str1 中的字符数，以便在删除其中任何一个后 str1 变为 str2

给定两个字符串str1和str2 ，任务是计算str1中的字符，以便在删除其中任何一个之后str1变得与str2相同。此外，打印这些字符的位置。如果不可能，则打印-1 。
例子：

Input: str1 = “abdrakadabra”, str2 = “abrakadabra” 
Output: 1 
The only valid character is at index 2 i.e. str1[2]
Input: str1 = “aa”, str2 = “a” 
Output: 2
Input: str1 = “geeksforgeeks”, str2 = “competitions” 
Output: 0 
 

方法：求两个字符串的最长公共前缀长度为l，最长公共后缀长度为r。解决方案显然是不可能的，如果

1. len(str) != len(str2) + 1
2. len(str1) + 1 < n – r

否则，有效索引是从 max(len(str1) – r, 1) 到 min(l + 1, len(str1))
下面是上述方法的实现：

C++

// Below is C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count
// of required indices
int Find_Index(string str1, string str2)
{
    int n = str1.size();
    int m = str2.size();
    int l = 0;
    int r = 0;
 
    // Solution doesn't exist
    if (n != m + 1) {
        return -1;
    }
 
    // Find the length of the longest
    // common prefix of strings
    for (int i = 0; i < m; i++) {
        if (str1[i] == str2[i]) {
            l += 1;
        }
        else {
            break;
        }
    }
 
    // Find the length of the longest
    // common suffix of strings
    int i = n - 1;
    int j = m - 1;
    while (i >= 0 && j >= 0 && str1[i] == str2[j]) {
        r += 1;
        i -= 1;
        j -= 1;
    }
 
    // If solution does not exist
    if (l + r < m) {
        return -1;
    }
 
    // Return the count of indices
    else {
        i = max(n - r, 1);
        j = min(l + 1, n);
        return (j - i + 1);
    }
}
 
// Driver code
int main()
{
    string str1 = "aaa", str2 = "aa";
 
    cout << Find_Index(str1, str2);
 
    return 0;
}
 
// This code is contributed by PrinciRaj1992

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the count
    // of required indices
    static int Find_Index(String str1, String str2)
    {
        int n = str1.length();
        int m = str2.length();
        int l = 0;
        int r = 0;
 
        // Solution doesn't exist
        if (n != m + 1) {
            return -1;
        }
 
        // Find the length of the longest
        // common prefix of strings
        for (int i = 0; i < m; i++) {
            if (str1.charAt(i) == str2.charAt(i)) {
                l += 1;
            }
            else {
                break;
            }
        }
 
        // Find the length of the longest
        // common suffix of strings
        int i = n - 1;
        int j = m - 1;
        while (i >= 0 && j >= 0
               && str1.charAt(i) == str2.charAt(j)) {
            r += 1;
            i -= 1;
            j -= 1;
        }
 
        // If solution does not exist
        if (l + r < m) {
            return -1;
        }
 
        // Return the count of indices
        else {
            i = Math.max(n - r, 1);
            j = Math.min(l + 1, n);
            return (j - i + 1);
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str1 = "aaa", str2 = "aa";
        System.out.println(Find_Index(str1, str2));
    }
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach
 
# Function to return the count of required indices
 
 
def Find_Index(str1, str2):
 
    n = len(str1)
    m = len(str2)
    l = 0
    r = 0
 
    # Solution doesn't exist
    if(n != m + 1):
        return -1
 
    # Find the length of the longest
    # common prefix of strings
    for i in range(m):
        if str1[i] == str2[i]:
            l += 1
        else:
            break
 
    # Find the length of the longest
    # common suffix of strings
    i = n-1
    j = m-1
    while i >= 0 and j >= 0 and str1[i] == str2[j]:
        r += 1
        i -= 1
        j -= 1
 
    # If solution does not exist
    if l + r < m:
        return -1
 
    # Return the count of indices
    else:
        i = max(n-r, 1)
        j = min(l + 1, n)
        return (j-i + 1)
 
 
# Driver code
if __name__ == "__main__":
    str1 = "aaa"
    str2 = "aa"
    print(Find_Index(str1, str2))

C#

// Program to print the given pattern
using System;
 
class GFG {
 
    // Function to return the count
    // of required indices
    static int Find_Index(String str1, String str2)
    {
        int n = str1.Length;
        int m = str2.Length;
        int l = 0;
        int r = 0;
        int i, j;
 
        // Solution doesn't exist
        if (n != m + 1) {
            return -1;
        }
 
        // Find the length of the longest
        // common prefix of strings
        for (i = 0; i < m; i++) {
            if (str1[i] == str2[i]) {
                l += 1;
            }
            else {
                break;
            }
        }
 
        // Find the length of the longest
        // common suffix of strings
        i = n - 1;
        j = m - 1;
        while (i >= 0 && j >= 0 && str1[i] == str2[j]) {
            r += 1;
            i -= 1;
            j -= 1;
        }
 
        // If solution does not exist
        if (l + r < m) {
            return -1;
        }
 
        // Return the count of indices
        else {
            i = Math.Max(n - r, 1);
            j = Math.Min(l + 1, n);
            return (j - i + 1);
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String str1 = "aaa", str2 = "aa";
        Console.WriteLine(Find_Index(str1, str2));
    }
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript program for the above approach
 
// Function to return the count
// of required indices
function Find_index(str1,str2)
{
    var n = str1.length;
    var m = str2.length;
    var l = 0;
    var r = 0;
 
    // Solution doesn't exist
    if (n != m + 1)
    {
        return -1;
    }
 
    // Find the length of the longest
    // common prefix of strings
    for (var i = 0; i < m; i++)
    {
        if (str1[i] == str2[i])
        {
            l += 1;
        }
        else
        {
            break;
        }
    }
     
    // Find the length of the longest
    // common suffix of strings
    var i = n - 1;
    var j = m - 1;
    while (i >= 0 && j >= 0 &&
           str1[i] == str2[j])
    {
        r += 1;
        i -= 1;
        j -= 1;
    }
     
    // If solution does not exist
    if (l + r < m)
    {
        return -1;
    }
     
    // Return the count of indices
    else
    {
        i = Math.max(n - r, 1);
        j = Math.min(l + 1, n);
        return (j - i + 1);
    }
}
// Driver code  
 
// Given Strings
var str1 = "aaa";
var str2 = "aa";
  
  
// Function call
var result = Find_index(str1,str2);
  
// Print the answer
document.write(result);
 
// This code is contributed by rj13to

输出：

3