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

  • Last Updated : 08 Jun, 2021

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

  • First Element of ith row =\frac{i*(i-1)}{2} + 1
  • Every element is a Arithmetic progression increasing difference where common difference is 1.
  • Initial Difference Term = i + 1

Examples: 
 

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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 Rth row and Cth column.
Time Complexity: O(R*C) 
Auxiliary Space: O(R*C)
Efficient Approach: The idea is to find the first term of the Rth row using the formulae \frac{R*(R+1)}{2}  and then finally compute the Cth 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

 




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