You are given a square matrix (matrix[][]) of order n, where matrix[i][j] = i*j. Find the number of cells which have entry equal to a given number x.
NOte : Indexing of matrix starts from 1, i.e. 1<= i,j <= n.
Examples :
Input : matrix[][] = {1, 2, 3, 4,
2, 4, 6, 8,
3, 6, 9, 12,
4, 8, 12, 16}
x = 4
Output : 3
Input : matrix[][] = {1, 2, 3, 4,
2, 4, 6, 8,
3, 6, 9, 12,
4, 8, 12, 16}
x = 12
Output : 2
A simple Approach is to traverse the whole of matrix and check whether cell value is equal to given x and then increase count value accordingly. Time complexity in this approach is quite high and is equal to O(n^2).
// traverse and check whole matrix
for (int i=1; i<=n ; i++)
{
for (int j=1; j<=n; j++)
if (i * j == x)
count++;
}
// return count
return count;
An efficient approach is to only find the number of factors of given x in the range 0 to x and also those factors (including divisor and quotient ) must be less than or equal to n (order of matrix). In this case time complexity will be O(n).
// traverse and find the factors
for (int i=1; i<=n && i<=x ; i++)
{
// x%i == 0 means i is factor of x
// x/i <= n means i and j are <= n (for i*j=x)
if ( x/i <= n && x%i ==0)
count++;
}
// return count
return count;
C++
// CPP program for counting number of cell // equals to given x #include<bits/stdc++.h> using namespace std; // function to count factors as number of cell int count (int n, int x) { int count == 0; // traverse and find the factors for (int i=1; i<=n && i<=x ; i++) { // x%i == 0 means i is factor of x // x/i <= n means i and j are <= n (for i*j=x) if ( x/i <= n && x%i ==0) count++; } // return count return count; } // driver program int main() { int n = 8; // we can manually assume matrix of order 8*8 // where mat[i][j] = i*j , 0<i,j<=n int x = 24; cout << count(n, x); return 0; } |
Java
// Java program for counting number of // cell equals to given x class GFG { // function to count factors as // number of cell static int count (int n, int x) { int count = 0; // traverse and find the factors for (int i = 1; i <= n && i <= x ; i++) { // x%i == 0 means i is factor // of x. x/i <= n means i and // j are <= n (for i*j=x) if ( x / i <= n && x % i == 0) count++; } // return count return count; } // driver program public static void main(String args[]) { int n = 8; // we can manually assume matrix // of order 8*8 where // mat[i][j] = i*j , 0<i,j<=n int x = 24; System.out.println(count(n, x)); } } /*This code is contributed by Danish kaleem*/ |
Python3
# Python 3 program for counting # number of cell equals to given x # function to count factors # as number of cell def count(n, x): cnt = 0 # traverse and find the factors for i in range(1, n + 1): # // x%i == 0 means i is factor of x # x/i <= n means i and j are <= n (for i*j=x) if i <= x: if x // i <= n and x % i == 0: cnt += 1 return cnt # Driver code n = 8x = 24print(count(n, x)) # This code is contributed by Shrikant13 |
C#
// C# program for counting number // of cell equals to given x using System; class GFG { // function to count factors as // number of cell static int count (int n, int x) { int count = 0; // traverse and find the factors for (int i = 1; i <= n && i <= x ; i++) { // x%i == 0 means i is factor // of x. x/i <= n means i and // j are <= n (for i*j=x) if ( x / i <= n && x % i == 0) count++; } // return count return count; } // Driver Code public static void Main() { int n = 8; // we can manually assume matrix // of order 8*8 where // mat[i][j] = i*j , 0<i,j<=n int x = 24; Console.Write(count(n, x)); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP program for counting // number of cell equals to // given x // function to count factors // as number of cell function c_ount ( $n, $x) { $Count = 0; // traverse and find the factors for ( $i = 1; $i <= $n and $i <= $x ; $i++) { // x%i == 0 means i is // factor of x x/i <= n // means i and j are // <= n (for i*j=x) if ( $x / $i <= $n and $x % $i == 0) $Count++; } // return count return $Count; } // Driver Code $n = 8; // we can manually assume // matrix of order 8*8 // where mat[i][j] = i*j , // 0<i,j<=n $x = 24; echo c_ount($n, $x); // This code is contributed by anuj_67. ?> |
Output :
4
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