Reduce string to shortest length by deleting a pair of same adjacent characters

Given a string str of lowercase characters. The task is to count the number of deletions required to reduce the string to its shortest length. In each delete operation, you can select a pair of adjacent lowercase letters that match, and then delete them. The task is to print the count of deletions done.

Examples:

```Input: str = "aaabccddd"
Output: 3
Following are sequence of operations:
aaabccddd -> abccddd -> abddd -> abd

Input: str = "aa"
Output: 1
```

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

Approach:

1. Initialize count = 1 initially.
2. Iterate for every character, increase count if s[i]==s[i-1].
3. If s[i]!=s[i-1], add count/2 to the number of steps, and re-initialize count to 1.

If s[i]!=s[i-1], then the number of deletions is increased by count/2. If the count is even, number of pairs will be count/2. If count is odd, then the number of deletions will be (count-1)/2 which is the same as (int)count/2.

Below is the implementation of the above approach:

C++

 `// C++ program to count deletions ` `// to reduce the string to its shortest ` `// length by deleting a pair of ` `// same adjacent characters ` `#include ` `using` `namespace` `std; ` ` `  `// Function count the operations ` `int` `reduceString(string s, ``int` `l) ` `{ ` ` `  `    ``int` `count = 1, steps = 0; ` ` `  `    ``// traverse in the string ` `    ``for` `(``int` `i = 1; i < l; i++) { ` `        ``// if adjacent characters are same ` `        ``if` `(s[i] == s[i - 1]) ` `            ``count += 1; ` ` `  `        ``else` `{ ` `            ``// if same adjacent pairs are more than 1 ` `         `  `                ``steps += (count / 2); ` ` `  `            ``count = 1; ` `        ``} ` `    ``} ` ` `  `     `  `        ``steps += count / 2; ` `    ``return` `steps; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``string s = ``"geeksforgeeks"``; ` `     `  `    ``int` `l = s.length(); ` `    ``cout << reduceString(s, l) << endl; ` `    ``return` `0; ` `} `

Java

 `// Java program to count deletions ` `// to reduce the string to its  ` `// shortest length by deleting a  ` `// pair of same adjacent characters ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG ` `{ ` `     `  `// Function count ` `// the operations ` `static` `int` `reduceString(String s,  ` `                        ``int` `l) ` `{ ` ` `  `    ``int` `count = ``1``, steps = ``0``; ` ` `  `    ``// traverse in the string ` `    ``for` `(``int` `i = ``1``; i < l; i++) ` `    ``{ ` `        ``// if adjacent characters  ` `        ``// are same ` `        ``if` `(s.charAt(i) == s.charAt(i - ``1``)) ` `            ``count += ``1``; ` ` `  `        ``else` `        ``{ ` `            ``// if same adjacent pairs  ` `            ``// are more than 1 ` `            ``steps += (count / ``2``); ` ` `  `            ``count = ``1``; ` `        ``} ` `    ``} ` `        ``steps += count / ``2``; ` `    ``return` `steps; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String s = ``"geeksforgeeks"``; ` `     `  `    ``int` `l = s.length(); ` `    ``System.out.print(reduceString(s, l) + ``"\n"``); ` `} ` `} `

Python3

 `# Python3 program to count  ` `# deletions to reduce  ` `# the string to its  ` `# shortest length by  ` `# deleting a pair of  ` `# same adjacent characters ` `  `  `# Function count  ` `# the operations ` `def` `reduceString(s, l): ` `    ``count ``=` `1``; ` `    ``steps ``=` `0``; ` `  `  `    ``# traverse in  ` `    ``# the string ` `    ``for` `i ``in` `range``(``1``,l): ` `        ``# if adjacent  ` `        ``# characters are same ` `        ``if` `(s[i] ``is` `s[i ``-` `1``]): ` `            ``count ``+``=` `1``; ` `  `  `        ``else``: ` `            ``# if same adjacent pairs  ` `            ``# are more than 1 ` `            ``steps ``+``=``(``int``)(count ``/` `2``); ` `  `  `            ``count ``=` `1``; ` `        ``steps ``+``=``(``int``)(count ``/` `2``); ` `    ``return` `steps; ` ` `  `  `  `# Driver Code ` `s ``=` `"geeksforgeeks"``; ` `  `  `l ``=` `len``(s); ` `print``(reduceString(s, l)); ` ` `  ` `  `# This code contributed by Rajput-Ji `

C#

 `// C# program to count deletions ` `// to reduce the string to its  ` `// shortest length by deleting a  ` `// pair of same adjacent characters ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function count  ` `// the operations ` `static` `int` `reduce(``string` `s,  ` `                  ``int` `l) ` `{ ` ` `  `    ``int` `count = 1, step = 0; ` ` `  `    ``// traverse in  ` `    ``// the string ` `    ``for` `(``int` `i = 1; i < l; i++)  ` `    ``{ ` `        ``// if adjacent characters ` `        ``// are same ` `        ``if` `(s[i] == s[i - 1]) ` `            ``count += 1; ` ` `  `        ``else`  `        ``{ ` `            ``// if same adjacent pairs ` `            ``// are more than 1 ` `            ``step += (count / 2); ` `            ``count = 1; ` `        ``} ` `    ``} ` `        ``step += count / 2; ` `    ``return` `step; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``string` `s = ``"geeksforgeeks"``; ` `     `  `    ``int` `l = s.Length; ` `    ``Console.WriteLine(reduce(s, l)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Akanksha Rai(Abby_akku) `

PHP

 ` `

Output:

```2
```

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