Ways of filling matrix such that product of all rows and all columns are equal to unity
Last Updated :
25 Jul, 2022
We are given three values n , m and k where n is number of rows in matrix, m is number of columns in the matrix and k is the number that can have only two values -1 and 1. Our aim is to find the number of ways of filling the matrix of n \times m such that the product of all the elements in each row and each column is equal to k . Since the number of ways can be large we will output ans \mod{1000000007}
Examples:
Input : n = 2, m = 4, k = -1
Output : 8
Following configurations satisfy the conditions:-
Input : n = 2, m = 1, k = -1
Output : The number of filling the matrix
are 0
From the above conditions, it is clear that the only elements that can be entered in the matrix are 1 and -1. Now we can easily deduce some of the corner cases
- If k = -1, then the sum of number of rows and columns cannot be odd because -1 will be present odd
number of times in each row and column therefore if the sum is odd then answer is 0 . - If n = 1 or m = 1 then there is only one way of filling the matrix therefore answer is 1.
- If none of the above cases are applicable then we fill the first n-1 rows and the first m-1 columns with 1 and -1. Then the remaining numbers can be uniquely identified since the product of each row an each column is already known therefore the answer is 2 ^ {(n-1) \times (m-1)} .
Implementation:
C++
// CPP program to find number of ways to fill
// a matrix under given constraints
#include <bits/stdc++.h>
using namespace std;
#define mod 100000007
/* Returns a raised power t under modulo mod */
long long modPower(long long a, long long t)
{
long long now = a, ret = 1;
// Counting number of ways of filling the matrix
while (t) {
if (t & 1)
ret = now * (ret % mod);
now = now * (now % mod);
t >>= 1;
}
return ret;
}
// Function calculating the answer
long countWays(int n, int m, int k)
{
// if sum of numbers of rows and columns is odd
// i.e (n + m) % 2 == 1 and k = -1 then there
// are 0 ways of filiing the matrix.
if (k == -1 && (n + m) % 2 == 1)
return 0;
// If there is one row or one column then there
// is only one way of filling the matrix
if (n == 1 || m == 1)
return 1;
// If the above cases are not followed then we
// find ways to fill the n - 1 rows and m - 1
// columns which is 2 ^ ((m-1)*(n-1)).
return (modPower(modPower((long long)2, n - 1),
m - 1) % mod);
}
// Driver function for the program
int main()
{
int n = 2, m = 7, k = 1;
cout << countWays(n, m, k);
return 0;
}
// Java program to find number of ways to fill
// a matrix under given constraints
import java.io.*;
class Example {
final static long mod = 100000007;
/* Returns a raised power t under modulo mod */
static long modPower(long a, long t, long mod)
{
long now = a, ret = 1;
// Counting number of ways of filling the
// matrix
while (t > 0) {
if (t % 2 == 1)
ret = now * (ret % mod);
now = now * (now % mod);
t >>= 1;
}
return ret;
}
// Function calculating the answer
static long countWays(int n, int m, int k)
{
// if sum of numbers of rows and columns is
// odd i.e (n + m) % 2 == 1 and k = -1,
// then there are 0 ways of filiing the matrix.
if (n == 1 || m == 1)
return 1;
// If there is one row or one column then
// there is only one way of filling the matrix
else if ((n + m) % 2 == 1 && k == -1)
return 0;
// If the above cases are not followed then we
// find ways to fill the n - 1 rows and m - 1
// columns which is 2 ^ ((m-1)*(n-1)).
return (modPower(modPower((long)2, n - 1, mod),
m - 1, mod) % mod);
}
// Driver function for the program
public static void main(String args[]) throws IOException
{
int n = 2, m = 7, k = 1;
System.out.println(countWays(n, m, k));
}
}
# Python program to find number of ways to
# fill a matrix under given constraints
# Returns a raised power t under modulo mod
def modPower(a, t):
now = a;
ret = 1;
mod = 100000007;
# Counting number of ways of filling
# the matrix
while (t):
if (t & 1):
ret = now * (ret % mod);
now = now * (now % mod);
t >>= 1;
return ret;
# Function calculating the answer
def countWays(n, m, k):
mod= 100000007;
