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
- Construct a matrix of size n*n.
- Fill the value with M(i, j)=i+j.(recall that here base index is 1)
- Iteratively traverse the matrix and count the frequency of a given element.
This method is not that much efficient because if the matrix size is very large it’s will result in a Time limit exceeded. Time complexity will be O(n^2).
Efficient method
In this method, we avoid creating a 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 the 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 the frequency of k is k-1
Else frequency will be 2*n+1-k
Implementation:
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 Codeint 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+jimport 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+jimport mathdef find( n, k): if (n + 1 >= k): return (k - 1) else: return (2 * n + 1 - k) # Driver Coden = 4k = 7freq = 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+jusing 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+jfunction 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.?> |
Javascript
<script>// Javascript program to find the frequency of k // in matrix where m(i, j)=i+jfunction find(n, k){ if (n + 1 >= k) return (k - 1); else return (2 * n + 1 - k);}// Driver Codevar n = 4, k = 7;var freq = find(n, k);if (freq < 0) document.write( " element not exist <br>");else document.write( " Frequency of " + k + " is " + freq + "<br>");</script> |
Frequency of 7 is 2
Time Complexity: O(1)
Auxiliary Space: O(1)
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