Minimize the count of characters to be added or removed to make String repetition of same substring

Last Updated : 23 Jul, 2025

Given a string S consisting of N characters, the task is to modify the string S by performing the minimum number of following operations such that the modified string S is the concatenation of its half.

Insert any new character at any index in the string.
Remove any character from the string S.
Replace any character with any other character in the string S.

Examples:

Input: S = "aabbaabb"
Output: 0
Explanation:
The given string S = "aabbaabb" is of the form A = B + B, where B = "aabb". Therefore, the minimum number of operations required is 0.

Input: S = "aba"
Output: 1

Approach: The given problem can be solved by traversing the given string and perform the given operations at every possible index recursively and then find the minimum operations required after traversing the string. Follow the steps below to solve the given problem:

Initialize a variable, say minSteps as INT_MAX that stores the minimum number of operations required.
Traverse the given string S using the variable i and perform the following steps:
- Find the substrings S1 as S[0, i] and S2 as S[i, N].
- Now find the minimum number of steps required to convert S1 into S2 as store it in the variable count using the approach discussed in this article as the operations are similar to this article.
- Update the value of minSteps to the minimum of minSteps and count.
After completing the above steps, print the value of minSteps as the resultant minimum operation.

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 of
// the three numbers
int getMin(int x, int y, int z)
{
    return min(min(x, y), z);
}

// Function to find the minimum number
// operations required to convert string
// str1 to str2 using the operations
int editDistance(string str1, string str2,
                 int m, int n)
{
    // Stores the results of subproblems
    int dp[m + 1][n + 1];

    // Fill dp[][] in bottom up manner
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= n; j++) {

            // If str1 is empty, then
            // insert all characters
            // of string str2
            if (i == 0)

                // Minimum operations
                // is j
                dp[i][j] = j;

            // If str2 is empty, then
            // remove all characters
            // of string str2
            else if (j == 0)

                // Minimum operations
                // is i
                dp[i][j] = i;

            // If the last characters
            // are same, then ignore
            // last character
            else if (str1[i - 1] == str2[j - 1])
                dp[i][j] = dp[i - 1][j - 1];

            // If the last character
            // is different, then
            // find the minimum
            else {

                // Perform one of the
                // insert, remove and
                // the replace
                dp[i][j] = 1
                           + getMin(
                                 dp[i][j - 1],
                                 dp[i - 1][j],
                                 dp[i - 1][j - 1]);
            }
        }
    }

    // Return the minimum number of
    // steps required
    return dp[m][n];
}

// Function to find the minimum number
// of steps to modify the string such
// that first half and second half
// becomes the same
void minimumSteps(string& S, int N)
{
    // Stores the minimum number of
    // operations required
    int ans = INT_MAX;

    // Traverse the given string S
    for (int i = 1; i < N; i++) {

        string S1 = S.substr(0, i);
        string S2 = S.substr(i);

        // Find the minimum operations
        int count = editDistance(
            S1, S2, S1.length(),
            S2.length());

        // Update the ans
        ans = min(ans, count);
    }

    // Print the result
    cout << ans << '\n';
}

// Driver Code
int main()
{
    string S = "aabb";
    int N = S.length();
    minimumSteps(S, N);

    return 0;
}

Java

// Java program for the above approach
class GFG
{

// Function to find the minimum of
// the three numbers
static int getMin(int x, int y, int z)
{
    return Math.min(Math.min(x, y), z);
}

// Function to find the minimum number
// operations required to convert String
// str1 to str2 using the operations
static int editDistance(String str1, String str2,
                 int m, int n)
{
  
    // Stores the results of subproblems
    int [][]dp = new int[m + 1][n + 1];

    // Fill dp[][] in bottom up manner
    for (int i = 0; i <= m; i++) 
    {
        for (int j = 0; j <= n; j++)
        {

            // If str1 is empty, then
            // insert all characters
            // of String str2
            if (i == 0)

                // Minimum operations
                // is j
                dp[i][j] = j;

