检查矩阵是否可以叠加在给定的矩阵上
给定一个大小为N * M的矩阵letter[][] ，由'#'和'*'组成，另一个大小为X * Y的矩阵stamp[][]仅包含'$' 。任务是通过将印章矩阵叠加在字母矩阵上来查找是否可以将较大的“*”替换为“$” 。
注意：在叠加操作中，只能考虑所有字符为“*”或“$”的区域。
例子：
Input: N = 3, M =5, X = 2, Y=2
Letter Matrix:
#****
#****
#****
Stamp Matrix:
$$
$$
Output: Possible
Explanation:
1st Step: Superimpose letter matrix from (0,1) to (1,2) , So, letter will look like ( Remember the place where ‘#’ is placed can’t be imposed)
#$$**
#$$**
#****
2nd Step: Superimpose letter matrix from (0,3) to (1,4), Letter becomes
#$$$$
#$$$$
#****
3rd Step: Since superimpose over ‘$’ is also allowed, do overlapping. Hence stamping next from (1,1) to (2,3)
#$$$$
#$$$$
#$$**
Final step: Again do overlapping, and stamp from (1,3) to (3,4)
#$$$$
#$$$$
#$$$$
Input: N = 3, M =5, X = 2, Y=2
Letter Matrix:
#**#*
#****
#****
Stamp Matrix:
$$
$$
Output: Impossible
Explanation: The ‘#’ at (0,3) cannot be superimposed.
方法：可以通过检查两个“#”之间或行尾和行首的“*”字符数来解决问题。如果这些计数中的任何一个小于标记矩阵的行长度，则无法解决，同样适用于列。按照下面提到的步骤来实施该方法：
- 循环遍历完整的较大矩阵行
- 对于每一行，计算在行的开头或结尾之前或两个'#'之间的连续'*'的数量。如果此计数小于标记矩阵行长度，则它是不可能的，并返回不可能作为答案。
- 以这种方式检查整个字母矩阵。
- 也对列应用相同的过程。
- 如果字母矩阵可以叠加，则返回 true。否则，返回假。
下面是上述方法的实现
C++
// C++ code to implement the above approach
#include
using namespace std;
// Function to check if superimpose
// is possible or not
bool solution(int n, int m, int x, int y,
vector& letter)
{
// Looping over rows
for (int i = 0; i < n; i++) {
// Initializing a length variable
// for counting the number of *
int len = 0;
for (int j = 0; j < m; j++) {
// If character encountered
// is *, then we
// increment the length
if (letter[i][j] == '*')
len++;
// If its not then there are
// two cases
else {
// 1st case is length is
// smaller than number of
// rows in stamp matrix,
// that would be impossible
// to stamp so return false
if (len != 0 && len < x)
return false;
// 2nd case is if its
// possible, reset len
else
len = 0;
}
}
}
// Do similarly for columns
// Looping over columns
for (int i = 0; i < m; i++) {
// Initializing length variable
int len = 0;
for (int j = 0; j < n; j++) {
// If encountered character
// is *, increment length
if (letter[j][i] == '*')
len++;
else {
// If its not possible
// return false
if (len != 0 && len < y)
return false;
// Otherwise reset len
else
len = 0;
}
}
}
return true;
}
// Driver code
int main()
{
int n = 3, x = 2, y = 2;
vector letter = { "#***", "#***", "#***" };
int m = letter[0].size();
/*
Stamp Matrix:
$$
$$
A 2*2 matrix ( x*y)
*/
if (solution(n, m, x, y, letter))
cout << "Possible\n";
else
cout << "Impossible\n";
// 2nd case
vector letter2 = { "#***", "#*#*", "#***" };
m = letter2[0].size();
if (solution(n, m, x, y, letter2))
cout << "Possible\n";
else
cout << "Impossible\n";
return 0;
}
// This code is contributed by rakeshsahni Java
// Java code to implement the above approach
import java.util.*;
class GFG {
// Function to check if superimpose
// is possible or not
public static boolean solution(int n,
int m, int x,
int y,
String[] letter)
{
// Looping over rows
for (int i = 0; i < n; i++) {
// Initializing a length variable
// for counting the number of *
int len = 0;
for (int j = 0; j < m; j++) {
// If character encountered
// is *, then we
// increment the length
if (letter[i].charAt(j)
== '*')
len++;
// If its not then there are
// two cases
else {
// 1st case is length is
// smaller than number of
// rows in stamp matrix,
// that would be impossible
// to stamp so return false
if (len != 0 && len < x)
return false;
// 2nd case is if its
// possible, reset len
else
len = 0;
}
}
}
// Do similarly for columns
// Looping over columns
for (int i = 0; i < m; i++) {
// Initializing length variable
int len = 0;
for (int j = 0; j < n; j++) {
// If encountered character
// is *, increment length
if (letter[j].charAt(i)
== '*')
