Count of characters in str1 such that after deleting anyone of them str1 becomes str2
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: 2Input: 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
- len(str) != len(str2) + 1
- 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 |
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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 |
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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)) |
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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 |
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Output:
3
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