从两个字符串的开头删除最少的字符以使它们相等

给定两个字符串S和T ，任务是找到从这些字符串的开头删除的最少字符数，以使这两个字符串相同。

Two empty strings will always be an identical strings. 

例子：

Input: S = “geeksforgeeks” T = “peeks” 
Output: 10 
Explanation: 
Substring “geeksforg” from S and “p” from T are deleted to make both the strings equals to “eeks”
Input: S = “geeksforgeeks” T = “code” 
Output: 17 
Explanation: 
Both the strings had to be deleted completely. 
 

方法：
从结束的同时遍历两个字符串，并比较两个字符串的字符。第一指数i(S1的指数)和j(S2的指数)，其中两个字符串的字符不同的是长度可达其中要被删除S1和S2需要的字符。因此， i + j的最终值是所需的答案。
下面是上述方法的实现：

C++

// C++ Program to count minimum
// number of characters to be deleted
// from the beginning of the two strings
// to make the strings equal
#include 
using namespace std;
 
// Function that finds minimum
// character required to be deleted
int minDel(string s1, string s2)
{
    int i = s1.length();
    int j = s2.length();
 
    // Iterate in the strings
    while (i > 0 && j > 0) {
 
        // Check if the characters are
        // not equal
        if (s1[i - 1] != s2[j - 1]) {
            break;
        }
 
        i--;
        j--;
    }
 
    // Return the result
    return i + j;
}
 
// Driver Program
int main()
{
    string s1 = "geeksforgeeks",
           s2 = "peeks";
 
    cout << minDel(s1, s2) << endl;
}

Java

// Java Program to count minimum
// number of characters to be deleted
// from the beginning of the two strings
// to make the strings equal
import java.util.*;
class GFG{
 
// Function that finds minimum
// character required to be deleted
static int minDel(String s1, String s2)
{
    int i = s1.length();
    int j = s2.length();
 
    // Iterate in the strings
    while (i > 0 && j > 0)
    {
 
        // Check if the characters are
        // not equal
        if (s1.charAt(i - 1) != s2.charAt(j - 1))
        {
            break;
        }
 
        i--;
        j--;
    }
 
    // Return the result
    return i + j;
}
 
// Driver Code
public static void main(String args[])
{
    String s1 = "geeksforgeeks",
           s2 = "peeks";
 
    System.out.print(minDel(s1, s2));
}
}
 
// This code is contributed by Nidhi_biet

Python3

# Python3 program to count minimum
# number of characters to be deleted
# from the beginning of the two strings
# to make the strings equal
 
# Function that finds minimum
# character required to be deleted
def minDel(s1, s2):
     
    i = len(s1)
    j = len(s2)
 
    # Iterate in the strings
    while (i > 0 and j > 0):
         
        # Check if the characters are
        # not equal
        if (s1[i - 1] != s2[j - 1]):
            break
 
        i -= 1
        j -= 1
 
    # Return the result
    return i + j
 
# Driver code
if __name__ == '__main__':
     
    s1 = "geeksforgeeks"
    s2 = "peeks"
 
    print(minDel(s1, s2))
 
# This code is contributed by Surendra_Gangwar

C#

// C# Program to count minimum
// number of characters to be deleted
// from the beginning of the two strings
// to make the strings equal
using System;
class GFG{
 
// Function that finds minimum
// character required to be deleted
static int minDel(string s1, string s2)
{
    int i = s1.Length;
    int j = s2.Length;
 
    // Iterate in the strings
    while (i > 0 && j > 0)
    {
 
        // Check if the characters are
        // not equal
        if (s1[i - 1] != s2[j - 1])
        {
            break;
        }
 
        i--;
        j--;
    }
 
    // Return the result
    return i + j;
}
 
// Driver Code
public static void Main()
{
    string s1 = "geeksforgeeks",
           s2 = "peeks";
 
    Console.Write(minDel(s1, s2));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// JavaScript Program to count minimum
// number of characters to be deleted
// from the beginning of the two strings
// to make the strings equal
 
// Function that finds minimum
// character required to be deleted
function minDel(s1, s2)
{
    var i = s1.length;
    var j = s2.length;
 
    // Iterate in the strings
    while (i > 0 && j > 0) {
 
        // Check if the characters are
        // not equal
        if (s1[i - 1] != s2[j - 1]) {
            break;
        }
 
        i--;
        j--;
    }
 
    // Return the result
    return i + j;
}
 
// Driver Program
var s1 = "geeksforgeeks",
       s2 = "peeks";
document.write( minDel(s1, s2) );
 
 

输出：

10

时间复杂度： O( min(len(S1), len(S2)) )
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live