Given some rules to generate an N * N matrix mat[][] and two integers R and C, the task is to find the element at the Rth row and Cth column. The rules are as follows:
- First row is an AP series starting with 1 and d = 1 (Common Difference).
- For all the element at (i, j) such that i > j, their value is 0.
- Rest of the elements follow, Element(i, j) = Element(i – 1, j) + Element(i – 1, j – 1).
Examples:
Input: N = 4, R = 3, C = 4
Output: 12
mat[][] =
{1, 2, 3, 4},
{0, 3, 5, 7},
{0, 0, 8, 12},
{0, 0, 0, 20}
and the element in the third row and fourth column is 12.Input: N = 2, R = 2, C = 2
Output: 3
Approach:
- Basic Key is to observe the pattern, first observation is that every row will have an AP after the element at (i, i). Common difference for row number j is pow(2, j – 1).
- Now we need to find the first term of each row to find any element (R, C). If we consider only starting elements in each row (i.e element at diagonal), we can observe it is equal to (R + 1) * pow(2, R – 2) for every row R from 2 to N.
- So, If R > C then element is 0 else C – R is the position of required element in AP which is present in the Rth row for which we already know the starting term and the common difference. So, we can find element as start + d * (C – R).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the element in the rth row // and cth column from the required matrix int getElement(int N, int r, int c) { // Condition for lower half of matrix if (r > c) return 0; // Condition if element is in first row if (r == 1) { return c; } // Starting element of AP in row r int a = (r + 1) * pow(2, r - 2); // Common difference of AP in row r int d = pow(2, r - 1); // Position of element to find // in AP in row r c = c - r; int element = a + d * c; return element; } // Driver Code int main() { int N = 4, R = 3, C = 4; cout << getElement(N, R, C); return 0; } |
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Java
// Java implementation of the above approach import java.io.*; import java.util.*; class GFG { // Function to return the element // in the rth row and cth column // from the required matrix static int getElement(int N, int r, int c) { // Condition for lower half of matrix if (r > c) return 0; // Condition if element is in first row if (r == 1) { return c; } // Starting element of AP in row r int a = (r + 1) * (int)(Math.pow(2,(r - 2))); // Common difference of AP in row r int d = (int)(Math.pow(2,(r - 1))); // Position of element to find // in AP in row r c = c - r; int element = a + d * c; return element; } // Driver Code public static void main(String[] args) { int N = 4, R = 3, C = 4; System.out.println(getElement(N, R, C)); } } // This code is contributed by Krikti |
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Python3
# Python3 implementation of the approach # Function to return the element in the rth row # and cth column from the required matrix def getElement(N, r, c) : # Condition for lower half of matrix if (r > c) : return 0; # Condition if element is in first row if (r == 1) : return c; # Starting element of AP in row r a = (r + 1) * pow(2, r - 2); # Common difference of AP in row r d = pow(2, r - 1); # Position of element to find # in AP in row r c = c - r; element = a + d * c; return element; # Driver Code if __name__ == "__main__" : N = 4; R = 3; C = 4; print(getElement(N, R, C)); # This Code is contributed by AnkitRai01 |
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C#
// C# implementation of the above approach using System; class GFG { // Function to return the element // in the rth row and cth column // from the required matrix static int getElement(int N, int r, int c) { // Condition for lower half of matrix if (r > c) return 0; // Condition if element is in first row if (r == 1) { return c; } // Starting element of AP in row r int a = (r + 1) * (int)(Math.Pow(2,(r - 2))); // Common difference of AP in row r int d = (int)(Math.Pow(2,(r - 1))); // Position of element to find // in AP in row r c = c - r; int element = a + d * c; return element; } // Driver Code public static void Main(String[] args) { int N = 4, R = 3, C = 4; Console.WriteLine(getElement(N, R, C)); } } // This code is contributed by Princi Singh |
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Output:
12
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