Comment Vérifier si Deux Chaînes de Caractères sont à une Distance d'Édition L'une de l'Autre ? (avec Exemples de Code)

// Java program to check if given
// two strings are at distance one.
class GFG {

    // Returns true if edit distance
    // between s1 and s2 is one, else false
    static boolean isEditDistanceOne(String s1, String s2) {
        // Find lengths of given strings
        int m = s1.length(), n = s2.length();

        // If difference between lengths is
        // more than 1, then strings can't
        // be at one distance
        if (Math.abs(m - n) > 1)
            return false;

        int count = 0; // Count of edits

        int i = 0, j = 0;
        while (i < m && j < n) {
            // If current characters don't match
            if (s1.charAt(i) != s2.charAt(j)) {
                if (count == 1)
                    return false;

                // If length of one string is
                // more, then only possible edit
                // is to remove a character
                if (m > n)
                    i++;
                else if (m < n)
                    j++;
                else { // Iflengths of both strings
                    // is same
                    i++;
                    j++;
                }

                // Increment count of edits
                count++;
            }
            else { // If current characters match
                i++;
                j++;
            }
        }

        // If last character is extra
        // in any string
        if (i < m || j < n)
            count++;

        return count == 1;
    }

    // driver code
    public static void main (String[] args) {
        String s1 = "gfg";
        String s2 = "gf";

        if (isEditDistanceOne(s1, s2))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}
// This code is contributed by Anant Agarwal.

# Python program to check if given two strings are
# at distance one

# Returns true if edit distance between s1 and s2 is
# one, else false
def isEditDistanceOne(s1, s2):
    # Find lengths of given strings
    m = len(s1)
    n = len(s2)

    # If difference between lengths is more than 1,
    # then strings can't be at one distance
    if abs(m - n) > 1:
        return False

    count = 0  # Count of isEditDistanceOne

    i = 0
    j = 0
    while i < m and j < n:
        # If current characters dont match
        if s1[i] != s2[j]:
            if count == 1:
                return False

            # If length of one string is
            # more, then only possible edit
            # is to remove a character
            if m > n:
                i += 1
            elif m < n:
                j += 1
            else:  # If lengths of both strings is same
                i += 1
                j += 1
            # Increment count of edits
            count += 1
        else:    # if current characters match
            i += 1
            j += 1

    # if last character is extra in any string
    if i < m or j < n:
        count += 1

    return count == 1

# Driver program
s1 = "gfg"
s2 = "gf"
if isEditDistanceOne(s1, s2):
    print("Yes")
else:
    print("No")

# This code is contributed by Bhavya Jain

// C# program to check if given
// two strings are at distance one.
using System;

class GFG
{

    // Returns true if edit distance
    // between s1 and s2 is one, else false
    static bool isEditDistanceOne(String s1,
                                        String s2)
    {
        
        // Find lengths of given strings
        int m = s1.Length, n = s2.Length;

        // If difference between lengths is
        // more than 1, then strings can't
        // be at one distance
        if (Math.Abs(m - n) > 1)
            return false;
        
        // Count of edits
        int count = 0;
        int i = 0, j = 0;
        
        while (i < m && j < n)
        {
            // If current characters
            // don't match
            if (s1[i] != s2[j])
            {
                if (count == 1)
                    return false;
                
                // If length of one string is
                // more, then only possible edit
                // is to remove a character
                if (m > n)
                    i++;
                else if (m< n)
                    j++;
                    
                // If lengths of both
                // strings is same
                else
                {
                    i++;
                    j++;
                }
                
                // Increment count of edits
                count++;
            }

            // If current characters match
            else
            {
                i++;
                j++;
            }
        }

        // If last character is extra
        // in any string
        if (i < m || j < n)
            count++;

        return count == 1;
    }

    // Driver code
    public static void Main()
    {
        String s1 = "gfg";
        String s2 = "gf";
        if (isEditDistanceOne(s1, s2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by Sam007.

    // Javascript program to check if given
    // two strings are at distance one.
    
    // Returns true if edit distance
    // between s1 and s2 is one, else false
    function isEditDistanceOne(s1, s2)
    {
        // Find lengths of given strings
        let m = s1.length, n = s2.length;
        // If difference between lengths is
        // more than 1, then strings can't
        // be at one distance
        if (Math.abs(m - n) > 1)
            return false ;
            
            // Count of edits
            let count = 0;
            let i = 0, j = 0;
        while (i < m && j < n)
        {
            // If current characters
            // don't match
            if (s1[i] != s2[j])
            {
                if (count == 1)
                    return false;
                    
                    // If length of one string is
                    // more, then only possible edit
                    // is to remove a character
                    if (m > n)
                    i++;
                    else if (m< n)
                    j++;
                    // If lengths of both
                    // strings is same
                    else
                    {
                        i++;
                        j++;
                    }
                    // Increment count of edits
                    count++;
                }
                // If current characters match
                else
                {
                    i++;
                    j++;
                }
            }
            // If last character is extra
            // in any string
            if (i < m || j < n)
            count++;
            return count == 1;
        }