# Operations required to make the string empty

Given a string **str**, the task is to make the string empty with the given operation. In a single operation, you can pick some characters of the string (each of the picked characters should have the same frequency) and remove them from the string. Print the total operations required to make the string empty.

**Examples:**

Input:str = “aabbccc”

Output:2

In one operation, characters ‘a’ and ‘b’ can be removed since both have the same frequency.

Second operation can remove character ‘c’ having frequency 3.

Total 2 operations are required.

Input:str = “geeksforgeeks”

Output:3

**Approach:** Find unique frequencies of the characters of the string. Total count of unique frequencies will be the number of operations required to make the string empty.

For **str = “aaabbbcccc”**, unique frequencies are **3** and **4**. Total count of unique frequencies is **2**.

HashMap can be used to store the characters and their frequencies then HashSet can be used to find the count of unique frequencies which is the number of operations required.

Below is the implementation of the above approach:

## Java

`// Java implementation of the approach ` `import` `java.util.*; ` `class` `GFG { ` ` ` ` ` `// Function to return the count of operations required ` ` ` `static` `int` `totalOperations(String str, ` `int` `len) ` ` ` `{ ` ` ` ` ` `// HashMap to store characters and their frequencies ` ` ` `HashMap<Character, Integer> h = ` `new` `HashMap<Character, Integer>(); ` ` ` `for` `(` `int` `i = ` `0` `; i < len; i++) { ` ` ` ` ` `// If already contains the character then ` ` ` `// increment its frequency by 1 ` ` ` `if` `(h.containsKey(str.charAt(i))) ` ` ` `h.put(str.charAt(i), h.get(str.charAt(i)) + ` `1` `); ` ` ` ` ` `// Else add the character to the HashMap with frequency 1 ` ` ` `else` ` ` `h.put(str.charAt(i), ` `1` `); ` ` ` `} ` ` ` ` ` `// Set to iterate over HashMap ` ` ` `Set<Map.Entry<Character, Integer> > set = h.entrySet(); ` ` ` ` ` `// HashSet to store unique frequency ` ` ` `HashSet<Integer> hs = ` `new` `HashSet<Integer>(); ` ` ` ` ` `// Insert frequencies into HashSet ` ` ` `for` `(Map.Entry<Character, Integer> me : set) ` ` ` `hs.add(me.getValue()); ` ` ` ` ` `// Count of unique frequencies ` ` ` `return` `hs.size(); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `String str = ` `"geeksforgeeks"` `; ` ` ` `int` `len = str.length(); ` ` ` `System.out.println(totalOperations(str, len)); ` ` ` `} ` `} ` |

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## Python3

`# Python implementation of the approach ` ` ` `# Function to return the count of operations required ` `def` `totalOperations(st, length): ` ` ` ` ` `# Dictionary to store characters and their frequencies ` ` ` `d ` `=` `{} ` ` ` `for` `i ` `in` `range` `(length): ` ` ` ` ` `# If already contains the character then ` ` ` `# increment its frequency by 1 ` ` ` `if` `st[i] ` `in` `d: ` ` ` `d[st[i]] ` `+` `=` `1` ` ` ` ` `# Else add the character to the HashMap with frequency 1 ` ` ` `else` `: ` ` ` `d[st[i]] ` `=` `1` ` ` ` ` `# Set to Store unique frequency ` ` ` `valueSet ` `=` `set` `() ` ` ` ` ` `# Insert frequencies into HashSet ` ` ` `for` `key ` `in` `d.keys(): ` ` ` `valueSet.add(d[key]) ` ` ` ` ` `# Count of unique frequencies ` ` ` `return` `len` `(valueSet) ` ` ` `# Driver Code ` `st ` `=` `"geeksforgeeks"` `l ` `=` `len` `(st) ` `print` `(totalOperations(st, l)) ` ` ` `# This code is contributed by Vivekkumar Singh ` |

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**Output:**

3

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