Minimum operation require to make first and last character same

Given a string S. You are allowed two types of operations:

• Remove a character from the front of the string.
• Remove a character from the end of the string.

The task is to find the minimum operations required to make the first and last character of the S same. In case, it is not possible, print “-1”.

Examples:

```Input : S = "bacdefghipalop"
Output : 4
Remove 'b' from the front and remove 'p', 'o',
'l' from the end of the string S.

Input : S = "pqr"
Output : -1
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approaches:

Recursive: Call a recursive function passing four arguments string, starting index, ending index and count of the number of eliminations still now.

Below is the implementation of the above approach:

C++

 `// CPP program to minimum operation require  ` `// to make first and last character same ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = INT_MAX; ` ` `  `// Recursive function call ` `int` `minOperation(string &s,``int` `i,``int` `j,``int` `count) ` `{ ` `    ``if``((i>=s.size() && j<0) || (i == j)) ` `        ``return` `MAX; ` `    ``if``(s[i] == s[j]) ` `        ``return` `count; ` ` `  `    ``// Decrement ending index only ` `    ``if``(i >=s.size()) ` `        ``return` `minOperation(s,i,j-1,count+1); ` `         `  `    ``// Increment starting index only ` `    ``else` `if``(j<0) ` `        ``return` `minOperation(s,i+1,j,count+1); ` `     `  `    ``// Increment starting index and decrement index ` `    ``else` `        ``return` `min(minOperation(s,i,j-1,count+1),minOperation(s,i+1,j,count+1)); ` ` `  ` `  `} ` ` `  `// Driver code ` `int` `main() { ` `     `  `    ``string s = ``"bacdefghipalop"``; ` `     `  `    ``// Function call ` `    ``int` `ans = minOperation(s,0,s.size()-1,0); ` `     `  `    ``if``(ans == MAX) ` `        ``cout<<-1; ` `    ``else` `        ``cout<

Java

 `// Java program to minimum operation require  ` `// to make first and last character same ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `final` `int` `MAX = Integer.MAX_VALUE; ` ` `  `    ``// Recursive function call ` `    ``static` `int` `minOperation(String s, ``int` `i, ``int` `j, ``int` `count)  ` `    ``{ ` `        ``if` `((i >= s.length() && j < ``0``) || (i == j)) ` `            ``return` `MAX; ` `         `  `        ``if` `(s.charAt(i) == s.charAt(j)) ` `            ``return` `count; ` ` `  `        ``// Decrement ending index only ` `        ``if` `(i >= s.length()) ` `            ``return` `minOperation(s, i, j - ``1``, count + ``1``); ` ` `  `        ``// Increment starting index only ` `        ``else` `if` `(j < ``0``) ` `            ``return` `minOperation(s, i + ``1``, j, count + ``1``); ` ` `  `        ``// Increment starting index and decrement index ` `        ``else` `            ``return` `Math.min(minOperation(s, i, j - ``1``, count + ``1``),  ` `                            ``minOperation(s, i + ``1``, j, count + ``1``)); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``String s = ``"bacdefghipalop"``; ` ` `  `        ``// Function call ` `        ``int` `ans = minOperation(s, ``0``, s.length() - ``1``, ``0``); ` ` `  `        ``if` `(ans == MAX) ` `            ``System.out.println(-``1``); ` `        ``else` `            ``System.out.println(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by ` `// sanjeev2552 `

Output:

```4
```

Dynamic Programming:
The idea is to prevent making further recursive calls for count>= Min once we found the Min during every recursive call. It saves a lot of time and is almost comparable to tabulation method in terms of time complexity.

 `// CPP program to find minimum operation require  ` `// to make first and last character same ` `#include ` `using` `namespace` `std; ` `const` `int` `MAX = INT_MAX; ` ` `  `// To store the visited strings ` `map m; ` ` `  `int` `Min = INT_MAX;  ` ` `  `// Function to find minimum operation require  ` `// to make first and last character same ` `int` `minOperation(string &s,``int` `i,``int` `j,``int` `count) ` `{ ` `    ``// Base conditions ` `    ``if``((i>=s.size() && j<0) || (i == j)) ` `        ``return` `MAX; ` `         `  `    ``// If answer found ` `    ``if``(s[i] == s[j] || (count >= Min)) ` `        ``return` `count; ` ` `  `    ``string str = to_string(i) + ``"|"``+to_string(j); ` `     `  `    ``// If string is already visited ` `    ``if``(m.find(str) == m.end()) ` `    ``{ ` `        ``// Decrement ending index only ` `        ``if``(i >=s.size()) ` `        ``m[str]= minOperation(s,i,j-1,count+1); ` `         `  `        ``// Increment starting index only ` `        ``else` `if``(j<0) ` `        ``m[str]= minOperation(s,i+1,j,count+1); ` `         `  `        ``// Increment starting index and decrement index ` `        ``else` `        ``m[str]= min(minOperation(s,i,j-1,count+1),minOperation(s,i+1,j,count+1)); ` `    ``} ` `     `  `    ``// Store the minimum value ` `    ``if``(m[str] < Min) ` `            ``Min = m[str]; ` `    ``return` `m[str]; ` ` `  `} ` ` `  `// Driver code ` `int` `main()  ` `{ ` `    ``string s = ``"bacdefghipalop"``; ` `     `  `    ``// Function call ` `    ``int` `ans = minOperation(s,0,s.size()-1,0); ` `     `  `    ``if``(ans == MAX) ` `        ``cout<<-1; ` `    ``else` `        ``cout<

Output:

