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

• Last Updated : 26 Aug, 2021

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 using namespace std; // Function to find the minimum of// the three numbersint 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 operationsint 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 samevoid 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 Codeint main(){    string S = "aabb";    int N = S.length();    minimumSteps(S, N);     return 0;}

Java

 // Java program for the above approachclass GFG{ // Function to find the minimum of// the three numbersstatic 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 operationsstatic 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 samestatic 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 Codepublic 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 numbersdef 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 operationsdef 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 samedef 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 CodeS = "aabb"N = len(S)minimumSteps(S, N) # This code is contributed by gfgking

C#

 // C# program for the above approachusing System; public class GFG{ // Function to find the minimum of// the three numbersstatic 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 operationsstatic 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 samestatic 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 Codepublic static void Main(string[] args){    string S = "aabb";    int N = S.Length;    minimumSteps(S, N);}} // This code is contributed by AnkThon

Javascript


Output:
2

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

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