# Number of ways to divide string in sub-strings such to make them in lexicographically increasing sequence

Given a string S, the task is to find the number of ways to divide/partition the given string in sub-strings S1, S2, S3, …, Sk such that S1 < S2 < S3 < … < Sk (Lexicographically).

Examples:

Input: S = “aabc”
Output: 6
Following are the allowed partitions:
{“aabc”}, {“aa”, “bc”}, {“aab”, “c”}, {“a”, “abc”},
{“a, “ab”, “c”} and {“aa”, “b”, “c”}.

Input: S = “za”
Output: 1
Only possible partition is {“za”}.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using dynamic programming.

• Define DP[i][j] as the number of ways to divide the sub-string S[0…j] such that S[i, j] is the last partition.
• Now, the recurrence relations will be DP[i][j] = Summation of (DP[k][i – 1]) for all k ≥ 0 and i ≤ N – 1 where N is the length of the string.
• Final answer will be the summation of (DP[i][N – 1]) for all i between 0 to N – 1 as these sub-strings will become the last partition in some possible way of partitioning.
• So, here for all the sub-strings S[i][j], find the sub-string S[k][i – 1] such that S[k][i – 1] is lexicographically less than S[i][j] and add DP[k][i – 1] to DP[i][j].

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the number of ` `// ways of partioning ` `int` `ways(string s, ``int` `n) ` `{ ` ` `  `    ``int` `dp[n][n]; ` ` `  `    ``// Initialize DP table ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``for` `(``int` `j = 0; j < n; j++) { ` `            ``dp[i][j] = 0; ` `        ``} ` ` `  `    ``// Base Case ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``dp[i] = 1; ` ` `  `    ``for` `(``int` `i = 1; i < n; i++) { ` ` `  `        ``// To store sub-string S[i][j] ` `        ``string temp; ` `        ``for` `(``int` `j = i; j < n; j++) { ` `            ``temp += s[j]; ` ` `  `            ``// To store sub-string S[k][i-1] ` `            ``string test; ` `            ``for` `(``int` `k = i - 1; k >= 0; k--) { ` `                ``test += s[k]; ` `                ``if` `(test < temp) { ` `                    ``dp[i][j] += dp[k][i - 1]; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// Add all the ways where S[i][n-1] ` `        ``// will be the last partition ` `        ``ans += dp[i][n - 1]; ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"aabc"``; ` `    ``int` `n = s.length(); ` ` `  `    ``cout << ways(s, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the above approach  ` `class` `GFG  ` `{ ` `    ``// Function to return the number of  ` `    ``// ways of partioning  ` `    ``static` `int` `ways(String s, ``int` `n)  ` `    ``{  ` `        ``int` `dp[][] = ``new` `int``[n][n];  ` `     `  `        ``// Initialize DP table  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``for` `(``int` `j = ``0``; j < n; j++) ` `            ``{  ` `                ``dp[i][j] = ``0``;  ` `            ``}  ` `     `  `        ``// Base Case  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``dp[``0``][i] = ``1``;  ` `     `  `        ``for` `(``int` `i = ``1``; i < n; i++) ` `        ``{  ` `     `  `            ``// To store sub-string S[i][j]  ` `            ``String temp = ``""``;  ` `            ``for` `(``int` `j = i; j < n; j++)  ` `            ``{  ` `                ``temp += s.charAt(j);  ` `     `  `                ``// To store sub-string S[k][i-1]  ` `                ``String test = ``""``;  ` `                ``for` `(``int` `k = i - ``1``; k >= ``0``; k--) ` `                ``{  ` `                    ``test += s.charAt(k);  ` `                    ``if` `(test.compareTo(temp) < ``0``)  ` `                    ``{  ` `                        ``dp[i][j] += dp[k][i - ``1``];  ` `                    ``}  ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``int` `ans = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{  ` `            ``// Add all the ways where S[i][n-1]  ` `            ``// will be the last partition  ` `            ``ans += dp[i][n - ``1``];  ` `        ``}  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``String s = ``"aabc"``;  ` `        ``int` `n = s.length();  ` `     `  `        ``System.out.println(ways(s, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

 `# Python3 implementation of the approach ` ` `  `# Function to return the number of ` `# ways of partioning ` `def` `ways(s, n): ` ` `  `    ``dp ``=` `[[``0` `for` `i ``in` `range``(n)] ` `             ``for` `i ``in` `range``(n)] ` ` `  `    ``# Base Case ` `    ``for` `i ``in` `range``(n): ` `        ``dp[``0``][i] ``=` `1` ` `  `    ``for` `i ``in` `range``(``1``, n): ` ` `  `        ``# To store sub-S[i][j] ` `        ``temp ``=` `"" ` `        ``for` `j ``in` `range``(i, n): ` `            ``temp ``+``=` `s[j] ` ` `  `            ``# To store sub-S[k][i-1] ` `            ``test ``=` `"" ` `            ``for` `k ``in` `range``(i ``-` `1``, ``-``1``, ``-``1``): ` `                ``test ``+``=` `s[k] ` `                ``if` `(test < temp): ` `                    ``dp[i][j] ``+``=` `dp[k][i ``-` `1``] ` ` `  `    ``ans ``=` `0` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Add all the ways where S[i][n-1] ` `        ``# will be the last partition ` `        ``ans ``+``=` `dp[i][n ``-` `1``] ` ` `  `    ``return` `ans ` ` `  `# Driver code ` `s ``=` `"aabc"` `n ``=` `len``(s) ` ` `  `print``(ways(s, n)) ` ` `  `# This code is contributed by Mohit Kumarv `

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``// Function to return the number of  ` `    ``// ways of partioning  ` `    ``static` `int` `ways(String s, ``int` `n)  ` `    ``{  ` `        ``int` `[,]dp = ``new` `int``[n, n];  ` `     `  `        ``// Initialize DP table  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``for` `(``int` `j = 0; j < n; j++) ` `            ``{  ` `                ``dp[i, j] = 0;  ` `            ``}  ` `     `  `        ``// Base Case  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``dp[0, i] = 1;  ` `     `  `        ``for` `(``int` `i = 1; i < n; i++) ` `        ``{  ` `     `  `            ``// To store sub-string S[i,j]  ` `            ``String temp = ``""``;  ` `            ``for` `(``int` `j = i; j < n; j++)  ` `            ``{  ` `                ``temp += s[j];  ` `     `  `                ``// To store sub-string S[k,i-1]  ` `                ``String test = ``""``;  ` `                ``for` `(``int` `k = i - 1; k >= 0; k--) ` `                ``{  ` `                    ``test += s[k];  ` `                    ``if` `(test.CompareTo(temp) < 0)  ` `                    ``{  ` `                        ``dp[i, j] += dp[k, i - 1];  ` `                    ``}  ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``int` `ans = 0;  ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{  ` `            ``// Add all the ways where S[i,n-1]  ` `            ``// will be the last partition  ` `            ``ans += dp[i, n - 1];  ` `        ``}  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args)  ` `    ``{  ` `        ``String s = ``"aabc"``;  ` `        ``int` `n = s.Length;  ` `     `  `        ``Console.WriteLine(ways(s, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:
```6
```

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :