Given two strings str1 and str2, where str2 is a subsequence of str1, the task is to find the length of the longest substring of str1 which when removed, makes the strings str2 and str1 equal.
Examples:
Input: str1 = “programmingbloods”, str2 = “ibloods”
Output: 8
Explanation:
Substrings to be removed from str1 are [“programm”, “ng”]. Therefore, the length of the longest deleted substring is 8.Input: str1=“GeeksforGeeks”, str2=“forks”
Output: 5
Naive Approach: The simplest approach is to generate all possible substrings of str1 and for each substring, remove it from str1 and check if the resulting string becomes equal to str2 or not. Print the length of longest such substrings.
Time Complexity: O(2N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to find the occurrence of str2 in str1. Traverse both strings from left to right and check if both characters are equal or not. If found to be true, proceed to the right in both strings. Otherwise, proceed to the right only in str2. Similarly, to find the last occurrence of str2 in str1, traverse both strings from right to left and proceed similarly. Follow the steps below to solve the problem:
- Initialize a variable, res=0 to store the length of the longest deleted substring.
- Create an array, pos[] to store the position of the first occurrence of str2 in str1.
- Traverse both strings from left to right and store the position of the first occurrence of str2 by deleting some characters of str1.
- Initialize a variable, lastPos = length(str1)-1 to store the position of the current character in the last occurrence of str2 by deleting some characters of str1.
- Traverse both strings from right to left and check if both characters match then find the position of the current character of str2 in pos[] else continue.
- If res > (lastPos – pos[i-1]-1) then update the res = lastPos – pos[i-1]-1.
- Finally, return the res.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to print the length // of longest substring to be deleted int longDelSub(string str1, string str2) { // Stores the length of string int N = str1.size(); int M = str2.size(); // Store the position of // previous matched // character of str1 int prev_pos = 0; // Store the position of first // occurrence of str2 in str1 int pos[M]; // Find the position of the // first occurrence of str2 for (int i = 0; i < M; i++) { // Store the index of str1 int index = prev_pos; // If both characters not matched while (index < N && str1[index] != str2[i]) { index++; } pos[i] = index; prev_pos = index + 1; } // Store the length of the // longest deleted substring int res = N - prev_pos; prev_pos = N - 1; // Store the position of last // occurrence of str2 in str1 for (int i = M - 1; i >= 0; i--) { int index = prev_pos; // If both characters not matched while (index >= 0 && str1[index] != str2[i]) { index--; } // Update res if (i != 0) { res = max( res, index - pos[i - 1] - 1); } prev_pos = index - 1; } // Update res. res = max(res, prev_pos + 1); return res; } // Driver Code int main() { // Given string string str1 = "GeeksforGeeks"; string str2 = "forks"; // Function Call cout << longDelSub(str1, str2); return 0; } |
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Java
// Java program to implement // the above approach import java.io.*; class GFG{ // Function to print the length // of longest substring to be deleted static int longDelSub(String str1, String str2) { // Stores the length of string int N = str1.length(); int M = str2.length(); // Store the position of // previous matched // character of str1 int prev_pos = 0; // Store the position of first // occurrence of str2 in str1 int pos[] = new int[M]; // Find the position of the // first occurrence of str2 for(int i = 0; i < M; i++) { // Store the index of str1 int index = prev_pos; // If both characters not matched while (index < N && str1.charAt(index) != str2.charAt(i)) { index++; } pos[i] = index; prev_pos = index + 1; } // Store the length of the // longest deleted substring int res = N - prev_pos; prev_pos = N - 1; // Store the position of last // occurrence of str2 in str1 for(int i = M - 1; i >= 0; i--) { int index = prev_pos; // If both characters not matched while (index >= 0 && str1.charAt(index) != str2.charAt(i)) { index--; } // Update res if (i != 0) { res = Math.max(res, index - pos[i - 1] - 1); } prev_pos = index - 1; } // Update res. res = Math.max(res, prev_pos + 1); return res; } // Driver Code public static void main (String[] args) { // Given string String str1 = "GeeksforGeeks"; String str2 = "forks"; // Function call System.out.print(longDelSub(str1, str2)); } } // This code is contributed by code_hunt |
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Python3
# Python3 program to implement # the above approach # Function to prthe length # of longest substring to be deleted def longDelSub(str1, str2): # Stores the length of string N = len(str1) M = len(str2) # Store the position of # previous matched # character of str1 prev_pos = 0 # Store the position of first # occurrence of str2 in str1 pos = [0] * M # Find the position of the # first occurrence of str2 for i in range(M): # Store the index of str1 index = prev_pos # If both characters not matched while (index < N and str1[index] != str2[i]): index += 1 pos[i] = index prev_pos = index + 1 # Store the length of the # longest deleted substring res = N - prev_pos prev_pos = N - 1 # Store the position of last # occurrence of str2 in str1 for i in range(M - 1, -1, -1): index = prev_pos # If both characters not matched while (index >= 0 and str1[index] != str2[i]): index -= 1 # Update res if (i != 0) : res = max(res, index - pos[i - 1] - 1) prev_pos = index - 1 # Update res. res = max(res, prev_pos + 1) return res # Driver Code # Given string str1 = "GeeksforGeeks"str2 = "forks" # Function call print(longDelSub(str1, str2)) # This code is contributed by code_hunt |
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C#
// C# program to implement // the above approach using System; class GFG{ // Function to print the length // of longest substring to be deleted static int longDelSub(string str1, string str2) { // Stores the length of string int N = str1.Length; int M = str2.Length; // Store the position of // previous matched // character of str1 int prev_pos = 0; // Store the position of first // occurrence of str2 in str1 int[] pos = new int[M]; // Find the position of the // first occurrence of str2 for(int i = 0; i < M; i++) { // Store the index of str1 int index = prev_pos; // If both characters not matched while (index < N && str1[index] != str2[i]) { index++; } pos[i] = index; prev_pos = index + 1; } // Store the length of the // longest deleted substring int res = N - prev_pos; prev_pos = N - 1; // Store the position of last // occurrence of str2 in str1 for(int i = M - 1; i >= 0; i--) { int index = prev_pos; // If both characters not matched while (index >= 0 && str1[index] != str2[i]) { index--; } // Update res if (i != 0) { res = Math.Max(res, index - pos[i - 1] - 1); } prev_pos = index - 1; } // Update res. res = Math.Max(res, prev_pos + 1); return res; } // Driver code public static void Main() { // Given string string str1 = "GeeksforGeeks"; string str2 = "forks"; // Function call Console.Write(longDelSub(str1, str2)); } } // This code is contributed by code_hunt |
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Output:
5
Time Complexity: O(N + M)
Auxiliary Space: O(M)
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