            // If str2 is empty, then
            // remove all characters
            // of String str2
            else if (j == 0)

                // Minimum operations
                // is i
                dp[i][j] = i;

            // If the last characters
            // are same, then ignore
            // last character
            else if (str1.charAt(i - 1) == str2.charAt(j - 1))
                dp[i][j] = dp[i - 1][j - 1];

            // If the last character
            // is different, then
            // find the minimum
            else {

                // Perform one of the
                // insert, remove and
                // the replace
                dp[i][j] = 1
                           + getMin(
                                 dp[i][j - 1],
                                 dp[i - 1][j],
                                 dp[i - 1][j - 1]);
            }
        }
    }

    // Return the minimum number of
    // steps required
    return dp[m][n];
}

// Function to find the minimum number
// of steps to modify the String such
// that first half and second half
// becomes the same
static void minimumSteps(String S, int N)
{
  
    // Stores the minimum number of
    // operations required
    int ans = Integer.MAX_VALUE;

    // Traverse the given String S
    for (int i = 1; i < N; i++) {

        String S1 = S.substring(0, i);
        String S2 = S.substring(i);

        // Find the minimum operations
        int count = editDistance(
            S1, S2, S1.length(),
            S2.length());

        // Update the ans
        ans = Math.min(ans, count);
    }

    // Print the result
    System.out.print(ans);
}

// Driver Code
public static void main(String[] args)
{
    String S = "aabb";
    int N = S.length();
    minimumSteps(S, N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python program for the above approach;

# Function to find the minimum of
# the three numbers
def getMin(x, y, z):
    return min(min(x, y), z)


# Function to find the minimum number
# operations required to convert string
# str1 to str2 using the operations
def editDistance(str1, str2, m, n):

    # Stores the results of subproblems
    dp = [[0 for i in range(n + 1)] for j in range(m + 1)]

    # Fill dp[][] in bottom up manner
    for i in range(0, m + 1):
        for j in range(0, n + 1):

            # If str1 is empty, then
            # insert all characters
            # of string str2
            if (i == 0):

                # Minimum operations
                # is j
                dp[i][j] = j

            # If str2 is empty, then
            # remove all characters
            # of string str2
            elif (j == 0):

                # Minimum operations
                # is i
                dp[i][j] = i

            # If the last characters
            # are same, then ignore
            # last character
            elif (str1[i - 1] == str2[j - 1]):
                dp[i][j] = dp[i - 1][j - 1]

            # If the last character
            # is different, then
            # find the minimum
            else:

                # Perform one of the
                # insert, remove and
                # the replace
                dp[i][j] = 1 + getMin( dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1])

    # Return the minimum number of
    # steps required
    return dp[m][n]


# Function to find the minimum number
# of steps to modify the string such
# that first half and second half
# becomes the same
def minimumSteps(S, N):
    # Stores the minimum number of
    # operations required
    ans = 10**10

    # Traverse the given string S
    for i in range(1, N):
        S1 = S[:i]
        S2 = S[i:]


        # Find the minimum operations
        count = editDistance(S1, S2, len(S1), len(S2))

        # Update the ans
        ans = min(ans, count)

    # Print the result
    print(ans)


# Driver Code
S = "aabb"
N = len(S)
minimumSteps(S, N)

# This code is contributed by gfgking

// C# program for the above approach
using System;

public class GFG
{

// Function to find the minimum of
// the three numbers
static int getMin(int x, int y, int z)
{
    return Math.Min(Math.Min(x, y), z);
}

// Function to find the minimum number
// operations required to convert String
// str1 to str2 using the operations
static int editDistance(string str1, string str2,
                 int m, int n)
{
  
    // Stores the results of subproblems
    int [,]dp = new int[m + 1,n + 1];

    // Fill dp[,] in bottom up manner
    for (int i = 0; i <= m; i++) 
    {
        for (int j = 0; j <= n; j++)
        {

            // If str1 is empty, then
            // insert all characters
            // of String str2
            if (i == 0)

                // Minimum operations
                // is j
                dp[i,j] = j;

