Find Number of Even cells in a Zero Matrix after Q queries
Last Updated :
08 Jun, 2021
Given a Zero matrix of N x N size, the task is to find the numbers of cells containing even numbers after performing Q queries. Each query will be in the form of (X, Y) such that for a given cell (X, Y), an increment operation has to be performed on all cells in the X'th row and Y'th column.
Note: The initial 0's are also to be taken as even numbers in the count.
Examples:
Input: N = 2, Q = { {1, 1}, {1, 2}, {2, 1} }
Output: 2
Explanation:
In the first query, we add 1
to all elements of row 1 and column 1.
Our matrix become
2 1
1 0
In the second query, we add 1
to all elements of row 1 and column 2.
Our matrix become
3 3
1 1
In the last query, we add 1
to all elements of row 2 and column 1.
Our matrix finally become
4 3
3 2
In the final updated matrix, there
are 2 even numbers 4 and 3
respectively, hence ans is 2
Input: N = 2, Q = { {1, 1} }
Output: 1
Naive Approach: The Naive approach would be to update each and every cell in the matrix as per the query. Then traverse the matrix to find out the even cells.
Time Complexity: O(N2)
Efficient Approach:
- Maintain two arrays, one for rows operation and one for column operation.
- The row[i] will store the number of operations on i+1th row and col[i] will store the number of operations on i+1th column.
- If 1 is to be added to ith row, we will do row[i-1]++, assuming 0 based indexing.
- Same for columns.
- Now, we need to observe that odd + odd = even and even + even = even.
- Hence if row[i] and col[j] are both odd or even, then that cell will have even value.
- So we need to just count odd and even values in both arrays and multiply them.
Below is the implementation of above approach:
C++
// C++ program find Number of Even cells
// in a Zero Matrix after Q queries
#include <bits/stdc++.h>
using namespace std;
// Function to find the number of
// even cell in a 2D matrix
int findNumberOfEvenCells(int n, int q[][2], int size)
{
// Maintain two arrays, one for rows operation
// and one for column operation
int row[n] = { 0 };
int col[n] = { 0 };
for (int i = 0; i < size; i++) {
int x = q[i][0];
int y = q[i][1];
// Increment operation on row[i]
row[x - 1]++;
// Increment operation on col[i]
col[y - 1]++;
}
int r1 = 0, r2 = 0;
int c1 = 0, c2 = 0;
// Count odd and even values in
// both arrays and multiply them
for (int i = 0; i < n; i++) {
// Count of rows having even numbers
if (row[i] % 2 == 0) {
r1++;
}
// Count of rows having odd numbers
if (row[i] % 2 == 1) {
r2++;
}
// Count of columns having even numbers
if (col[i] % 2 == 0) {
c1++;
}
// Count of columns having odd numbers
if (col[i] % 2 == 1) {
c2++;
}
}
int count = r1 * c1 + r2 * c2;
return count;
}
// Driver code
int main()
{
int n = 2;
int q[][2] = { { 1, 1 },
{ 1, 2 },
{ 2, 1 } };
int size = sizeof(q) / sizeof(q[0]);
cout << findNumberOfEvenCells(n, q, size);
return 0;
}
// Java program find Number of Even cells
// in a Zero Matrix after Q queries
class GFG
{
// Function to find the number of
// even cell in a 2D matrix
static int findNumberOfEvenCells(int n, int q[][], int size)
{
// Maintain two arrays, one for rows operation
// and one for column operation
int row[] = new int[n] ;
int col[] = new int[n] ;
for (int i = 0; i < size; i++)
{
int x = q[i][0];
int y = q[i][1];
// Increment operation on row[i]
row[x - 1]++;
// Increment operation on col[i]
col[y - 1]++;
}
int r1 = 0, r2 = 0;
int c1 = 0, c2 = 0;
// Count odd and even values in
// both arrays and multiply them
for (int i = 0; i < n; i++)
{
// Count of rows having even numbers
if (row[i] % 2 == 0)
{
r1++;
}
// Count of rows having odd numbers
if (row[i] % 2 == 1)
{
r2++;
}
// Count of columns having even numbers
if (col[i] % 2 == 0)
{
c1++;
}
