# 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:
Following are the allowed partitions:
{“aabc”}, {“aa”, “bc”}, {“aab”, “c”}, {“a”, “abc”},
{“a, “ab”, “c”} and {“aa”, “b”, “c”}.

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

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++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the number of` `// ways of partitioning` `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[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 < 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

 `// Java implementation of the above approach ` `class` `GFG ` `{` `    ``// Function to return the number of ` `    ``// ways of partitioning ` `    ``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

 `# Python3 implementation of the approach`   `# Function to return the number of` `# ways of partitioning` `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#

 `// C# implementation of the above approach ` `using` `System;`   `class` `GFG ` `{` `    ``// Function to return the number of ` `    ``// ways of partitioning ` `    ``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`

## Javascript

 ``

Output:

`6`

Time Complexity: O(n2)

Auxiliary Space: O(n2)

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