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]

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 the above idea:

C++

// CPP program to print the
// Minimal moves to form a string
// by appending string and adding characters
#include <bits/stdc++.h>
using namespace std;

// function 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 stored 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();

    // function call to return minimal number of moves
    cout << minimalSteps(s, n);

    return 0;
}

Java

// Java program to print the
// Minimal moves to form a string
// by appending string and adding characters
import java.util.*;

class GFG
{

// function to return the minimal number of moves
static int minimalSteps(String s, int n)
{

    int []dp = new int[n];

    // initializing dp[i] to INT_MAX
    for (int i = 0; i < n; i++)
        dp[i] = Integer.MAX_VALUE;

    // initialize both strings to null
    String s1 = "", s2 = "";

    // base case
    dp[0] = 1;

    s1 += s.charAt(0);

    for (int i = 1; i < n; i++)
    {
        s1 += s.charAt(i);

        // check if it can be appended
        s2 = s.substring(i + 1, i + 1);

        // addition of character takes one step
        dp[i] = Math.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 stored in i*2+1
        if (s1 == s2)
            dp[i * 2 + 1] = Math.min(dp[i] + 1,
                                   dp[i * 2 + 1]);
    }
    return dp[n - 1];
}

// Driver Code
public static void main(String args[])
{

    String s = "aaaaaaaa";
    int n = s.length();

    // function call to return minimal number of moves
    System.out.println(minimalSteps(s, n)/2);
}
}

// This code is contributed by
// Shashank_Sharma

Python3

# Python program to print the
# Minimal moves to form a string
# by appending string and adding characters

INT_MAX = 100000000

# function to return the
# minimal number of moves
def minimalSteps(s, n):
    dp = [INT_MAX for i in range(n)]

    # initialize both strings to null
    s1 = ""
    s2 = ""

    # base case
    dp[0] = 1
    s1 += s[0]

    for i in range(1, n):
        s1 += s[i]

        # check if it can be appended
        s2 = s[i + 1: 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 stored 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 =len(s)

# function call to return
# minimal number of moves
print( minimalSteps(s, n) )

# This code is contributed
# by sahilshelangia

C#

// C# program to print the
// Minimal moves to form a string
// by appending string and adding characters
using System;

class GFG
{

// function to return the minimal number of moves
static int minimalSteps(String s, int n)
{

    int []dp = new int[n];

    // initializing dp[i] to INT_MAX
    for (int i = 0; i < n; i++)
        dp[i] = int.MaxValue;

    // 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.Substring(i , 1);

        // addition of character takes one step
        dp[i] = Math.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 stored in i*2+1
        if (s1 == s2)
            dp[i * 2 + 1] = Math.Min(dp[i] + 1,
                                dp[i * 2 + 1]);
    }
    return dp[n - 1];
}

// Driver Code
public static void Main(String []args)
{

    String s = "aaaaaaaa";
    int n = s.Length;

    // function call to return minimal number of moves
    Console.Write(minimalSteps(s, n)/2);
}
}

// This code has been contributed by 29AjayKumar

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 stored 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
?>

Javascript

// Javascript 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)
{
    let dp = [n]

    // initializing dp[i] to INT_MAX
    for (let i = 0; i < n; i++)
        dp[i] = Number.MAX_SAFE_INTEGER

    // initialize both strings to null
    let s1 = ""
    let s2 = ""

    // base case
    dp[0] = 1

    s1 += s[0]

    for (let 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] = Math.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 stored in i*2+1
        if (s1 == s2)
            dp[i * 2 + 1] = Math.min(dp[i] + 1, dp[i * 2 + 1])
    }

    return dp[n - 1]
}

// Driver Code
let s = "aaaaaaaa"
let n = s.length

// function call to return minimal number of moves
console.log(minimalSteps(s, n))

// This code is contributed by Samim Hossain Mondal.

Output

4

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