            // If str2 is empty, then
            // remove all characters
            // of String str2
            else if (j == 0)

                // Minimum operations
                // is i
                dp[i,j] = i;

            // If the last characters
            // are same, then ignore
            // last character
            else if (str1[i - 1] == str2[j - 1])
                dp[i,j] = dp[i - 1,j - 1];

            // If the last character
            // is different, then
            // find the minimum
            else {

                // Perform one of the
                // insert, remove and
                // the replace
                dp[i,j] = 1
                           + getMin(
                                 dp[i,j - 1],
                                 dp[i - 1,j],
                                 dp[i - 1,j - 1]);
            }
        }
    }

    // Return the minimum number of
    // steps required
    return dp[m,n];
}

// Function to find the minimum number
// of steps to modify the String such
// that first half and second half
// becomes the same
static void minimumSteps(string S, int N)
{
  
    // Stores the minimum number of
    // operations required
    int ans = int.MaxValue;

    // Traverse the given String S
    for (int i = 1; i < N; i++) {

        string S1 = S.Substring(0, i);
        string S2 = S.Substring(i);

        // Find the minimum operations
        int count = editDistance(
            S1, S2, S1.Length,
            S2.Length);

        // Update the ans
        ans = Math.Min(ans, count);
    }

    // Print the result
    Console.Write(ans);
}

// Driver Code
public static void Main(string[] args)
{
    string S = "aabb";
    int N = S.Length;
    minimumSteps(S, N);
}
}

// This code is contributed by AnkThon

JavaScript

<script>

        // JavaScript program for the above approach;

        // Function to find the minimum of
        // the three numbers
        function getMin(x, y, z) {
            return Math.min(Math.min(x, y), z);
        }

        // Function to find the minimum number
        // operations required to convert string
        // str1 to str2 using the operations
        function editDistance(str1, str2, m, n)
        {
        
            // Stores the results of subproblems
            let dp = new Array(m + 1).fill(new Array(n + 1));

            // Fill dp[][] in bottom up manner
            for (let i = 0; i <= m; i++) {
                for (let j = 0; j <= n; j++) {

                    // If str1 is empty, then
                    // insert all characters
                    // of string str2
                    if (i == 0)

                        // Minimum operations
                        // is j
                        dp[i][j] = j;

                    // If str2 is empty, then
                    // remove all characters
                    // of string str2
                    else if (j == 0)

                        // Minimum operations
                        // is i
                        dp[i][j] = i;

                    // If the last characters
                    // are same, then ignore
                    // last character
                    else if (str1[i - 1] == str2[j - 1])
                        dp[i][j] = dp[i - 1][j - 1];

                    // If the last character
                    // is different, then
                    // find the minimum
                    else {

                        // Perform one of the
                        // insert, remove and
                        // the replace
                        dp[i][j] = 1
                            + getMin(
                                dp[i][j - 1],
                                dp[i - 1][j],
                                dp[i - 1][j - 1]);
                    }
                }
            }

            // Return the minimum number of
            // steps required
            return dp[m][n];
        }

        // Function to find the minimum number
        // of steps to modify the string such
        // that first half and second half
        // becomes the same
        function minimumSteps(S, N) {
            // Stores the minimum number of
            // operations required
            let ans = Number.MAX_VALUE;

            // Traverse the given string S
            for (let i = 1; i < N; i++) {

                let S1 = S.substring(0, i);
                let S2 = S.substring(i);

                // Find the minimum operations
                let count = editDistance(
                    S1, S2, S1.length,
                    S2.length);

                // Update the ans
                ans = Math.min(ans, count);
            }

            // Print the result
            document.write(ans - 1);
        }

        // Driver Code
        let S = "aabb";
        let N = S.length;
        minimumSteps(S, N);

   // This code is contributed by Potta Lokesh
    </script>

Output:

Time Complexity: O(N³)
Auxiliary Space: O(N²)

Efficient approach : Space optimization

In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

Create a 1D vector dp of size n+1.
Set a base case by initializing the values of DP .
Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
Now Create a temporary variable prev and temp to store previous values .
After every iteration assign the value of temp to prev for further iteration.
At last return and print the final answer stored in dp[n].

