Given a string **s**. Let **k** be the maximum number of partitions possible of the given string with each partition starts with distinct character. The task is to find the number of ways string s can be split into **k** partition (non-empty) such that each partition start with distinct character.

**Examples:**

Input : s = "abb" Output : 2 "abb" can be maximum split into 2 partitions {a, bb} with distinct starting character, so k = 2. And, number of ways to split "abb" into 2 partition with distinct starting character is 2 that are {a, bb} and {ab, b}. Input : s = "acbbcc" Output : 6

First we need to find the value of k. Observe that k will be equal to number of distinct characters in the string because only that number of partitions can be maximum such that each partition have distinct starting element.

Now, to find the number of ways to split string into k parts with each partition starts with distinct character. First observe that first partition will always have first character of string fixed, no matter how much it is long. Now, we need to deal with all other character except the first one.

Let’s take an example, say s = “acbbcc”, we have discussed about first character ‘a’ above. Now, to deal with ‘b’ and ‘c’, observe ‘b’ occurs at two positions in the string whereas ‘c’ at three. If we select any positions out of two for ‘b’ and any one position out of three for ‘c’, then, we can partition the string at those positions. Note that number of parts will be equal to 3 (equals to number of distinct characters in s i.e k).

So generalizing the observation, let count_{i} be the number of occurrences of character i in s. So our answer will be product of count_{i} of all i’s such that i occurs in the string and i is not equal to first character of the string.

Below is the implementation of this approach:

`// CPP Program to find number of way ` `// to split string such that each partition ` `// starts with distinct character with ` `// maximum number of partitions. ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; `
` ` `// Returns the number of we can split ` `// the string ` `int` `countWays(string s) `
`{ ` ` ` `int` `count[26] = { 0 }; `
` ` ` ` `// Finding the frequency of each `
` ` `// character. `
` ` `for` `(` `char` `x : s) `
` ` `count[x - ` `'a'` `]++; `
` ` ` ` `// making frequency of first character `
` ` `// of string equal to 1. `
` ` `count[s[0] - ` `'a'` `] = 1; `
` ` ` ` `// Finding the product of frequency `
` ` `// of occurrence of each character. `
` ` `int` `ans = 1; `
` ` `for` `(` `int` `i = 0; i < 26; ++i) `
` ` `if` `(count[i] != 0) `
` ` `ans *= count[i]; `
` ` ` ` `return` `ans; `
`} ` ` ` `// Driven Program ` `int` `main() `
`{ ` ` ` `string s = ` `"acbbcc"` `; `
` ` `cout << countWays(s) << endl; `
` ` `return` `0; `
`} ` |

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`// Java Program to find number ` `// of way to split string such ` `// that each partition starts ` `// with distinct character with ` `// maximum number of partitions. ` `import` `java.util.*; `
`import` `java.lang.*; `
`import` `java.io.*; `
` ` `class` `GFG `
`{ ` ` ` `// Returns the number of we ` `// can split the string ` `static` `int` `countWays(String s) `
`{ ` ` ` `int` `count[] = ` `new` `int` `[` `26` `]; `
` ` ` ` `// Finding the frequency of `
` ` `// each character. `
` ` `for` `(` `int` `i = ` `0` `; i < s.length(); i++) `
` ` `count[s.charAt(i) - ` `'a'` `]++; `
` ` ` ` `// making frequency of first `
` ` `// character of string equal to 1. `
` ` `count[s.charAt(` `0` `) - ` `'a'` `] = ` `1` `; `
` ` ` ` `// Finding the product of frequency `
` ` `// of occurrence of each character. `
` ` `int` `ans = ` `1` `; `
` ` `for` `(` `int` `i = ` `0` `; i < ` `26` `; ++i) `
` ` `if` `(count[i] != ` `0` `) `
` ` `ans *= count[i]; `
` ` ` ` `return` `ans; `
`} ` ` ` `// Driver Code ` `public` `static` `void` `main(String ags[]) `
`{ ` ` ` `String s = ` `"acbbcc"` `; `
` ` `System.out.println(countWays(s)); `
`} ` `} ` ` ` `// This code is contributed ` `// by Subhadeep ` |

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`# Python3 Program to find number of way ` `# to split string such that each partition ` `# starts with distinct character with ` `# maximum number of partitions. ` ` ` `# Returns the number of we can split ` `# the string ` `def` `countWays(s): `
` ` `count ` `=` `[` `0` `] ` `*` `26` `; `
` ` ` ` `# Finding the frequency of each `
` ` `# character. `
` ` `for` `x ` `in` `s: `
` ` `count[` `ord` `(x) ` `-` ` ` `ord` `(` `'a'` `)] ` `=` `(count[` `ord` `(x) ` `-` ` ` `ord` `(` `'a'` `)]) ` `+` `1` `; `
` ` ` ` `# making frequency of first character `
` ` `# of string equal to 1. `
` ` `count[` `ord` `(s[` `0` `]) ` `-` `ord` `(` `'a'` `)] ` `=` `1` `; `
` ` ` ` `# Finding the product of frequency `
` ` `# of occurrence of each character. `
` ` `ans ` `=` `1` `; `
` ` `for` `i ` `in` `range` `(` `26` `): `
` ` `if` `(count[i] !` `=` `0` `): `
` ` `ans ` `*` `=` `count[i]; `
` ` ` ` `return` `ans; `
` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: `
` ` `s ` `=` `"acbbcc"` `; `
` ` `print` `(countWays(s)); `
` ` `# This code is contributed by Rajput-Ji ` |

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`// C# Program to find number ` `// of way to split string such ` `// that each partition starts ` `// with distinct character with ` `// maximum number of partitions. ` ` ` `using` `System; `
` ` `class` `GFG `
`{ ` ` ` `// Returns the number of we ` `// can split the string ` `static` `int` `countWays(` `string` `s) `
`{ ` ` ` `int` `[] count = ` `new` `int` `[26]; `
` ` ` ` `// Finding the frequency of `
` ` `// each character. `
` ` `for` `(` `int` `i = 0; i < s.Length; i++) `
` ` `count[s[i] - ` `'a'` `]++; `
` ` ` ` `// making frequency of first `
` ` `// character of string equal to 1. `
` ` `count[s[0] - ` `'a'` `] = 1; `
` ` ` ` `// Finding the product of frequency `
` ` `// of occurrence of each character. `
` ` `int` `ans = 1; `
` ` `for` `(` `int` `i = 0; i < 26; ++i) `
` ` `if` `(count[i] != 0) `
` ` `ans *= count[i]; `
` ` ` ` `return` `ans; `
`} ` ` ` `// Driver Code ` `public` `static` `void` `Main() `
`{ ` ` ` `string` `s = ` `"acbbcc"` `; `
` ` `Console.WriteLine(countWays(s)); `
`} ` `} ` |

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**Output:**

6

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