给定两个字符串str1和str2 ，任务是通过复制字符串的任何字符及其相邻字符来检查两个字符串是否相等。注意，该操作可以执行多次。
例子：
Input: str1 = “abc”, str2 = “def”
Output: No
As all the characters in both the string are different.
So, there is no way they can be made equal.
Input: str1 = “abc”, str2 = “fac”
Output: Yes
str1 = “abc” -> “aac”
str2 = “fac” -> “aac”
的方法：在将被制造为使字符串与所述给定的操作相同，它们必须是相等长度的并且必须有至少一个字符是在两个字符串的常见。要检查，创建一个频率数组频率[]将存储STR1的所有字符的频率，然后对STR2的每个字符，如果其在STR1频率大于0，那么它可以使两个字符串相等。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 26
// Function that returns true if both
// the strings can be made equal
// with the given operation
bool canBeMadeEqual(string str1, string str2)
{
int len1 = str1.length();
int len2 = str2.length();
// Lengths of both the strings
// have to be equal
if (len1 == len2) {
// To store the frequency of the
// characters of str1
int freq[MAX];
for (int i = 0; i < len1; i++) {
freq[str1[i] - 'a']++;
}
// For every character of str2
for (int i = 0; i < len2; i++) {
// If current character of str2
// also appears in str1
if (freq[str2[i] - 'a'] > 0)
return true;
}
}
return false;
}
// Driver code
int main()
{
string str1 = "abc", str2 = "defa";
if (canBeMadeEqual(str1, str2))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the above approach
class GFG
{
static int MAX = 26;
// Function that returns true if both
// the strings can be made equal
// with the given operation
static boolean canBeMadeEqual(String str1,
String str2)
{
int len1 = str1.length();
int len2 = str2.length();
// Lengths of both the strings
// have to be equal
if (len1 == len2)
{
// To store the frequency of the
// characters of str1
int freq[] = new int[MAX];
for (int i = 0; i < len1; i++)
{
freq[str1.charAt(i) - 'a']++;
}
// For every character of str2
for (int i = 0; i < len2; i++)
{
// If current character of str2
// also appears in str1
if (freq[str2.charAt(i) - 'a'] > 0)
return true;
}
}
return false;
}
// Driver code
public static void main (String[] args)
{
String str1 = "abc", str2 = "defa";
if (canBeMadeEqual(str1, str2))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the approach
MAX = 26
# Function that returns true if both
# the strings can be made equal
# with the given operation
def canBeMadeEqual(str1, str2):
len1 = len(str1)
len2 = len(str2)
# Lengths of both the strings
# have to be equal
if (len1 == len2):
# To store the frequency of the
# characters of str1
freq = [0 for i in range(MAX)]
for i in range(len1):
freq[ord(str1[i]) - ord('a')] += 1
# For every character of str2
for i in range(len2):
# If current character of str2
# also appears in str1
if (freq[ord(str2[i]) - ord('a')] > 0):
return True
return False
# Driver code
str1 = "abc"
str2 = "defa"
if (canBeMadeEqual(str1, str2)):
print("Yes")
else:
print("No")
# This code is contributed by Mohit KumarC#
// C# implementation of the above approach
using System;
class GFG
{
static int MAX = 26;
// Function that returns true if both
// the strings can be made equal
// with the given operation
static Boolean canBeMadeEqual(String str1,
String str2)
{
int len1 = str1.Length;
int len2 = str2.Length;
// Lengths of both the strings
// have to be equal
if (len1 == len2)
{
// To store the frequency of the
// characters of str1
int []freq = new int[MAX];
for (int i = 0; i < len1; i++)
{
freq[str1[i] - 'a']++;
}
// For every character of str2
for (int i = 0; i < len2; i++)
{
// If current character of str2
// also appears in str1
if (freq[str2[i] - 'a'] > 0)
return true;
}
}
return false;
}
// Driver code
public static void Main (String []args)
{
String str1 = "abc", str2 = "defa";
if (canBeMadeEqual(str1, str2))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Arnab Kundu输出：
No