给定尺寸N * N * N的3D矩阵mat [] [] []由正整数组成,任务是计算存在于主对角线上的所有矩阵元素的按位XOR。
例子:
Input: arr[][][] = {{{1, 2}, {3, 4}}, {{5, 6}, {7, 8}}}
Output: 12
Explanation: The major diagonal elements are {1, 8, 2, 7}. Therefore, the Bitwise XOR of all these elements is 12.
Input: arr[][][]={{{1}}}
Output: 0
Explanation: There 2 major diagonals of the matrix are {1}, {1}. Therefore, the Bitwise XOR of the major diagonal elements is 0.
天真的方法:最简单的解决问题的方法是使用三个嵌套循环遍历给定的3D矩阵mat [] [] [] ,使用变量i , j和k并计算mat [i] [的按位XOR如果i , j和k的值相等,则j] [k]和mat [i] [j] [N – k – 1] 。完成矩阵遍历后,打印获得的按位XOR值。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate Bitwise XOR of
// major diagonal elements of 3D matrix
void findXOR(
vector > >& mat,
int N)
{
// Stores the Bitwise XOR of
// the major diagonal elements
int XOR = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
for (int k = 0; k < N; k++) {
// If element is part
// of major diagonal
if ((i == j && j == k)) {
XOR ^= mat[i][j][k];
XOR ^= mat[i][j][N - k - 1];
}
}
}
}
// Print the resultant Bitwise XOR
cout << XOR << "\n";
}
// Driver Code
int main()
{
vector > > mat
= { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.size();
findXOR(mat, N);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to calculate Bitwise XOR of
// major diagonal elements of 3D matrix
static void findXOR(int mat[][][], int N)
{
// Stores the Bitwise XOR of
// the major diagonal elements
int XOR = 0;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
for(int k = 0; k < N; k++)
{
// If element is part
// of major diagonal
if ((i == j && j == k))
{
XOR ^= mat[i][j][k];
XOR ^= mat[i][j][N - k - 1];
}
}
}
}
// Print the resultant Bitwise XOR
System.out.println(XOR);
}
// Driver Code
public static void main(String[] args)
{
int mat[][][] = { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.length;
findXOR(mat, N);
}
}
// This code is contributed by Kingash
Python3
# Python3 program for the above approach
# Function to calculate Bitwise XOR of
# major diagonal elements of 3D matrix
def findXOR(mat, N):
# Stores the Bitwise XOR of
# the major diagonal elements
XOR = 0
for i in range(N):
for j in range(N):
for k in range(N):
# If element is part
# of major diagonal
if ((i == j and j == k)):
XOR ^= mat[i][j][k]
XOR ^= mat[i][j][N - k - 1]
# Print the resultant Bitwise XOR
print(XOR)
# Driver Code
mat = [ [ [ 1, 2 ], [ 3, 4 ] ],
[ [ 5, 6 ], [ 7, 8 ] ] ]
N = len(mat)
findXOR(mat, N)
# This code is contributed by splevel62
C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate Bitwise XOR of
// major diagonal elements of 3D matrix
static void findXOR(int[, , ] mat, int N)
{
// Stores the Bitwise XOR of
// the major diagonal elements
int XOR = 0;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
for(int k = 0; k < N; k++)
{
// If element is part
// of major diagonal
if ((i == j && j == k))
{
XOR ^= mat[i, j, k];
XOR ^= mat[i, j, N - k - 1];
}
}
}
}
// Print the resultant Bitwise XOR
Console.WriteLine(XOR);
}
// Driver Code
public static void Main(string[] args)
{
int[,,] mat = { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.GetLength(0);
findXOR(mat, N);
}
}
// This code is contributed by ukasp
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the Bitwise XOR of
// both diagonal elements of 3D matrix
void findXOR(
vector > >& mat,
int N)
{
// Stores the Bitwise XOR of the
// major diagonal elements
int XOR = 0;
for (int i = 0; i < N; i++) {
XOR ^= mat[i][i][i];
XOR ^= mat[i][i][N - i - 1];
}
// Print the resultant Bitwise XOR
cout << XOR << "\n";
}
// Driver Code
int main()
{
vector > > mat
= { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.size();
findXOR(mat, N);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to calculate Bitwise XOR of
// major diagonal elements of 3D matrix
