Minimal moves to form a string by adding characters or appending string itself

Given a string S, we need to write a program to check if it is possible to construct the given string S by performing any of the below operations any number of times. In each step, we can:

  • Add any character at the end of the string.
  • or, append the string to the string itself.

The above steps can be applied any number of times. We need to write a program to print the minimum steps required to form the string.

Examples:

Input : aaaaaaaa
Output : 4 
Explanation: move 1: add 'a' to form "a"
move 2: add 'a' to form "aa"
move 3: append "aa" to form "aaaa" 
move 4: append "aaaa" to form "aaaaaaaa" 

Input: aaaaaa
Output: 4 
Explanation: move 1: add 'a' to form "a"
move 2: add 'a' to form "aa"
move 3: add 'a' to form "aaa" 
move 4: append "aaa" to form "aaaaaa" 

Input: abcabca
Output: 5  


The idea to solve this problem is to use Dynamic Programming to count the minimum number of moves. Create an array named dp of size n, where n is the length of the input string. dp[i] stores the minimum number of moves that are required to make substring (0…i). According to the question there are two moves that are possible:

  1. dp[i] = min(dp[i], dp[i-1] + 1) which signifies addition of characters.
  2. dp[i*2+1] = min(dp[i]+1, dp[i*2+1]), appending of string is done if s[0…i]==s[i+1..i*2+1]
  3. The answer will be stored in dp[n-1] as we need to form the string(0..n-1) index-wise.

    Below is the implementation of above idea:

    CPP

    // CPP program to print the
    // Minimal moves to form a string
    // by appending string and adding characters
    #include <bits/stdc++.h>
    using namespace std;
    
    // fucntion to return the minimal number of moves
    int minimalSteps(string s, int n)
    {
    
        int dp[n];
    
        // initializing dp[i] to INT_MAX
        for (int i = 0; i < n; i++)
            dp[i] = INT_MAX;
    
        // initialize both strings to null
        string s1 = "", s2 = "";
        
        // base case
        dp[0] = 1;
    
        s1 += s[0];
    
        for (int i = 1; i < n; i++) {
            s1 += s[i];
    
            // check if it can be appended
            s2 = s.substr(i + 1, i + 1);
    
            // addition of character takes one step
            dp[i] = min(dp[i], dp[i - 1] + 1);
    
            // appending takes 1 step, and we directly
            // reach index i*2+1 after appending
            // so the number of steps is stord in i*2+1
            if (s1 == s2)
                dp[i * 2 + 1] = min(dp[i] + 1, dp[i * 2 + 1]);
        }
    
        return dp[n - 1];
    }
    
    // Driver Code
    int main()
    {
    
        string s = "aaaaaaaa";
        int n = s.length();
    
        // fucntion call to return minimal number of moves
        cout << minimalSteps(s, n);
    
        return 0;
    }
    

    PHP

    
    <?php
    // php program to print the
    // Minimal moves to form a 
    // string by appending string
    // and adding characters
    
    // function to return the
    // minimal number of moves
    function minimalSteps($s,$n)
    {
    
        // initializing dp[i] to INT_MAX
        for ($i = 0; $i < $n; $i++)
            $dp[$i] = PHP_INT_MAX;
    
        // initialize both 
        // strings to null
        $s1 = "";
        $s2 = "";
        
        // base case
        $dp[0] = 1;
    
        $s1=$s1.$s[0];
    
        for ($i = 1; $i < $n; $i++) 
        {
            $s1 = $s1.$s[$i];
    
            // check if it can 
            // be appended
            $s2 = substr($s, $i + 1, $i + 1);
    
            // addition of character
            // takes one step
            $dp[$i] = min($dp[$i], 
                      $dp[$i - 1] + 1);
    
            // appending takes 1 step,
            // and we directly
            // reach index i*2+1 
            // after appending
            // so the number of steps 
            // is stord in i*2+1
            if ($s1 == $s2)
                $dp[$i * 2 + 1] = min($dp[$i] + 1,
                                  $dp[$i * 2 + 1]);
        }
    
        return $dp[$n - 1];
    }
    
        // Driver Code
        $s = "aaaaaaaa";
        $n = strlen($s);
    
        // function call to return
        //minimal number of moves
        echo minimalSteps($s, $n);
    
    // This code is contributed by mits 
    ?>
    
    


    Output:

    4
    

    Time Complexity : O(n2), where n is the length of input string.



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    Improved By : Mithun Kumar




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