Count of strings that become equal to one of the two strings after one removal

Given two strings str1 and str2, the task is to count all the valid strings. An example of a valid string is given below: 
If str1 = "toy" and str2 = "try". Then S = "tory" is a valid string because when a single character is removed from it i.e. S = "tory" = "try" it becomes equal to str1. This property must also be valid with str2 i.e. S = "tory" = "toy" = str2. 
The task is to print the count of all possible valid strings.

Examples: 

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. 

Approach: Calculate A as a longest common prefix of str1 and str2 and C as a longest common suffix of str1 and str2. If both the string are equal then 26 * (n + 1) strings are possible. Otherwise, set count = 0 and l equal to the first index in that is not a part of the common prefix and r is the rightmost index which is not a part of the common suffix.

Now, if str1[l+1 ... r] = str2[l ... r-1] then update count = count + 1. 
And if str1[l ... r-1] = str2[l+1 ... r] then update count = count + 1. 
Print the count in the end.

Below is the implementation of the approach: 

// C++ implementation of the approach
#include <bits/stdc++.h>
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 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_Gangwar

# 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 29

// 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
// shs

<script>

// Javascript implementation of the approach

    // Function to return the count of the
    // required strings
    function findAnswer( str1,  str2 , n) 
    {
        var l = 0, r = 0;
        var ans = 2;

        // Searching index after longest common
        // prefix ends
        for (i = 0; i < n; ++i) {
            if (str1.charAt(i) != str2.charAt(i)) 
            {
                l = i;
                break;
            }
        }

        // Searching index before longest common
        // suffix ends
        for (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 (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 (i = l + 1; i <= r; i++) {
                if (str1.charAt(i - 1) != str2.charAt(i)) 
                {
                    ans--;
                    break;
                }
            }

            return ans;
        }
    }

    // Driver code
    
        var str1 = "toy", str2 = "try";
        var n = str1.length;
        document.write(findAnswer(str1, str2, n));


// This code contributed by gauravrajput1

</script>

<?php
// PHP implementation of the above approach 

// Function to return the count of  
// the required strings 
function findAnswer($str1, $str2, $n) 
{ 
    $ans = 2; 

    // Searching index after longest  
    // common prefix ends 
    for ($i = 0; $i < $n; ++$i)
    { 
        if ($str1[$i] != $str2[$i]) 
        { 
            $l = $i; 
            break; 
        } 
    } 

    // Searching index before longest 
    // common suffix ends 
    for ($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 ($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
?>

2

Complexity Analysis:

• Time Complexity: O(N), The function findAnswer() consists of several loops with linear time complexity, and some basic conditional statements with constant time complexity. Therefore, the overall time complexity of the function is O(n), where n is the length of the input strings.
• Auxiliary Space: O(1), The function findAnswer() uses a few integer variables to store the left and right indices of the non-matching characters, as well as an integer variable to store the answer. It also uses two string variables to store the input strings. Therefore, the overall space complexity of the function is O(1), which is constant space complexity.