给定一个大小为M * N的布尔矩阵mat[][] ,任务是打印矩阵中索引的计数,其对应的行和列包含相同数量的设置位。
例子:
Input; mat[][] = {{0, 1}, {1, 1}}
Output: 2
Explanation:
Position (0, 0) contains 1 set bit in corresponding row {(0, 0), (0, 1)} and column {(0, 0), (1, 0)}.
Position (1, 1) contains 2 set bits in corresponding row {(1, 0), (1, 1)} and column {(0, 1), (1, 1)}.
Input: mat[][] = {{0, 1, 1, 0}, {0, 1, 0, 0}, {1, 0, 1, 0}}
Output: 5
朴素的方法:最简单的解决方法是计算和存储给定矩阵的每一行和每一列的元素中存在的设置位的数量。然后,遍历矩阵,对于每个位置,检查行和列的设置位计数是否相等。对于发现上述条件为真的每个索引,增加count 。完整遍历矩阵后打印count的最终值。
时间复杂度: O(N 2 )
辅助空间: O(N 2 )
高效的方法:要优化上述方法,请按照以下步骤操作:
- 分别初始化大小为N和M 的两个临时数组row[]和col[] 。
- 遍历矩阵mat[][] 。检查每个索引(i, j) ,如果mat[i][j]等于1或不。如果发现为真,则将row[i]和col[j]增加1 。
- 再次遍历矩阵mat[][] 。检查每个索引(i, j) ,如果row[i]和col[j]相等。如果发现为真,则增加计数。
- 打印count的最终值。
下面是上述方法的实现:
C++14
// C++14 program to implement
// above approach
#include
using namespace std;
// Function to return the count of indices in
// from the given binary matrix having equal
// count of set bits in its row and column
int countPosition(vector> mat)
{
int n = mat.size();
int m = mat[0].size();
// Stores count of set bits in
// corresponding column and row
vector row(n);
vector col(m);
// Traverse matrix
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
// Since 1 contains a set bit
if (mat[i][j] == 1)
{
// Update count of set bits
// for current row and col
col[j]++;
row[i]++;
}
}
}
// Stores the count of
// required indices
int count = 0;
// Traverse matrix
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
// If current row and column
// has equal count of set bits
if (row[i] == col[j])
{
count++;
}
}
}
// Return count of
// required position
return count;
}
// Driver Code
int main()
{
vector> mat = { { 0, 1 },
{ 1, 1 } };
cout << (countPosition(mat));
}
// This code is contributed by mohit kumar 29
Java
// Java Program to implement
// above approach
import java.util.*;
import java.lang.*;
class GFG {
// Function to return the count of indices in
// from the given binary matrix having equal
// count of set bits in its row and column
static int countPosition(int[][] mat)
{
int n = mat.length;
int m = mat[0].length;
// Stores count of set bits in
// corresponding column and row
int[] row = new int[n];
int[] col = new int[m];
// Traverse matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Since 1 contains a set bit
if (mat[i][j] == 1) {
// Update count of set bits
// for current row and col
col[j]++;
row[i]++;
}
}
}
// Stores the count of
// required indices
int count = 0;
// Traverse matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// If current row and column
// has equal count of set bits
if (row[i] == col[j]) {
count++;
}
}
}
// Return count of
// required position
return count;
}
// Driver Code
public static void main(
String[] args)
{
int mat[][] = { { 0, 1 },
{ 1, 1 } };
System.out.println(
countPosition(mat));
}
}
Python3
# Python3 program to implement
# the above approach
# Function to return the count
# of indices in from the given
# binary matrix having equal
# count of set bits in its
# row and column
def countPosition(mat):
n = len(mat)
m = len(mat[0])
# Stores count of set bits in
# corresponding column and row
row = [0] * n
col = [0] * m
# Traverse matrix
for i in range(n):
for j in range(m):
# Since 1 contains a set bit
if (mat[i][j] == 1):
# Update count of set bits
# for current row and col
col[j] += 1
row[i] += 1
# Stores the count of
# required indices
count = 0
# Traverse matrix
for i in range(n):
for j in range(m):
# If current row and column
# has equal count of set bits
if (row[i] == col[j]):
count += 1
# Return count of
# required position
return count
# Driver Code
mat = [ [ 0, 1 ],
[ 1, 1 ] ]
print(countPosition(mat))
# This code is contributed by sanjoy_62
C#
// C# Program to implement
// above approach
using System;
class GFG{
// Function to return the count of indices in
// from the given binary matrix having equal
// count of set bits in its row and column
static int countPosition(int[,] mat)
{
int n = mat.GetLength(0);
int m = mat.GetLength(1);
// Stores count of set bits in
// corresponding column and row
int[] row = new int[n];
int[] col = new int[m];
// Traverse matrix
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Since 1 contains a set bit
if (mat[i, j] == 1)
{
// Update count of set bits
// for current row and col
col[j]++;
row[i]++;
}
}
}
// Stores the count of
// required indices
int count = 0;
// Traverse matrix
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// If current row and column
// has equal count of set bits
if (row[i] == col[j])
{
count++;
}
}
}
// Return count of
// required position
return count;
}
// Driver Code
public static void Main(String[] args)
{
int [,]mat = {{0, 1},
{1, 1}};
Console.WriteLine(countPosition(mat));
}
}
// This code is contributed by Rajput-Ji
Javascript
2
时间复杂度: O(N * M)
辅助空间: O(N+M)
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