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:
- dp[i] = min(dp[i], dp[i-1] + 1) which signifies addition of characters.
- 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 movesint 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 Codeint 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 charactersimport java.util.*;class GFG{// function to return the minimal number of movesstatic 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 Codepublic 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 charactersusing System; class GFG{// function to return the minimal number of movesstatic 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 Codepublic 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 movesfunction 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 movesfunction 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 Codelet s = "aaaaaaaa"let n = s.length// function call to return minimal number of movesconsole.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)
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