Dada una array bidimensional arr[][] , la tarea es verificar si es posible hacer que todos los elementos de la array sean iguales a un número k dado si, en una operación, se puede elegir cualquier elemento y la diagonal circundante los elementos pueden hacerse iguales a él.
Ejemplos:
Input:
arr[][] = 1 8 3
1 2 2
4 1 9
k = 2
Output: Yes
Explanation:
In first operation choose element at (2, 2)
New array = 2 8 2
1 2 2
2 1 2
In second operation choose element at (2, 3)
New array = 2 2 2
1 2 2
2 2 2
In third operation choose element at (1, 2)
New array = 2 2 2
2 2 2
2 2 2
Input:
arr[][] = 3 1 2 3
2 1 8 6
9 7 9 9
k = 4
Output:
No
Acercarse:
- La array se puede considerar como un tablero de ajedrez con casillas en blanco y negro.
- Si se elige cualquier elemento en la caja negra que sea igual al número dado, entonces todos los elementos de las cajas negras pueden hacerse iguales usando la operación dada,
- Del mismo modo, se puede comprobar por las casillas blancas. Por lo tanto, debe haber al menos un elemento igual al elemento dado en los cuadros blanco y negro.
- Entonces necesitamos iterar sobre todos los elementos usando un contador. Si el valor de la ficha es impar, se puede considerar una caja negra y para valores pares, se puede considerar una caja blanca.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ implementation of the above approach.
#include <iostream>
using namespace std;
// Function to check if all
// elements can be equal or not
void checkEqualMatrix(int arr[][3], int n,
int m, int k)
{
int c = 0, cnt1 = 0, cnt2 = 0;
// Iterate over the matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (c % 2 == 0) {
// Update the counter for odd values
// if array element at that position is k
if (arr[i][j] == k) {
cnt1++;
}
}
else {
// Update the counter for even values
// if array element at that position is k
if (arr[i][j] == k) {
cnt2++;
}
}
c = c + 1;
}
}
// To check if there is at least one
// element at both even and odd indices.
if (cnt1 >= 1 && cnt2 >= 1) {
cout << "Yes";
}
else {
cout << "No";
}
}
// Driver code
int main()
{
int arr[3][3] = { { 1, 8, 3 },
{ 1, 2, 2 },
{ 4, 1, 9 } };
int k = 2;
// Function calling
checkEqualMatrix(arr, 3, 3, k);
}
Java
// Java implementation of the above approach.
class GFG
{
// Function to check if all
// elements can be equal or not
static void checkEqualMatrix(int arr[][], int n,
int m, int k)
{
int c = 0, cnt1 = 0, cnt2 = 0;
// Iterate over the matrix
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (c % 2 == 0)
{
// Update the counter for odd values
// if array element at that position is k
if (arr[i][j] == k)
{
cnt1++;
}
}
else
{
// Update the counter for even values
// if array element at that position is k
if (arr[i][j] == k)
{
cnt2++;
}
}
c = c + 1;
}
}
// To check if there is at least one
// element at both even and odd indices.
if (cnt1 >= 1 && cnt2 >= 1)
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
// Driver code
public static void main (String[] args)
{
int arr[][] = { { 1, 8, 3 },
{ 1, 2, 2 },
{ 4, 1, 9 } };
int k = 2;
// Function calling
checkEqualMatrix(arr, 3, 3, k);
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the above approach.
# Function to check if all
# elements can be equal or not
def checkEqualMatrix(arr, n, m, k) :
c = 0; cnt1 = 0; cnt2 = 0;
# Iterate over the matrix
for i in range(n) :
for j in range(m) :
if (c % 2 == 0) :
# Update the counter for odd values
# if array element at that position is k
if (arr[i][j] == k) :
cnt1 += 1;
else :
# Update the counter for even values
# if array element at that position is k
if (arr[i][j] == k) :
cnt2 += 1;
c = c + 1;
# To check if there is at least one
# element at both even and odd indices.
if (cnt1 >= 1 and cnt2 >= 1) :
print("Yes");
else :
print("No");
# Driver code
if __name__ == "__main__" :
arr = [
[ 1, 8, 3 ],
[ 1, 2, 2 ],
[ 4, 1, 9 ]
];
k = 2;
# Function calling
checkEqualMatrix(arr, 3, 3, k);
# This code is contributed by AnkitRai01
C#
// C# implementation of the above approach.
using System;
class GFG
{
// Function to check if all
// elements can be equal or not
static void checkEqualMatrix(int [,]arr, int n,
int m, int k)
{
int c = 0, cnt1 = 0, cnt2 = 0;
// Iterate over the matrix
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (c % 2 == 0)
{
// Update the counter for odd values
// if array element at that position is k
if (arr[i,j] == k)
{
cnt1++;
}
}
else
{
// Update the counter for even values
// if array element at that position is k
if (arr[i,j] == k)
{
cnt2++;
}
}
c = c + 1;
}
}
// To check if there is at least one
// element at both even and odd indices.
if (cnt1 >= 1 && cnt2 >= 1)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
// Driver code
public static void Main()
{
int [,]arr = { { 1, 8, 3 },
{ 1, 2, 2 },
{ 4, 1, 9 } };
int k = 2;
// Function calling
checkEqualMatrix(arr, 3, 3, k);
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// Javascript implementation of the above approach.
// Function to check if all
// elements can be equal or not
function checkEqualMatrix(arr, n, m, k)
{
var c = 0, cnt1 = 0, cnt2 = 0;
// Iterate over the matrix
for (var i = 0; i < n; i++) {
for (var j = 0; j < m; j++) {
if (c % 2 == 0) {
// Update the counter for odd values
// if array element at that position is k
if (arr[i][j] == k) {
cnt1++;
}
}
else {
// Update the counter for even values
// if array element at that position is k
if (arr[i][j] == k) {
cnt2++;
}
}
c = c + 1;
}
}
// To check if there is at least one
// element at both even and odd indices.
if (cnt1 >= 1 && cnt2 >= 1) {
document.write( "Yes");
}
else {
document.write( "No");
}
}
// Driver code
var arr = [ [ 1, 8, 3 ],
[ 1, 2, 2 ],
[ 4, 1, 9 ] ];
var k = 2;
// Function calling
checkEqualMatrix(arr, 3, 3, k);
// This code is contributed by importantly.
</script>
Producción:
Yes
Publicación traducida automáticamente
Artículo escrito por AmanGupta65 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA