Skip to content
Related Articles

Related Articles

Minimum operations required to make the string satisfy the given condition
  • Last Updated : 26 Nov, 2019

Given a string str, the task is to make the string start and end at the same character with the minimum number of given operations. In a single operation, any character of the string can be removed. Note that the length of the resultant string must be greater than 1 and it is not possible then print -1.

Examples:

Input: str = “geeksforgeeks”
Output: 3
Remove the first and the last two characters
and the string becomes “eeksforgee”

Input: str = “abcda”
Output: 0

Approach: If the string needs to start and end at a character say ch then an optimal way is to remove all the characters before the first occurrence of ch and all the characters after the last occurrence of ch. Find the number of characters that need to be removed for every possible value of ch starting from ‘a’ to ‘z’ and choose the one with the minimum number of delete operations.



Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 26;
  
// Function to return the minimum
// operations required
int minOperation(string str, int len)
{
  
    // To store the first and the last
    // occurrence of all the characters
    int first[MAX], last[MAX];
  
    // Set the first and the last occurrence
    // of all the characters to -1
    for (int i = 0; i < MAX; i++) {
        first[i] = -1;
        last[i] = -1;
    }
  
    // Update the occurrences of the characters
    for (int i = 0; i < len; i++) {
  
        int index = (str[i] - 'a');
  
        // Only set the first occurrence if
        // it hasn't already been set
        if (first[index] == -1)
            first[index] = i;
  
        last[index] = i;
    }
  
    // To store the minimum operations
    int minOp = -1;
  
    for (int i = 0; i < MAX; i++) {
  
        // If the frequency of the current
        // character in the string
        // is less than 2
        if (first[i] == -1 || first[i] == last[i])
            continue;
  
        // Count of characters to be
        // removed so that the string
        // starts and ends at the
        // current character
        int cnt = len - (last[i] - first[i] + 1);
  
        if (minOp == -1 || cnt < minOp)
            minOp = cnt;
    }
  
    return minOp;
}
  
// Driver code
int main()
{
    string str = "abcda";
    int len = str.length();
  
    cout << minOperation(str, len);
  
    return 0;
}

Java




// Java implementation of the approach 
class GFG
{
    final static int MAX = 26
      
    // Function to return the minimum 
    // operations required 
    static int minOperation(String str, int len) 
    
      
        // To store the first and the last 
        // occurrence of all the characters 
        int first[] = new int[MAX];
        int last[] = new int[MAX]; 
      
        // Set the first and the last occurrence 
        // of all the characters to -1 
        for (int i = 0; i < MAX; i++)
        
            first[i] = -1
            last[i] = -1
        
      
        // Update the occurrences of the characters 
        for (int i = 0; i < len; i++)
        
            int index = (str.charAt(i) - 'a'); 
      
            // Only set the first occurrence if 
            // it hasn't already been set 
            if (first[index] == -1
                first[index] = i; 
      
            last[index] = i; 
        
      
        // To store the minimum operations 
        int minOp = -1
      
        for (int i = 0; i < MAX; i++) 
        
      
            // If the frequency of the current 
            // character in the string 
            // is less than 2 
            if (first[i] == -1 || 
                first[i] == last[i]) 
                continue
      
            // Count of characters to be 
            // removed so that the string 
            // starts and ends at the 
            // current character 
            int cnt = len - (last[i] - first[i] + 1); 
      
            if (minOp == -1 || cnt < minOp) 
                minOp = cnt; 
        
        return minOp; 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        String str = "abcda"
        int len = str.length(); 
      
        System.out.println(minOperation(str, len)); 
    
}
  
// This code is contributed by AnkitRai01

Python3




# Python implementation of the approach
MAX = 26;
  
# Function to return the minimum
# operations required
def minOperation(str, len):
  
    # To store the first and the last
    # occurrence of all the characters
    first, last = [0] * MAX, [0] * MAX;
  
    # Set the first and the last occurrence
    # of all the characters to -1
    for i in range(MAX):
        first[i] = -1;
        last[i] = -1;
  
    # Update the occurrences of the characters
    for i in range(len):
  
        index = (ord(str[i]) - ord('a'));
  
        # Only set the first occurrence if
        # it hasn't already been set
        if (first[index] == -1):
            first[index] = i;
  
        last[index] = i;
  
    # To store the minimum operations
    minOp = -1;
  
    for i in range(MAX):
  
        # If the frequency of the current
        # character in the string
        # is less than 2
        if (first[i] == -1 or first[i] == last[i]):
            continue;
  
        # Count of characters to be
        # removed so that the string
        # starts and ends at the
        # current character
        cnt = len - (last[i] - first[i] + 1);
  
        if (minOp == -1 or cnt < minOp):
            minOp = cnt;
    return minOp;
  
# Driver code
str = "abcda";
len = len(str);
  
print( minOperation(str, len));
  
# This code is contributed by 29AjayKumar

C#




// C# implementation of the approach 
using System;
      
class GFG
{
    readonly static int MAX = 26; 
      
    // Function to return the minimum 
    // operations required 
    static int minOperation(String str, int len) 
    
      
        // To store the first and the last 
        // occurrence of all the characters 
        int []first = new int[MAX];
        int []last = new int[MAX]; 
      
        // Set the first and the last occurrence 
        // of all the characters to -1 
        for (int i = 0; i < MAX; i++)
        
            first[i] = -1; 
            last[i] = -1; 
        
      
        // Update the occurrences of the characters 
        for (int i = 0; i < len; i++)
        
            int index = (str[i] - 'a'); 
      
            // Only set the first occurrence if 
            // it hasn't already been set 
            if (first[index] == -1) 
                first[index] = i; 
      
            last[index] = i; 
        
      
        // To store the minimum operations 
        int minOp = -1; 
      
        for (int i = 0; i < MAX; i++) 
        
      
            // If the frequency of the current 
            // character in the string 
            // is less than 2 
            if (first[i] == -1 || 
                first[i] == last[i]) 
                continue
      
            // Count of characters to be 
            // removed so that the string 
            // starts and ends at the 
            // current character 
            int cnt = len - (last[i] - first[i] + 1); 
      
            if (minOp == -1 || cnt < minOp) 
                minOp = cnt; 
        
        return minOp; 
    
      
    // Driver code 
    public static void Main (String[] args) 
    
        String str = "abcda"
        int len = str.Length; 
      
        Console.WriteLine(minOperation(str, len)); 
    
}
  
// This code is contributed by 29AjayKumar
Output:
0

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 :