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 Arthmetic progression increasing difference where common difference is 1.
- Intial Difference Term =

**Examples:**

Input:R = 4, C = 4Output:25Explanation:

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 = 3Output:13Explanation:

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` |

**Output:**

25

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