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# 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:
Explanation:
There can be four different formed by using the operations, the four strings are
“0110000022”, “030000022”, “03000004”, “011000004”
Input: S = “111”
Output:

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 = dp = 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 = dp = 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 ``#include ``#include ``void` `differentStrings(``char``* s)``{``  ` `    ``// Computing the length of``    ``// the given string``    ``int` `n = ``strlen``(s);` `    ``int` `dp[n + 1];` `    ``// Base case``    ``dp = dp = 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 = dp = 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

 ``

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(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 ``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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