```4
```

Dynamic programming with Tabulation:
The idea is to find the first and last occurrences of each character in the string. The total amount of operations needed will be simply “number of operations needed to remove the first occurrence” plus “number of operations needed to remove the last occurrence”. So, do this for each character in the string and the answer will be minimum of such operations performed on each character.

For example, S = “zabcdefghaabbbb”, calculate the operations required to have character ‘a’ at both the front and the end, meaning to say the string “a….a”. For the minimum number of operations, we will form the string “abcdefghaa” i.e we will remove one character ‘z’ from front and 4 characters ‘bbbb’ from back. Hence total 5 operations will be required.
So, apply the above algorithm for each character and hence we can then find the minimum of those operations.

Below is implementation of this approach:

C++

 `// C++ program to find minimum operation  ` `// require to make first and last character same ` `#include ` `using` `namespace` `std; ` `#define MAX 256 ` ` `  `// Return the minimum operation require  ` `// to make string first and last character same. ` `int` `minimumOperation(string s) ` `{ ` `    ``int` `n = s.length(); ` `     `  `    ``// Store indexes of first occurrences of characters. ` `    ``vector<``int``> first_occ(MAX, -1);  ` ` `  `    ``// Initialize result ` `    ``int` `res = INT_MAX;  ` ` `  `    ``// Traverse through all characters ` `    ``for` `(``int` `i=0; i

Java

 `// Java program to find minimum  ` `// operation require to make  ` `// first and last character same ` ` `  `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` `  `class` `GFG{ ` ` `  `static` `final` `int` `MAX=``256``; ` ` `  `// Return the minimum operation requires to  ` `// make string first and last character same. ` `static` `int` `minimumOperation(String s) ` `{ ` `    ``int` `n = s.length(); ` `     `  `    ``// Store indexes of first occurrences of characters. ` `    ``Vector first_occ=``new` `Vector(); ` `     `  `    ``//Initialize all the elements to -1 ` `    ``for``(``int` `i=``0``;i

Python3

 `# Python3 program to find minimum operation  ` `# require to make first and last character same  ` `MAX` `=` `256` ` `  `# Return the minimum operation require to  ` `# make string first and last character same.  ` `def` `minimumOperation(s):  ` ` `  `    ``n ``=` `len``(s)  ` `     `  `    ``# Store indexes of first  ` `    ``# occurrences of characters.  ` `    ``first_occ ``=` `[``-``1``] ``*` `MAX` ` `  `    ``# Initialize result  ` `    ``res ``=` `float``(``'inf'``)  ` ` `  `    ``# Traverse through all characters  ` `    ``for` `i ``in` `range``(``0``, n):  ` `     `  `        ``# Find first occurrence  ` `        ``x ``=` `s[i]  ` `        ``if` `first_occ[``ord``(x)] ``=``=` `-``1``:  ` `            ``first_occ[``ord``(x)] ``=` `i  ` ` `  `        ``# Update result for subsequent occurrences  ` `        ``else``: ` `            ``last_occ ``=` `n ``-` `i ``-` `1` `            ``res ``=` `min``(res, first_occ[``ord``(x)] ``+` `last_occ)  ` `     `  `    ``return` `res  ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``s ``=` `"bacdefghipalop"` `    ``print``(minimumOperation(s))  ` `     `  `# This code is contributed by Rituraj Jain `

C#

 `// C# program to find minimum  ` `// operation require to make  ` `// first and last character same ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG ` `{ ` ` `  `static` `int` `MAX = 256; ` ` `  `// Return the minimum operation requires to  ` `// make string first and last character same. ` `static` `int` `minimumOperation(String s) ` `{ ` `    ``int` `n = s.Length; ` `     `  `    ``// Store indexes of first occurrences of characters. ` `    ``List<``int``> first_occ = ``new` `List<``int``>(); ` `     `  `    ``//Initialize all the elements to -1 ` `    ``for``(``int` `i = 0; i < MAX; i++) ` `    ``first_occ.Insert(i,-1); ` `     `  `    ``// Initialize result ` `    ``int` `res = ``int``.MaxValue;  ` ` `  `    ``// Traverse through all characters ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``// Find first occurrence ` `        ``int` `x = (``int``)s[i]; ` `        ``if` `(first_occ[x] == -1) ` `        ``first_occ.Insert(x,i); ` ` `  `        ``// Update result for subsequent occurrences ` `        ``else` `        ``{ ` `            ``int` `last_occ = (n - i - 1); ` `            ``res = Math.Min(res, first_occ[x] + last_occ); ` `        ``} ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``String s = ``"bacdefghipalop"``; ` `    ``Console.WriteLine(minimumOperation(s)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

Output:

```4
```

Time Complexity : O(n)

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