len++;
else {
// If its not possible
// return false
if (len != 0 && len < y)
return false;
// Otherwise reset len
else
len = 0;
}
}
}
return true;
}
// Driver code
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n = 3, x = 2, y = 2;
String[] letter = { "#***",
"#***",
"#***" };
int m = letter[0].length();
/*
Stamp Matrix:
$$
$$
A 2*2 matrix ( x*y)
*/
if (solution(n, m, x, y, letter))
System.out.println("Possible");
else
System.out.println("Impossible");
// 2nd case
String[] letter2 = { "#***",
"#*#*",
"#***" };
m = letter2[0].length();
if (solution(n, m, x, y, letter2))
System.out.println("Possible");
else
System.out.println("Impossible");
}
}Python3
# Python code to implement the above approach
# Function to check if superimpose
# is possible or not
def solution(n, m, x, y, letter):
# Looping over rows
for i in range(n):
# Initializing a length variable
# for counting the number of *
len = 0
for j in range(m):
# If character encountered
# is *, then we
# increment the length
if (letter[i][j] == '*'):
len += 1
# If its not then there are
# two cases
else:
# 1st case is length is
# smaller than number of
# rows in stamp matrix,
# that would be impossible
# to stamp so return false
if (len != 0 and len < x):
return False
# 2nd case is if its
# possible, reset len
else:
len = 0
# Do similarly for columns
# Looping over columns
for i in range(m):
# Initializing length variable
len = 0
for j in range(n):
# If encountered character
# is *, increment length
if (letter[j][i] == '*'):
len += 1
else:
# If its not possible
# return false
if (len != 0 and len < y):
return False
# Otherwise reset len
else:
len = 0
return True
# Driver code
n = 3
x = 2
y = 2
letter = ["#***", "#***", "#***"]
m = len(letter[0])
'''
Stamp Matrix:
$$
$$
A 2*2 matrix ( x*y)
'''
if (solution(n, m, x, y, letter)):
print("Possible")
else:
print("Impossible")
# 2nd case
letter2 = ["#***", "#*#*", "#***"]
m = len(letter2[0])
if (solution(n, m, x, y, letter2)):
print("Possible")
else:
print("Impossible")
# This code is contributed by Shubham SinghC#
// C# code to implement the above approach
using System;
class GFG{
// Function to check if superimpose
// is possible or not
public static bool solution(int n, int m, int x, int y,
string[] letter)
{
// Looping over rows
for(int i = 0; i < n; i++)
{
// Initializing a length variable
// for counting the number of *
int len = 0;
for(int j = 0; j < m; j++)
{
// If character encountered
// is *, then we
// increment the length
if (letter[i][j] == '*')
len++;
// If its not then there are
// two cases
else
{
// 1st case is length is
// smaller than number of
// rows in stamp matrix,
// that would be impossible
// to stamp so return false
if (len != 0 && len < x)
return false;
// 2nd case is if its
// possible, reset len
else
len = 0;
}
}
}
// Do similarly for columns
// Looping over columns
for(int i = 0; i < m; i++)
{
// Initializing length variable
int len = 0;
for(int j = 0; j < n; j++)
{
// If encountered character
// is *, increment length
if (letter[j][i] == '*')
len++;
else
{
// If its not possible
// return false
if (len != 0 && len < y)
return false;
// Otherwise reset len
else
len = 0;
}
}
}
return true;
}
// Driver code
public static void Main(string[] args)
{
int n = 3, x = 2, y = 2;
string[] letter = { "#***", "#***", "#***" };
int m = letter.GetLength(0);
/*
Stamp Matrix:
$$
$$
A 2*2 matrix ( x*y)
*/
if (solution(n, m, x, y, letter))
Console.WriteLine("Possible");
else
Console.WriteLine("Impossible");
// 2nd case
String[] letter2 = { "#***", "#*#*", "#***" };
m = letter2.GetLength(0);
if (solution(n, m, x, y, letter2))
Console.WriteLine("Possible");
else
Console.WriteLine("Impossible");
}
}
// This code is contributed by ukaspJavascript
输出
Impossible时间复杂度： O(N*M)
辅助空间： O(1)