Skip to content
Related Articles

Related Articles

Count of characters in str1 such that after deleting anyone of them str1 becomes str2
  • Last Updated : 16 Sep, 2019

Given two strings str1 and str2, the task is to count the characters in str1 such that after removing any one of them str1 becomes identical to str2. Also, print positions of these characters. If it is not possible then print -1.

Examples:

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



Approach: Find the length of longest common prefix let it be l and the length of the longest common suffix let it be r of two strings. The solution is clearly not possible if

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

Otherwise, the valid indices are from max(len(str1) – r, 1) to min(l + 1, len(str1))

Below is the implementation of the above approach:

C++




// Below is C++ implementation of the approach
#include<bits/stdc++.h>
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
Output:
3

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :