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Minimum operations to make a numeric string palindrome by removing at most 2 unique character occurrences

  • Last Updated : 17 Dec, 2021

Given numeric string str, the task is to find the minimum operations needed to make the string palindrome. An operation is defined as: 

  • Select one character and remove any of its occurrences.
  • For all the operations, the number of unique characters chosen must be less than 2

Examples:

Input: str = “11221”
Output: 1
Explanation: Select ‘1’ and remove anyone character out of the first two characters of the given string. After the operation, the string becomes “1221” which is a palindrome.

Input: str = “1123211”
Output: 0
Explanation: The given string is already a palindrome.

Input: str = “1332”
Output: -1
Explanation: The string can’t be made palindrome after selecting a single character and removing the occurrences of the selected character only. 

 

Approach: This problem can be solved with the help of the two-pointer approach. 
Follow the steps below to solve the problem:

  1. Initialize a variable answer as n that stores our final answer (n is the length of the string).
  2. Iterate over ‘0’ to ‘9’ and let’s call it currentCharacter.
  3. Initialize toBeRemoved as 0. It stores the minimum number of characters to be removed of currentCharacter type from the string to make it a palindrome.
  4. Initialize i and j as 0 and n – 1 respectively.
  5. Initialize a possible as true. It helps us to determine whether it is possible to make the string palindrome by removing the occurrences of the character c only.
  6. Now run a nested while loop till i is less than j.
    • Case 1: str[i] is equal to str[j], It means that we are not required to remove any character at the present step so we can increment i and decrement j.
    • Case 2: str[i] is not equal to str[j] but str[i] is equal to currentCharacter. Increment i = i + 1 and toBeRemoved  = toBeRemoved + 1 as we are removing this character.
    • Case 3: str[i] is not equal to str[j] but str[j] is equal to currentCharacter. Increment j = j – 1 and toBeRemoved  = toBeRemoved + 1 as we are removing this character
    • Case 4: str[i] is not equal to str[j]. It means that we can’t make our string palindrome by removing the currentCharacter.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum operations
void solve(string str)
{
    // Length of the string
    int n = str.length();
 
    // answer variable for final answer
    // initializing it with INF
    int answer = n;
 
    // Iterating over characters
    // from '0' to '9' (string
    // consist of those characters only)
    for (char currentCharacter = '0';
         currentCharacter <= '9';
         currentCharacter++) {
 
        // i and j variables to check corresponding
        // characters from the start and the end
        int i = 0, j = n - 1;
 
        // toBeRemoved store the character of type
        // currentCharacter to be removed
        // to make the string palindrome
 
        int toBeRemoved = 0;
        bool possible = true;
        while (i < j) {
            // If corresponding currentCharacters are
            // already same
            if (str[i] == str[j]) {
                i++, j--;
            }
 
            // If corresponding currentCharacters are
            // not same and str[i] is equal to
            // currentCharacter
            // remove it and check for str[i + 1] and
            // str[j] in the next iteration
            else if (str[i] != str[j]
                     && str[i] == currentCharacter) {
                toBeRemoved++;
                i++;
            }
 
            // If corresponding currentCharacters are
            // not same and str[j] is equal to
            // currentCharacter
            // remove it and check for str[j - 1] and
            // str[i] in the next iteration
            else if (str[i] != str[j]
                     && str[j] == currentCharacter) {
                toBeRemoved++;
                j--;
            }
 
            // Character of currentCharacter type
            // can't be removed
            // to make the str palindrome.
            // Hence, we mark possible to false
            // break out of the loop
            else {
                possible = false;
                break;
            }
        }
 
        // If it is possible to remove
        // some occurences of currentCharacters
        // we will update our answer
        if (possible)
 
            // Take the minimum value out
            // of answer and toBeRemoved
            answer = min(answer, toBeRemoved);
    }
 
    // Print the final answer
    if (answer == n)
        cout << "-1";
    else
        cout << answer;
}
 
// Driver Code
int main()
{
    // Given string
    string str = "11221";
    solve(str);
}

Java




// Java code for the above approach
import java.io.*;
 
class GFG
{
   
  // Function to find the minimum operations
  static void solve(String str)
  {
    // Length of the string
    int n = str.length();
 
    // answer variable for final answer
    // initializing it with INF
    int answer = n;
 
    // Iterating over characters
    // from '0' to '9' (string
    // consist of those characters only)
    for (char currentCharacter = '0';
         currentCharacter <= '9';
         currentCharacter++) {
 
      // i and j variables to check corresponding
      // characters from the start and the end
      int i = 0, j = n - 1;
 
      // toBeRemoved store the character of type
      // currentCharacter to be removed
      // to make the string palindrome
 
      int toBeRemoved = 0;
      boolean possible = true;
      while (i < j) {
        // If corresponding currentCharacters are
        // already same
        if (str.charAt(i) == str.charAt(j)) {
          i++; j--;
        }
 
        // If corresponding currentCharacters are
        // not same and str[i] is equal to
        // currentCharacter
        // remove it and check for str[i + 1] and
        // str[j] in the next iteration
        else if (str.charAt(i) != str.charAt(j)
                 && str.charAt(i) == currentCharacter) {
          toBeRemoved++;
          i++;
        }
 
        // If corresponding currentCharacters are
        // not same and str[j] is equal to
        // currentCharacter
        // remove it and check for str[j - 1] and
        // str[i] in the next iteration
        else if (str.charAt(i) != str.charAt(j)
                 && str.charAt(j)  == currentCharacter) {
          toBeRemoved++;
          j--;
        }
 
        // Character of currentCharacter type
        // can't be removed
        // to make the str palindrome.
        // Hence, we mark possible to false
        // break out of the loop
        else {
          possible = false;
          break;
        }
      }
 
      // If it is possible to remove
      // some occurences of currentCharacters
      // we will update our answer
      if (possible)
 
        // Take the minimum value out
        // of answer and toBeRemoved
        answer = Math.min(answer, toBeRemoved);
    }
 
    // Print the final answer
    if (answer == n)
      System.out.println("-1");
    else
      System.out.println(answer);
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Given string
    String str = "11221";
    solve(str);
  }
}
 
// This code is contributed by lokeshpotta20.

Python3




# Python program for the above approach
 
# Function to find the minimum operations
def solve(str):
   
    # Length of the string
    n = len(str)
 
    # answer variable for final answer
    # initializing it with INF
    answer = n;
    currentCharacter = '0'
     
    # Iterating over characters
    # from '0' to '9' (string
    # consist of those characters only)
    while(currentCharacter <= '9'):
 
        # i and j variables to check corresponding
        # characters from the start and the end
        i = 0
        j = n - 1
 
        # toBeRemoved store the character of type
        # currentCharacter to be removed
        # to make the string palindrome
        toBeRemoved = 0
        possible = True
        while (i < j):
           
            # If corresponding currentCharacters are
            # already same
            if (str[i] == str[j]):
                i += 1
                j -= 1
             
            # If corresponding currentCharacters are
            # not same and str[i] is equal to
            # currentCharacter
            # remove it and check for str[i + 1] and
            # str[j] in the next iteration
            elif (str[i] != str[j] and str[i] == currentCharacter):
                toBeRemoved += 1   
                i += 1
             
            # If corresponding currentCharacters are
            # not same and str[j] is equal to
            # currentCharacter
            # remove it and check for str[j - 1] and
            # str[i] in the next iteration
            elif (str[i] != str[j] and str[j] == currentCharacter):
                toBeRemoved += 1
                j -= 1
             
            # Character of currentCharacter type
            # can't be removed
            # to make the str palindrome.
            # Hence, we mark possible to false
            # break out of the loop
            else:
                possible = False;
                break;
 
            currentCharacter = f"{int(currentCharacter) + 1}"
             
        # If it is possible to remove
        # some occurences of currentCharacters
        # we will update our answer
        if (possible):
 
            # Take the minimum value out
            # of answer and toBeRemoved
            answer = min(answer, toBeRemoved);
     
    # Print the final answer
    if (answer == n):
        print("-1");
    else:
        print(answer);
 
# Driver Code
 
# Given string
str = "11221";
solve(str);
 
# This code is contributed by Saurabh Jaiswal

C#




// C# program for above approach
using System;
class GFG
{
 
// Function to find the minimum operations
static void solve(string str)
{
    // Length of the string
    int n = str.Length;
 