// Count of columns having odd numbers
if (col[i] % 2 == 1)
{
c2++;
}
}
int count = r1 * c1 + r2 * c2;
return count;
}
// Driver code
public static void main (String[] args)
{
int n = 2;
int q[][] = { { 1, 1 },
{ 1, 2 },
{ 2, 1 } };
int size = q.length;
System.out.println(findNumberOfEvenCells(n, q, size));
}
}
// This code is contributed by AnkitRai01
# Python3 program find Number of Even cells
# in a Zero Matrix after Q queries
# Function to find the number of
# even cell in a 2D matrix
def findNumberOfEvenCells(n, q, size) :
# Maintain two arrays, one for rows operation
# and one for column operation
row = [0]*n ;
col = [0]*n
for i in range(size) :
x = q[i][0];
y = q[i][1];
# Increment operation on row[i]
row[x - 1] += 1;
# Increment operation on col[i]
col[y - 1] += 1;
r1 = 0;
r2 = 0;
c1 = 0;
c2 = 0;
# Count odd and even values in
# both arrays and multiply them
for i in range(n) :
# Count of rows having even numbers
if (row[i] % 2 == 0) :
r1 += 1;
# Count of rows having odd numbers
if (row[i] % 2 == 1) :
r2 += 1;
# Count of columns having even numbers
if (col[i] % 2 == 0) :
c1 +=1;
# Count of columns having odd numbers
if (col[i] % 2 == 1) :
c2 += 1;
count = r1 * c1 + r2 * c2;
return count;
# Driver code
if __name__ == "__main__" :
n = 2;
q = [ [ 1, 1 ],
[ 1, 2 ],
[ 2, 1 ] ];
size = len(q);
print(findNumberOfEvenCells(n, q, size));
# This code is contributed by AnkitRai01
// C# program find Number of Even cells
// in a Zero Matrix after Q queries
using System;
class GFG
{
// Function to find the number of
// even cell in a 2D matrix
static int findNumberOfEvenCells(int n, int [,]q, int size)
{
// Maintain two arrays, one for rows operation
// and one for column operation
int []row = new int[n] ;
int []col = new int[n] ;
for (int i = 0; i < size; i++)
{
int x = q[i, 0];
int y = q[i, 1];
// Increment operation on row[i]
row[x - 1]++;
// Increment operation on col[i]
col[y - 1]++;
}
int r1 = 0, r2 = 0;
int c1 = 0, c2 = 0;
// Count odd and even values in
// both arrays and multiply them
for (int i = 0; i < n; i++)
{
// Count of rows having even numbers
if (row[i] % 2 == 0)
{
r1++;
}
// Count of rows having odd numbers
if (row[i] % 2 == 1)
{
r2++;
}
// Count of columns having even numbers
if (col[i] % 2 == 0)
{
c1++;
}
// Count of columns having odd numbers
if (col[i] % 2 == 1)
{
c2++;
}
}
int count = r1 * c1 + r2 * c2;
return count;
}
// Driver code
public static void Main ()
{
int n = 2;
int [,]q = { { 1, 1 },
{ 1, 2 },
{ 2, 1 } };
int size = q.GetLength(0);
Console.WriteLine(findNumberOfEvenCells(n, q, size));
}
}
// This code is contributed by AnkitRai01
<script>
// JavaScript program find Number of Even cells
// in a Zero Matrix after Q queries
// Function to find the number of
// even cell in a 2D matrix
function findNumberOfEvenCells(n, q, size)
{
// Maintain two arrays, one for rows operation
// and one for column operation
let row = new Array(n);
row.fill(0);
let col = new Array(n);
col.fill(0);
for (let i = 0; i < size; i++)
{
let x = q[i][0];
let y = q[i][1];
// Increment operation on row[i]
row[x - 1]++;
// Increment operation on col[i]
col[y - 1]++;
}
let r1 = 0, r2 = 0;
let c1 = 0, c2 = 0;
// Count odd and even values in
// both arrays and multiply them
for (let i = 0; i < n; i++)
{
// Count of rows having even numbers
if (row[i] % 2 == 0)
{
r1++;
}
// Count of rows having odd numbers
if (row[i] % 2 == 1)
{
r2++;
}
// Count of columns having even numbers
if (col[i] % 2 == 0)
{
c1++;
}
// Count of columns having odd numbers
if (col[i] % 2 == 1)
{
c2++;
}
}
let count = r1 * c1 + r2 * c2;
return count;
}
let n = 2;
let q = [ [ 1, 1 ],
[ 1, 2 ],
[ 2, 1 ] ];
let size = q.length;
document.write(findNumberOfEvenCells(n, q, size));
</script>
Time Complexity: O(N)
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