# if sum of numbers of rows and columns
# is odd i.e (n + m) % 2 == 1 and k = -1
# then there are 0 ways of filiing the matrix.
if (k == -1 and ((n + m) % 2 == 1)):
return 0;
# If there is one row or one column then
# there is only one way of filling the matrix
if (n == 1 or m == 1):
return 1;
# If the above cases are not followed then we
# find ways to fill the n - 1 rows and m - 1
# columns which is 2 ^ ((m-1)*(n-1)).
return (modPower(modPower(2, n - 1),
m - 1) % mod);
# Driver Code
n = 2;
m = 7;
k = 1;
print(countWays(n, m, k));
# This code is contributed
# by Shivi_Aggarwal
// C# program to find number of ways to fill
// a matrix under given constraints
using System;
class Example
{
static long mod = 100000007;
// Returns a raised power t
// under modulo mod
static long modPower(long a, long t,
long mod)
{
long now = a, ret = 1;
// Counting number of ways
// of filling the
// matrix
while (t > 0)
{
if (t % 2 == 1)
ret = now * (ret % mod);
now = now * (now % mod);
t >>= 1;
}
return ret;
}
// Function calculating the answer
static long countWays(int n, int m,
int k)
{
// if sum of numbers of rows
// and columns is odd i.e
// (n + m) % 2 == 1 and
// k = -1, then there are 0
// ways of filiing the matrix.
if (n == 1 || m == 1)
return 1;
// If there is one row or one
// column then there is only
// one way of filling the matrix
else if ((n + m) % 2 == 1 && k == -1)
return 0;
// If the above cases are not
// followed then we find ways
// to fill the n - 1 rows and
// m - 1 columns which is
// 2 ^ ((m-1)*(n-1)).
return (modPower(modPower((long)2, n - 1,
mod), m - 1, mod) % mod);
}
// Driver Code
public static void Main()
{
int n = 2, m = 7, k = 1;
Console.WriteLine(countWays(n, m, k));
}
}
// This code is contributed by vt_m.
<?php
// PHP program to find number
// of ways to fill a matrix under
// given constraints
$mod = 100000007;
// Returns a raised power t
// under modulo mod
function modPower($a, $t)
{
global $mod;
$now = $a; $ret = 1;
// Counting number of ways
// of filling the matrix
while ($t)
{
if ($t & 1)
$ret = $now * ($ret % $mod);
$now = $now * ($now % $mod);
$t >>= 1;
}
return $ret;
}
// Function calculating the answer
function countWays($n, $m, $k)
{
global $mod;
// if sum of numbers of rows
// and columns is odd i.e
// (n + m) % 2 == 1 and k = -1
// then there are 0 ways of
// filiing the matrix.
if ($k == -1 and ($n + $m) % 2 == 1)
return 0;
// If there is one row or
// one column then there
// is only one way of
// filling the matrix
if ($n == 1 or $m == 1)
return 1;
// If the above cases are
// not followed then we
// find ways to fill the
// n - 1 rows and m - 1
// columns which is
// 2 ^ ((m-1)*(n-1)).
return (modPower(modPower(2, $n - 1),
$m - 1) % $mod);
}
// Driver Code
$n = 2;
$m = 7;
$k = 1;
echo countWays($n, $m, $k);
// This code is contributed by anuj_67.
?>
<script>
// JavaScript program to find number of
// ways to fill a matrix under given
// constraints
let mod = 100000007;
// Returns a raised power t under modulo mod
function modPower(a, t, mod)
{
let now = a, ret = 1;
// Counting number of ways of
// filling the matrix
while (t > 0)
{
if (t % 2 == 1)
ret = now * (ret % mod);
now = now * (now % mod);
t >>= 1;
}
return ret;
}
// Function calculating the answer
function countWays(n, m, k)
{
// If sum of numbers of rows and columns is
// odd i.e (n + m) % 2 == 1 and k = -1,
// then there are 0 ways of filiing the matrix.
if (n == 1 || m == 1)
return 1;
// If there is one row or one column then
// there is only one way of filling the matrix
else if ((n + m) % 2 == 1 && k == -1)
return 0;
// If the above cases are not followed then we
// find ways to fill the n - 1 rows and m - 1
// columns which is 2 ^ ((m-1)*(n-1)).
return (modPower(modPower(2, n - 1, mod),
m - 1, mod) % mod);
}
// Driver Code
let n = 2, m = 7, k = 1;
document.write(countWays(n, m, k));
// This code is contributed by code_hunt
</script>
Output:
64
Time complexity: .
Space Complexity : O(1).
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