 3, C = 3 
Output: 13 
Explanation:
Pattern of size 3 * 3 is – 
1 3 6 
2 5 9 
4 8 13 
Therefore, element at Pat[3][3] = 13 

Naive Approach: A simple solution is to generate the pattern matrix of size R * C and then finally return the element at the R^th row and C^th column.
Time Complexity: O(R*C) 
Auxiliary Space: O(R*C)
Efficient Approach: The idea is to find the first term of the R^th row using the formulae and then finally compute the C^th term of that column using the help of loop.
Below is the implementation of the above approach:

C++

// C++ implementation to compute the
// R'th row and C'th column of the
// given pattern

#include <bits/stdc++.h>
using namespace std;

// Function to compute the
// R'th row and C'th column of the
// given pattern
int findValue(int R, int C)
{

    // First element of a given row
    int k = (R * (R - 1)) / 2 + 1;

    int diff = R + 1;

    // Element in the given column
    for (int i = 1; i < C; i++) {
        k = (k + diff);
        diff++;
    }

    return k;
}

// Driver Code
int main()
{
    int R = 4;
    int C = 4;

    // Function call
    int k = findValue(R, C);

    cout << k;

    return 0;
}

Java

// Java implementation to compute the
// R'th row and C'th column of the
// given pattern
import java.io.*;

class GFG{

// Function to compute the R'th
// row and C'th column of the
// given pattern
static int findValue(int R, int C)
{

    // First element of a given row
    int k = (R * (R - 1)) / 2 + 1;

    int diff = R + 1;

    // Element in the given column
    for(int i = 1; i < C; i++)
    {
       k = (k + diff);
       diff++;
    }
    return k;
}

// Driver code
public static void main (String[] args)
{
    int R = 4;
    int C = 4;

    // Function call
    int k = findValue(R, C);

    System.out.println(k);
}
}

// This code is contributed by mohit kumar 29

Python3

# Python3 implementation to find the
# R'th row and C'th column value in
# the given pattern

# Function to find the
# R'th row and C'th column value in
# the given pattern
def findValue(R, C):

    # First element of a given row
    k = (R*(R-1))//2 + 1

    diff = R + 1

    # Element in the given column
    for i in range(1, C):
        k = (k + diff)
        diff+= 1

    return k

# Driver Code
if __name__ == "__main__":
    R = 4
    C = 4

    k = findValue(R, C)
    print(k)

C#

// C# implementation to compute the
// R'th row and C'th column of the
// given pattern
using System;
class GFG{

// Function to compute the R'th
// row and C'th column of the
// given pattern
static int findValue(int R, int C)
{

    // First element of a given row
    int k = (R * (R - 1)) / 2 + 1;

    int diff = R + 1;

    // Element in the given column
    for(int i = 1; i < C; i++)
    {
        k = (k + diff);
        diff++;
    }
    return k;
}

// Driver code
public static void Main()
{
    int R = 4;
    int C = 4;

    // Function call
    int k = findValue(R, C);

    Console.Write(k);
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// Javascript implementation to compute the
// R'th row and C'th column of the
// given pattern

// Function to compute the R'th
// row and C'th column of the
// given pattern
function findValue(R, C)
{

    // First element of a given row
    let k = (R * (R - 1)) / 2 + 1;

    let diff = R + 1;

    // Element in the given column
    for(let i = 1; i < C; i++)
    {
       k = (k + diff);
       diff++;
    }
    return k;
}

  // Driver Code

    let R = 4;
    let C = 4;

    // Function call
    let k = findValue(R, C);

     document.write(k);

</script>

Output:

25

Time complexity :  O(C), where C is the column number. The code iterates C times to find the value in the specified column.

Space complexity : O(1), as it only uses a few variables and arrays of constant size, which means the space required does not increase with increasing input size.