static void findXOR(int mat[][][], int N)
{
// Stores the Bitwise XOR of the
// major diagonal elements
int XOR = 0;
for(int i = 0; i < N; i++)
{
XOR ^= mat[i][i][i];
XOR ^= mat[i][i][N - i - 1];
}
// Print the resultant Bitwise XOR
System.out.println(XOR);
}
// Driver Code
public static void main(String[] args)
{
int mat[][][] = { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.length;
findXOR(mat, N);
}
}
// This code is contributed by Kingash
Python3
# Python3 program for the above approach
# Function to find the Bitwise XOR of
# both diagonal elements of 3D matrix
def findXOR(mat, N):
# Stores the Bitwise XOR of the
# major diagonal elements
XOR = 0
for i in range(N):
XOR ^= mat[i][i][i]
XOR ^= mat[i][i][N - i - 1]
# Print the resultant Bitwise XOR
print(XOR)
# Driver Code
mat = [ [ [ 1, 2 ], [ 3, 4 ] ],
[ [ 5, 6 ], [ 7, 8 ] ] ]
N = len(mat)
findXOR(mat, N)
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate Bitwise XOR of
// major diagonal elements of 3D matrix
static void findXOR(int[,,] mat, int N)
{
// Stores the Bitwise XOR of the
// major diagonal elements
int XOR = 0;
for(int i = 0; i < N; i++)
{
XOR ^= mat[i, i, i];
XOR ^= mat[i, i, N - i - 1];
}
// Print the resultant Bitwise XOR
Console.Write(XOR);
}
// Driver Code
static public void Main ()
{
int[,,] mat = { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.GetLength(0);
findXOR(mat, N);
}
}
// This code is contributed by avijitmondal1998
12
时间复杂度: O(N 3 )
辅助空间: O(1)
高效方法:可以通过使用索引i , j和k与对角元素相同的元素来优化上述方法。因此,这个想法是遍历索引的范围[0,N – 1]使用变量i,并计算所有处于指数主要对角线的一部分的元素(I,I,i)和按位XOR(ⅰ ,i,N – i – 1)在给定的矩阵mat [] [] []中分别作为mat [i] [i] [i]和mat [i] [i] [N – i – 1] 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the Bitwise XOR of
// both diagonal elements of 3D matrix
void findXOR(
vector > >& mat,
int N)
{
// Stores the Bitwise XOR of the
// major diagonal elements
int XOR = 0;
for (int i = 0; i < N; i++) {
XOR ^= mat[i][i][i];
XOR ^= mat[i][i][N - i - 1];
}
// Print the resultant Bitwise XOR
cout << XOR << "\n";
}
// Driver Code
int main()
{
vector > > mat
= { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.size();
findXOR(mat, N);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to calculate Bitwise XOR of
// major diagonal elements of 3D matrix
static void findXOR(int mat[][][], int N)
{
// Stores the Bitwise XOR of the
// major diagonal elements
int XOR = 0;
for(int i = 0; i < N; i++)
{
XOR ^= mat[i][i][i];
XOR ^= mat[i][i][N - i - 1];
}
// Print the resultant Bitwise XOR
System.out.println(XOR);
}
// Driver Code
public static void main(String[] args)
{
int mat[][][] = { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.length;
findXOR(mat, N);
}
}
// This code is contributed by Kingash
Python3
# Python3 program for the above approach
# Function to find the Bitwise XOR of
# both diagonal elements of 3D matrix
def findXOR(mat, N):
# Stores the Bitwise XOR of the
# major diagonal elements
XOR = 0
for i in range(N):
XOR ^= mat[i][i][i]
XOR ^= mat[i][i][N - i - 1]
# Print the resultant Bitwise XOR
print(XOR)
# Driver Code
mat = [ [ [ 1, 2 ], [ 3, 4 ] ],
[ [ 5, 6 ], [ 7, 8 ] ] ]
N = len(mat)
findXOR(mat, N)
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate Bitwise XOR of
// major diagonal elements of 3D matrix
static void findXOR(int[,,] mat, int N)
{
// Stores the Bitwise XOR of the
// major diagonal elements
int XOR = 0;
for(int i = 0; i < N; i++)
{
XOR ^= mat[i, i, i];
XOR ^= mat[i, i, N - i - 1];
}
// Print the resultant Bitwise XOR
Console.Write(XOR);
}
// Driver Code
static public void Main ()
{
int[,,] mat = { { { 1, 2 }, { 3, 4 } },
{ { 5, 6 }, { 7, 8 } } };
int N = mat.GetLength(0);
findXOR(mat, N);
}
}
// This code is contributed by avijitmondal1998
12
时间复杂度: O(N)
辅助空间: O(1)