给定两个字符串str1和str2 ，任务是计算所有有效字符串。有效字符串的示例如下：
如果str1 =“玩具”和str2 =“尝试” 。则S =“托里”是一个有效的字符串，因为当一个字符从它除去即S =“吨ÕRY” =“尝试”，它变成等于STR1。此属性还必须在str2中有效，即S =“ to r y” =“ toy” = str2。
该任务是打印所有可能的有效字符串的计数。
例子：
Input: str = “toy”, str2 = “try”
Output: 2
The given two words could be obtained from either word “tory” or word “troy”. So output is 2.
Input: str1 = “sweet”, str2 = “sheep”
Output: 0
The two given word couldn’t be obtained from the same word by removing one letter.
方法：计算作为str1和STR2赛车和C的最长公共前缀为str1和STR2的最长公共后缀。如果两个字符串相等，则可以使用26 *(n + 1)个字符串。否则，将count设置为0且l等于第一个索引，因为它不是公共前缀的一部分，而r是最右端的索引，这不是公共后缀的一部分。
现在，如果str1 [l + 1…r] = str2 [l…r-1]，则更新count = count +1 。
如果str1 [l…r-1] = str2 [l + 1…r]，则更新count = count + 1 。
最后打印计数。
下面是该方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the count of the
// required strings
int findAnswer(string str1, string str2, int n)
{
int l, r;
int ans = 2;
// Searching index after longest common
// prefix ends
for (int i = 0; i < n; ++i) {
if (str1[i] != str2[i]) {
l = i;
break;
}
}
// Searching index before longest common
// suffix ends
for (int i = n - 1; i >= 0; i--) {
if (str1[i] != str2[i]) {
r = i;
break;
}
}
// If str1 = str2
if (r < l)
return 26 * (n + 1);
// If only 1 character is different
// in both the strings
else if (l == r)
return ans;
else {
// Checking remaining part of string
// for equality
for (int i = l + 1; i <= r; i++) {
if (str1[i] != str2[i - 1]) {
ans--;
break;
}
}
// Searching in right of string h
// (g to h)
for (int i = l + 1; i <= r; i++) {
if (str1[i - 1] != str2[i]) {
ans--;
break;
}
}
return ans;
}
}
// Driver code
int main()
{
string str1 = "toy", str2 = "try";
int n = str1.length();
cout << findAnswer(str1, str2, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the count of the
// required strings
static int findAnswer(String str1, String str2, int n)
{
int l = 0, r = 0;
int ans = 2;
// Searching index after longest common
// prefix ends
for (int i = 0; i < n; ++i)
{
if (str1.charAt(i) != str2.charAt(i))
{
l = i;
break;
}
}
// Searching index before longest common
// suffix ends
for (int i = n - 1; i >= 0; i--)
{
if (str1.charAt(i) != str2.charAt(i))
{
r = i;
break;
}
}
// If str1 = str2
if (r < l)
return 26 * (n + 1);
// If only 1 character is different
// in both the strings
else if (l == r)
return ans;
else {
// Checking remaining part of string
// for equality
for (int i = l + 1; i <= r; i++)
{
if (str1.charAt(i) != str2.charAt(i - 1))
{
ans--;
break;
}
}
// Searching in right of string h
// (g to h)
for (int i = l + 1; i <= r; i++)
{
if (str1.charAt(i-1) != str2.charAt(i))
{
ans--;
break;
}
}
return ans;
}
}
// Driver code
public static void main(String args[])
{
String str1 = "toy", str2 = "try";
int n = str1.length();
System.out.println(findAnswer(str1, str2, n));
}
}
// This code is contributed by
// Surendra_GangwarPython3
# Python3 implementation of the approach
import math as mt
# Function to return the count of
# the required strings
def findAnswer(str1, str2, n):
l, r = 0, 0
ans = 2
# Searching index after longest
# common prefix ends
for i in range(n):
if (str1[i] != str2[i]):
l = i
break
# Searching index before longest
# common suffix ends
for i in range(n - 1, -1, -1):
if (str1[i] != str2[i]):
r = i
break
if (r < l):
return 26 * (n + 1)
# If only 1 character is different
# in both the strings
elif (l == r):
return ans
else:
# Checking remaining part of
# string for equality
for i in range(l + 1, r + 1):
if (str1[i] != str2[i - 1]):
ans -= 1
break
# Searching in right of string h
# (g to h)
for i in range(l + 1, r + 1):
if (str1[i - 1] != str2[i]):
ans -= 1
break
return ans
# Driver code
str1 = "toy"
str2 = "try"
n = len(str1)
print(findAnswer(str1, str2, n))
# This code is contributed
# by Mohit kumar 29C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count of the
// required strings
static int findAnswer(string str1, string str2, int n)
{
int l = 0, r = 0;
int ans = 2;
// Searching index after longest common
// prefix ends
for (int i = 0; i < n; ++i)
{
if (str1[i] != str2[i])
{
l = i;
break;
}
}
// Searching index before longest common
// suffix ends
for (int i = n - 1; i >= 0; i--)
{
if (str1[i] != str2[i])
{
r = i;
break;
}
}
// If str1 = str2
if (r < l)
return 26 * (n + 1);
// If only 1 character is different
// in both the strings
else if (l == r)
return ans;
else
{
// Checking remaining part of string
// for equality
for (int i = l + 1; i <= r; i++)
{
if (str1[i] != str2[i - 1])
{
ans--;
break;
}
}
// Searching in right of string h
// (g to h)
for (int i = l + 1; i <= r; i++)
{
if (str1[i-1] != str2[i])
{
ans--;
break;
}
}
return ans;
}
}
// Driver code
public static void Main()
{
String str1 = "toy", str2 = "try";
int n = str1.Length;
Console.WriteLine(findAnswer(str1, str2, n));
}
}
// This code is contributed by
// shsPHP
= 0; $i--)
{
if ($str1[$i] != $str2[$i])
{
$r = $i;
break;
}
}
// If str1 = str2
if ($r < $l)
return 26 * ($n + 1);
// If only 1 character is different
// in both the strings
else if ($l == $r)
return $ans;
else
{
// Checking remaining part of string
// for equality
for ($i = $l + 1; $i <= $r; $i++)
{
if ($str1[$i] != $str2[$i - 1])
{
$ans--;
break;
}
}
// Searching in right of string h
// (g to h)
for ($i = $l + 1; $i <= $r; $i++)
{
if ($str1[$i - 1] != $str2[$i])
{
$ans--;
break;
}
}
return $ans;
}
}
// Driver code
$str1 = "toy";
$str2 = "try";
$n = strlen($str1);
echo findAnswer($str1, $str2, $n);
// This code is contributed by Ryuga
?>Javascript
输出：
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