Minimum characters to be deleted from the beginning of two strings to make them equal

Given two strings S and T, the task is to find the minimum number of characters to be deleted from the beginning of these strings to make the two strings identical.

Two empty strings will always be an identical strings.

Examples:

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.



Approach:
Traverse both the strings simultaneously from the end and compare the characters of the two strings. The first indices i (index of S1) and j (index of S2) where the characters of the two strings differ is the length up to which the characters of S1 and S2 need to be deleted. Hence, the final value of i + j is the required answer.

Below is the implementation of the above approach:

C++

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// C++ Program to count minimum
// number of characters to be deleted
// from the beginning of the two strings
// to make the strings equal
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

10

Time Complexity: O( min(len(S1), len(S2)) )
Auxiliary Space: O(1)

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