Count of distinct possible strings after performing given operations
Given a numeric string S consisting of only three types of characters 0, 1, and 2 initially and following two operations:
- The occurrence of two consecutive 1 can be replaced by 3.
- The occurrence of two consecutive 2 can be replaced by 4.
The given task is to find the total number of different strings that can be formed by using the operations.
Examples:
Input: S = “0110000022”
Output: 4
Explanation:
There can be four different formed by using the operations, the four strings are
“0110000022”, “030000022”, “03000004”, “011000004”
Input: S = “111”
Output: 3
Approach:
In order to solve this problem, we are using a dynamic programming approach. We define an array dp to store the count of possible strings of length equal to its respective indices.
- Initialize dp[0] = dp[1] = 1.
- For any index i between [1, N-1], if the character at that index is ‘1’ or ‘2’ and is equal to that of its previous index, add the possible strings that can be formed of length i-1 and i-2. Otherwise, it is the same as that of length i-1.
dp[i + 1] = dp[i] + dp[i-1] if S[i] == S[i-1] where S[i] is either ‘1’ or ‘2’.
dp[i + 1] = dp[i] otherwise.
- Return dp[n] as the count of different strings possible.
Below is the implementation of the above approach:
C++
// C++ implementation of // the above approach #include using namespace std; // Function that prints the // number of different strings // that can be formed void differentStrings(string s) { // Computing the length of // the given string int n = s.length(); vector dp(n + 1); // Base case dp[0] = dp[1] = 1; // Traverse the given string for ( int i = 1; i < n; i++) { // If two consecutive 1's // or 2's are present if (s[i] == s[i - 1] && (s[i] == '1' || s[i] == '2' )) dp[i + 1] = dp[i] + dp[i - 1]; // Otherwise take // the previous value else dp[i + 1] = dp[i]; } cout << dp[n] << "\n" ; } // Driver Code int main() { string S = "0111022110" ; differentStrings(S); return 0; } |
C
#include <stdio.h> #include <stdlib.h> #include <string.h> void differentStrings( char * s) { // Computing the length of // the given string int n = strlen (s); int dp[n + 1]; // Base case dp[0] = dp[1] = 1; // Traverse the given string for ( int i = 1; i < n; i++) { // If two consecutive 1's // or 2's are present if (s[i] == s[i - 1] && (s[i] == '1' || s[i] == '2' )) dp[i + 1] = dp[i] + dp[i - 1]; // Otherwise take // the previous value else dp[i + 1] = dp[i]; } printf ( "%d\n" , dp[n]); } // Driver Code int main() { char S[] = "0111022110" ; differentStrings(S); return 0; } // This code is contributed by phalashi. |
Java
// Java implementation of the above approach import java.io.*; class GFG{ // Function that prints the // number of different strings // that can be formed static void differentStrings(String s) { // Computing the length of // the given string int n = s.length(); int [] dp = new int [n + 1 ]; // Base case dp[ 0 ] = dp[ 1 ] = 1 ; // Traverse the given string for ( int i = 1 ; i < n; i++) { // If two consecutive 1's // or 2's are present if (s.charAt(i) == s.charAt(i - 1 ) && (s.charAt(i) == '1' || s.charAt(i) == '2' )) dp[i + 1 ] = dp[i] + dp[i - 1 ]; // Otherwise take the // previous value else dp[i + 1 ] = dp[i]; } System.out.println(dp[n]); } // Driver code public static void main(String[] args) { String S = "0111022110" ; differentStrings(S); } } // This code is contributed by offbeat |
Python3
# Python3 implementation of # the above approach # Function that prints the # number of different strings # that can be formed def differentStrings(s): # Computing the length of # the given string n = len (s) dp = [ 0 ] * (n + 1 ) # Base case dp[ 0 ] = dp[ 1 ] = 1 # Traverse the given string for i in range ( 1 , n): # If two consecutive 1's # or 2's are present if (s[i] = = s[i - 1 ] and (s[i] = = '1' or s[i] = = '2' )): dp[i + 1 ] = dp[i] + dp[i - 1 ] # Otherwise take # the previous value else : dp[i + 1 ] = dp[i] print (dp[n]) # Driver Code if __name__ = = "__main__" : S = "0111022110" differentStrings(S) # This code is contributed by Chitranayal |
C#
