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

05 May, 2021
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
Approach:  
Initialize count = 1 initially.Iterate for every character, increase count if s[i]==s[i-1].If s[i]!=s[i-1], add count/2 to the number of steps, and re-initialize count to 1.
Initialize count = 1 initially.
Iterate for every character, increase count if s[i]==s[i-1].
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++
Java
Python3
C#
PHP
Javascript
// C++ program to count deletions// to reduce the string to its shortest// length by deleting a pair of// same adjacent characters#include <bits/stdc++.h>using namespace std; // Function count the operationsint 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 Codeint main(){     string s = "geeksforgeeks";         int l = s.length();    cout << reduceString(s, l) << endl;    return 0;}
// Java program to count deletions// to reduce the string to its// shortest length by deleting a// pair of same adjacent charactersimport java.io.*;import java.util.*;import java.lang.*; class GFG{     // Function count// the operationsstatic 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 Codepublic static void main(String[] args){    String s = "geeksforgeeks";         int l = s.length();    System.out.print(reduceString(s, l) + "\n");}}
# Python3 program to count# deletions to reduce# the string to its# shortest length by# deleting a pair of# same adjacent characters  # Function count# the operationsdef 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 Codes = "geeksforgeeks";  l = len(s);print(reduceString(s, l));  # This code contributed by Rajput-Ji
// C# program to count deletions// to reduce the string to its// shortest length by deleting a// pair of same adjacent charactersusing System; class GFG{     // Function count// the operationsstatic 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 Codepublic 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// PHP program to count// deletions to reduce// the string to its// shortest length by// deleting a pair of// same adjacent characters // Function count// the operationsfunction reduceString($s, $l){    $count = 1;    $steps = 0;     // traverse in    // the string    for ($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            $steps +=(int)($count / 2);             $count = 1;        }    }        $steps +=(int)($count / 2);    return $steps;} // Driver Code$s = "geeksforgeeks"; $l = strlen($s);echo reduceString($s, $l);     // This code is contributed by ajit?>
<script> // Javascript program to count deletions// to reduce the string to its// shortest length by deleting a// pair of same adjacent characters // Function count// the operationsfunction reduce(s, l){    let count = 1, step = 0;     // Traverse in the string    for(let 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 += parseInt(count / 2, 10);            count = 1;        }    }    step += parseInt(count / 2, 10);    return step;} // Driver codelet s = "geeksforgeeks";let l = s.length; document.write(reduce(s, l)); // This code is contributed by mukesh07   </script>
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