Implementation:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum of
// the three numbers
int getMin(int x, int y, int z)
{
    return min(min(x, y), z);
}

// Function to find the minimum number
// operations required to convert string
// str1 to str2 using the operations
int editDistance(string str1, string str2, int m, int n) {
    int dp[n+1];
    for (int j = 0; j <= n; j++) {
        dp[j] = j;
    }
    for (int i = 1; i <= m; i++) {
        int prev = dp[0];
        dp[0] = i;
        for (int j = 1; j <= n; j++) {
            int temp = dp[j];
            if (str1[i-1] == str2[j-1]) {
                dp[j] = prev;
            } else {
                dp[j] = 1 + min(prev, min(dp[j], dp[j-1]));
            }
            prev = temp;
        }
    }
    return dp[n];
}


// Function to find the minimum number
// of steps to modify the string such
// that first half and second half
// becomes the same
void minimumSteps(string& S, int N)
{
    // Stores the minimum number of
    // operations required
    int ans = INT_MAX;

    // Traverse the given string S
    for (int i = 1; i < N; i++) {

        string S1 = S.substr(0, i);
        string S2 = S.substr(i);

        // Find the minimum operations
        int count = editDistance(
            S1, S2, S1.length(),
            S2.length());

        // Update the ans
        ans = min(ans, count);
    }

    // Print the result
    cout << ans << '\n';
}

// Driver Code
int main()
{
    string S = "aabb";
    int N = S.length();
    minimumSteps(S, N);

    return 0;
}

// this code is contributed by bhardwajji

Java

import java.util.*;

public class Main {

    // Function to find the minimum of
    // the three numbers
    static int getMin(int x, int y, int z)
    {
        return Math.min(Math.min(x, y), z);
    }

    // Function to find the minimum number
    // operations required to convert string
    // str1 to str2 using the operations
    static int editDistance(String str1, String str2, int m,
                            int n)
    {
        int[] dp = new int[n + 1];
        for (int j = 0; j <= n; j++) {
            dp[j] = j;
        }
        for (int i = 1; i <= m; i++) {
            int prev = dp[0];
            dp[0] = i;
            for (int j = 1; j <= n; j++) {
                int temp = dp[j];
                if (str1.charAt(i - 1)
                    == str2.charAt(j - 1)) {
                    dp[j] = prev;
                }
                else {
                    dp[j]
                        = 1
                          + getMin(prev, dp[j], dp[j - 1]);
                }
                prev = temp;
            }
        }
        return dp[n];
    }

    // Function to find the minimum number
    // of steps to modify the string such
    // that first half and second half
    // becomes the same
    static void minimumSteps(String S, int N)
    {
        // Stores the minimum number of
        // operations required
        int ans = Integer.MAX_VALUE;

        // Traverse the given string S
        for (int i = 1; i < N; i++) {
            String S1 = S.substring(0, i);
            String S2 = S.substring(i);

            // Find the minimum operations
            int count = editDistance(S1, S2, S1.length(),
                                     S2.length());

            // Update the ans
            ans = Math.min(ans, count);
        }

        // Print the result
        System.out.println(ans);
    }

    // Driver Code
    public static void main(String[] args)
    {
        String S = "aabb";
        int N = S.length();
        minimumSteps(S, N);
    }
}

Python

import sys

# Function to find the minimum of the three numbers
def getMin(x, y, z):
    return min(min(x, y), z)

# Function to find the minimum number operations 
# required to convert string str1 to str2 using the operations
def editDistance(str1, str2, m, n):
    dp = [j for j in range(n+1)]
    for i in range(1, m+1):
        prev = dp[0]
        dp[0] = i
        for j in range(1, n+1):
            temp = dp[j]
            if str1[i-1] == str2[j-1]:
                dp[j] = prev
            else:
                dp[j] = 1 + getMin(prev, dp[j], dp[j-1])
            prev = temp
    return dp[n]

