Given a matrix of size n*n. Count the frequency of given element k in that matrix. Here base index is 1.
Examples:
Input : n = 4, k = 7 Output : 2 Explanation The matrix will be 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8 in the given matrix where M(i, j) = i+j, frequency of 7 is 2 Input : n = 5, k = 4 Output : 3 Explanation The matrix will be 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 Explanation In the given matrix where M(i, j) = i+j, frequency of 4 is 3
First method
1) Construct a matrix of size n*n.
2) Fill the value with M(i, j)=i+j.(recall that here base index is 1)
3) Iteratively traverse the matrix and and count the frequency of given element.
This method is not that much efficient because if the matrix size is very large it’s will result in Time limit exceed.time complexity will be O(n^2).
Efficient method
In this method we avoid creating matrix of size n*n.
for example
if n = 10 the matrix will be
2 3 4 5 6 7 8 9 10 11 3 4 5 6 7 8 9 10 11 12 4 5 6 7 8 9 10 11 12 13 5 6 7 8 9 10 11 12 13 14 6 7 8 9 10 11 12 13 14 15 7 8 9 10 11 12 13 14 15 16 8 9 10 11 12 13 14 15 16 17 9 10 11 12 13 14 15 16 17 18 10 11 12 13 14 15 16 17 18 19 11 12 13 14 15 16 17 18 19 20
Now, notice how the values are same in the secondary Diagonal, and we can also find a pattern in the count it increases like 1, 2, 3, 4,
here we can see that
if (n+1)>=k then frequency of k is k-1
else frequency will be 2*n+1-k
CPP
// CPP program to find the frequency of k // in matrix where m(i, j)=i+j #include <bits/stdc++.h> using namespace std; int find(int n, int k) { if (n + 1 >= k) return (k - 1); else return (2 * n + 1 - k); } // Driver Code int main() { int n = 4, k = 7; int freq = find(n, k); if (freq < 0) cout << " element not exist \n "; else cout << " Frequency of " << k << " is " << freq << "\n"; return 0; } |
Java
// Java program to find the // frequency of k in matrix // in matrix where m(i, j)=i+j import java.util.*; import java.lang.*; public class GfG{ public static int find(int n, int k) { if (n + 1 >= k) return (k - 1); else return (2 * n + 1 - k); } // Driver function public static void main(String argc[]) { int n = 4, k = 7; int freq = find(n, k); if (freq < 0) System.out.print(" element" + " not exist \n "); else System.out.print(" Frequency" + " of " + k + " is " + freq + "\n"); } } // This code is contributed by Sagar Shukla |
Python3
# Python program to find # the frequency of k # in matrix where # m(i, j)=i+j import math def find( n, k): if (n + 1 >= k): return (k - 1) else: return (2 * n + 1 - k) # Driver Code n = 4k = 7 freq = find(n, k) if (freq < 0): print ( " element not exist") else: print(" Frequency of " , k ," is " , freq ) # This code is contributed # by Gitanjali. |
C#
// C# program to find the // frequency of k in matrix // in matrix where m(i, j)=i+j using System; public class GfG{ public static int find(int n, int k) { if (n + 1 >= k) return (k - 1); else return (2 * n + 1 - k); } // Driver function public static void Main() { int n = 4, k = 7; int freq = find(n, k); if (freq < 0) Console.WriteLine(" element" + " not exist "); else Console.WriteLine(" Frequency" + " of " + k + " is " + freq ); } } // This code is contributed by vt_m |
PHP
<?php // PHP program to find the frequency of k // in matrix where m(i, j)=i+j function find($n, $k) { if ($n + 1 >= $k) return ($k - 1); else return (2 * $n + 1 - $k); } // Driver Code $n = 4; $k = 7; $freq = find($n, $k); if ($freq < 0) echo " element not exist \n "; else echo " Frequency of " , $k , " is " , $freq , "\n"; // This code is contributed by anuj_67. ?> |
Output:
Frequency of 7 is 2
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Improved By : vt_m