    // answer variable for final answer
    // initializing it with INF
    int answer = n;
 
    // Iterating over characters
    // from '0' to '9' (string
    // consist of those characters only)
    for (char currentCharacter = '0';
         currentCharacter <= '9';
         currentCharacter++) {
 
        // i and j variables to check corresponding
        // characters from the start and the end
        int i = 0, j = n - 1;
 
        // toBeRemoved store the character of type
        // currentCharacter to be removed
        // to make the string palindrome
 
        int toBeRemoved = 0;
        bool possible = true;
        while (i < j) {
            // If corresponding currentCharacters are
            // already same
            if (str[i] == str[j]) {
                i++;
                j--;
            }
 
            // If corresponding currentCharacters are
            // not same and str[i] is equal to
            // currentCharacter
            // remove it and check for str[i + 1] and
            // str[j] in the next iteration
            else if (str[i] != str[j]
                     && str[i] == currentCharacter) {
                toBeRemoved++;
                i++;
            }
 
            // If corresponding currentCharacters are
            // not same and str[j] is equal to
            // currentCharacter
            // remove it and check for str[j - 1] and
            // str[i] in the next iteration
            else if (str[i] != str[j]
                     && str[j] == currentCharacter) {
                toBeRemoved++;
                j--;
            }
 
            // Character of currentCharacter type
            // can't be removed
            // to make the str palindrome.
            // Hence, we mark possible to false
            // break out of the loop
            else {
                possible = false;
                break;
            }
        }
 
        // If it is possible to remove
        // some occurences of currentCharacters
        // we will update our answer
        if (possible)
 
            // Take the minimum value out
            // of answer and toBeRemoved
            answer = Math.Min(answer, toBeRemoved);
    }
 
    // Print the final answer
    if (answer == n)
        Console.Write("-1");
    else
        Console.Write(answer);
}
 
// Driver Code
public static void Main()
{
    // Given string
    string str = "11221";
    solve(str);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
// Javascript program for the above approach
 
// Function to find the minimum operations
function solve(str)
{
    // Length of the string
    let n = str.length;
 
    // answer variable for final answer
    // initializing it with INF
    let answer = n;
 
    // Iterating over characters
    // from '0' to '9' (string
    // consist of those characters only)
    for (let currentCharacter = '0';
         currentCharacter <= '9';
         currentCharacter++) {
 
        // i and j variables to check corresponding
        // characters from the start and the end
        let i = 0, j = n - 1;
 
        // toBeRemoved store the character of type
        // currentCharacter to be removed
        // to make the string palindrome
 
        let toBeRemoved = 0;
        let possible = true;
        while (i < j) {
            // If corresponding currentCharacters are
            // already same
            if (str[i] == str[j]) {
                i++;
                j--;
            }
 
            // If corresponding currentCharacters are
            // not same and str[i] is equal to
            // currentCharacter
            // remove it and check for str[i + 1] and
            // str[j] in the next iteration
            else if (str[i] != str[j]
                     && str[i] == currentCharacter) {
                toBeRemoved++;
                i++;
            }
 
            // If corresponding currentCharacters are
            // not same and str[j] is equal to
            // currentCharacter
            // remove it and check for str[j - 1] and
            // str[i] in the next iteration
            else if (str[i] != str[j]
                     && str[j] == currentCharacter) {
                toBeRemoved++;
                j--;
            }
 
            // Character of currentCharacter type
            // can't be removed
            // to make the str palindrome.
            // Hence, we mark possible to false
            // break out of the loop
            else {
                possible = false;
                break;
            }
        }
 
        // If it is possible to remove
        // some occurences of currentCharacters
        // we will update our answer
        if (possible)
 
            // Take the minimum value out
            // of answer and toBeRemoved
            answer = Math.min(answer, toBeRemoved);
    }
 
    // Print the final answer
    if (answer == n)
        document.write("-1");
    else
        document.write(answer);
}
 
// Driver Code
 
// Given string
let str = "11221";
solve(str);
 
// This code is contributed by Samim Hossain Mondal.
</script>
Output
1

Time complexity: O(10 * n) = O(n) where n is the length of the string str.
Auxiliary Space: O(1) 


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