# Function to find the minimum number of steps to modify the string such that first half and second half becomes the same
def minimumSteps(S, N):
    # Stores the minimum number of operations required
    ans = sys.maxsize

    # Traverse the given string S
    for i in range(1, N):
        S1 = S[:i]
        S2 = S[i:]

        # Find the minimum operations
        count = editDistance(S1, S2, len(S1), len(S2))

        # Update the ans
        ans = min(ans, count)

    # Print the result
    print(ans)

# Driver Code
if __name__ == "__main__":
    S = "aabb"
    N = len(S)
    minimumSteps(S, N)

using System;

class Program {

    // Function to find the minimum of the three numbers
    static int GetMin(int x, int y, int z)
    {
        return Math.Min(Math.Min(x, y), z);
    }

    // Function to find the minimum number operations
    // required to convert string str1 to str2 using the
    // operations
    static int EditDistance(string str1, string str2, int m,
                            int n)
    {
        int[] dp = new int[n + 1];
        for (int j = 0; j <= n; j++) {
            dp[j] = j;
        }
        for (int i = 1; i <= m; i++) {
            int prev = dp[0];
            dp[0] = i;
            for (int j = 1; j <= n; j++) {
                int temp = dp[j];
                if (str1[i - 1] == str2[j - 1]) {
                    dp[j] = prev;
                }
                else {
                    dp[j]
                        = 1
                          + GetMin(prev, dp[j], dp[j - 1]);
                }
                prev = temp;
            }
        }
        return dp[n];
    }

    // Function to find the minimum number of steps to
    // modify the string such that first half and second
    // half becomes the same
    static void MinimumSteps(string S, int N)
    {
        // Stores the minimum number of operations required
        int ans = Int32.MaxValue;

        // Traverse the given string S
        for (int i = 1; i < N; i++) {
            string S1 = S.Substring(0, i);
            string S2 = S.Substring(i);

            // Find the minimum operations
            int count = EditDistance(S1, S2, S1.Length,
                                     S2.Length);

            // Update the ans
            ans = Math.Min(ans, count);
        }

        // Print the result
        Console.WriteLine(ans);
    }

    // Driver Code
    static void Main(string[] args)
    {
        string S = "aabb";
        int N = S.Length;
        MinimumSteps(S, N);
    }
}

JavaScript

// Function to find the minimum of
// the three numbers
function getMin(x, y, z) {
  return Math.min(Math.min(x, y), z);
}

// Function to find the minimum number
// operations required to convert string
// str1 to str2 using the operations
function editDistance(str1, str2, m, n) {
  const dp = new Array(n + 1);
  for (let j = 0; j <= n; j++) {
    dp[j] = j;
  }
  for (let i = 1; i <= m; i++) {
    let prev = dp[0];
    dp[0] = i;
    for (let j = 1; j <= n; j++) {
      const temp = dp[j];
      if (str1[i - 1] === str2[j - 1]) {
        dp[j] = prev;
      } else {
        dp[j] = 1 + getMin(prev, dp[j], dp[j - 1]);
      }
      prev = temp;
    }
  }
  return dp[n];
}

// Function to find the minimum number
// of steps to modify the string such
// that first half and second half
// becomes the same
function minimumSteps(S, N) {
  // Stores the minimum number of
  // operations required
  let ans = Infinity;

  // Traverse the given string S
  for (let i = 1; i < N; i++) {
    const S1 = S.substr(0, i);
    const S2 = S.substr(i);

    // Find the minimum operations
    const count = editDistance(S1, S2, S1.length, S2.length);

    // Update the ans
    ans = Math.min(ans, count);
  }

  // Print the result
  console.log(ans);
}

// Driver Code
const S = "aabb";
const N = S.length;
minimumSteps(S, N);

Output:

Time Complexity: O(N^2)
Auxiliary Space: O(N)

Analysis of Algorithms

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Practice Tags :

Minimize the count of characters to be added or removed to make String repetition of same substring

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