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Minimal moves to form a string

Minimal moves to form a string: In this article, we are going to see a standard interview problem and its solution which can be featured in any company interview process.
Submitted by Radib Kar, on June 09, 2020

Problem statement

Given a string S, 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, you can:

1. Add any character at the end of the string.
2. Append the existing string to the string itself.

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

Input

Input String

Output

Minimum number of steps required to form that string.

Constraints:

1 <= string length <= 10^5

Example

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:
ijhijhi

Output:
5

Explanation:
Move 1: add 'i' to form "i"
Move 2: add 'j' to form "ij"
Move 3: add 'h' to form "ijh"
Move 4: append "ijh" to form "ijhijh"
Move 5: add 'i' to form "ijhijhi"

Solution Approach

So, we have two option

1. Append single character
2. Append the string itself

So, say the string is

x1x2 ... xn

For any sub-problem of size i,

x1x2 ... xi

Two things are possible,

1. Appends x_i to sub-problem x₁x₂ ... x_(i-1)
2. Append x₁..x_j to x₁..x_j is x_(j+1)..x_i is similar to x₁..x_j

So, we can formulate the problem recursively,

f(i) = minimum steps to form x1x2 ... xi  
f(i) = f(i-1)+1
f(i) = f(j)+1 if j=i/2 and the partitioned strings are same 

Now find the minimum.

The base value of f(i)=i obviously which is maximum value to form the string.

So, the above recursion can be transformed to DP.

1)  Declare int dp[n+1];
2)  Assign the base value.    
    for i=0 to n
        dp[i]=i;
3)  Fill the DP array
    for i=2 to n
        dp[i]=minimum(dp[i-1]+1,dp[i]); // normal insertion, recursion first case
        if i is even
            if s1s(i/2) = s(i/2+1)sn // if appending string is possible
                dp[i]=minimum(dp[i],dp[i/2]+1);
        end if
    end for
    
4)  return dp[n];

DP[n] is the final result

C++ Implementation

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

int minimumMoves(string s)
{
    int n = s.length();

    int dp[n + 1];
    for (int i = 0; i <= n; i++)
        dp[i] = i;

    for (int i = 2; i <= n; i++) {
        dp[i] = std::min(dp[i - 1] + 1, dp[i]);
        if (i % 2 == 0) {
            int flag = 0;
            for (int j = 0; j < i / 2; j++) {
                if (s[j] != s[j + i / 2]) {
                    flag = 1;
                    break;
                }
            }
            if (flag == 0)
                dp[i] = std::min(dp[i], dp[i / 2] + 1);
        }
    }

    return dp[n];
}

int main()
{
    cout << "Enter input string\n";
    string s;
    cin >> s;
    
    cout << "Minimum steps to form te string: " << minimumMoves(s) << endl;

    return 0;
}

Output

Enter input string
incinclude
Minimum steps to form te string: 8