// C# implementation of the above approach using System; class GFG{ // Function that prints the // number of different strings // that can be formed static void differentStrings( string s) { // Computing the length of // the given string int n = s.Length; int [] dp = new int [n + 1]; // Base case dp[0] = dp[1] = 1; // Traverse the given string for ( int i = 1; i < n; i++) { // If two consecutive 1's // or 2's are present if (s[i] == s[i - 1] && (s[i] == '1' || s[i] == '2' )) dp[i + 1] = dp[i] + dp[i - 1]; // Otherwise take the // previous value else dp[i + 1] = dp[i]; } Console.Write(dp[n]); } // Driver code public static void Main() { string S = "0111022110" ; differentStrings(S); } } // This code is contributed by Code_Mech |
Javascript
<script> // JavaScript implementation of // the above approach // Function that prints the // number of different strings // that can be formed function differentStrings(s) { // Computing the length of // the given string var n = s.length; var dp = Array(n + 1); // Base case dp[0] = dp[1] = 1; // Traverse the given string for ( var i = 1; i < n; i++) { // If two consecutive 1's // or 2's are present if (s[i] == s[i - 1] && (s[i] == '1' || s[i] == '2' )) dp[i + 1] = dp[i] + dp[i - 1]; // Otherwise take // the previous value else dp[i + 1] = dp[i]; } document.write( dp[n] + "<br>" ); } // Driver Code var S = "0111022110" ; differentStrings(S); </script> |
12
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(N), as we are using extra space for DP array.
Efficient approach : Space optimization O(1)
In previous approach we the current value dp[i] is only depend upon the previous 2 values i.e. dp[i-1] and dp[i+1]. So to optimize the space we can keep track of previous and next values by the help of three variables prev and next which will reduce the space complexity from O(N) to O(1).
Implementation Steps:
- Create 2 variables prev1 and prev2 to keep track o previous values of DP.
- Initialize base case prev = curr = 1.
- Create a variable curr to store current value.
- Iterate over subproblem using loop and update curr.
- Create variable next to keep track of next value of DP.
- After every iteration update prev and curr for further iterations.
- At last return curr.
Implementation:
C++
#include <iostream> using namespace std; // Function that prints the // number of different strings // that can be formed void differentStrings(string s) { // Computing the length of // the given string int n = s.length(); // Base case int prev = 1, curr = 1; // Traverse the given string for ( int i = 1; i < n; i++) { // If two consecutive 1's // or 2's are present if (s[i] == s[i - 1] && (s[i] == '1' || s[i] == '2' )) { int next = prev + curr; prev = curr; curr = next; } // Otherwise take // the previous value else { prev = curr; } } cout << curr << "\n" ; } // Driver Code int main() { string S = "0111022110" ; differentStrings(S); return 0; } |
Java
import java.util.*; public class Main { // Utility function to find the number of different strings // that can be formed public static void differentStrings(String s) { // Computing the length of the given string int n = s.length(); // Base case int prev = 1 , curr = 1 ; // Traverse the given string for ( int i = 1 ; i < n; i++) { // If two consecutive 1's or 2's are present if (s.charAt(i) == s.charAt(i - 1 ) && (s.charAt(i) == '1' || s.charAt(i) == '2' )) { int next = prev + curr; prev = curr; curr = next; } // Otherwise take the previous value else { prev = curr; } } System.out.println(curr); } // Driver Code public static void main(String[] args) { String S = "0111022110" ; differentStrings(S); } } |
Python3
def different_strings(s: str ) - > None : # Computing the length of the given string n = len (s) # Base case prev, curr = 1 , 1 # Traverse the given string for i in range ( 1 , n): # If two consecutive 1's or 2's are present if s[i] = = s[i - 1 ] and (s[i] = = '1' or s[i] = = '2' ): next = prev + curr prev = curr curr = next # Otherwise take the previous value else : prev = curr print (curr) # Driver Code if __name__ = = '__main__' : S = "0111022110" different_strings(S) |
Output:
12
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(1)
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