Given a string S, the task is to find the resultant smallest string possible, formed by repeatedly removing pairs of adjacent characters which are equal.
Examples:
Input: S = “aaabccddd”
Output: abd
Explanation: Following sequence of removal of pairs of adjacent characters minimizes the length of the string:
aaabccddd ? abccddd ? abddd ? abd.Input: S = aa
Output: Empty String
Explanation: Following sequence of removal of pairs of adjacent characters minimizes the length of the string:
aa ? “”.
Approach: The main idea to solve the given problem is to recursively delete all pairs of adjacent characters which are equal, one by one. Follow the steps to solve the given problem:
- Initialize an empty string, say ans, to store the string of minimum length after deleting all pairs of equal adjacent characters.
- Initialize a string, say pre, to store the updated string after every removal of equal adjacent characters.
- Now, iterate until the string ans and pre are unequal and perform the following steps:
- Update the value of the string ans by removing the first adjacent same character using the function removeAdjacent().
- If the value of the string ans is the same as the string pre, then break out of the loop. Otherwise, update the value of string pre as the string ans.
- After completing the above steps, print the string ans as the resultant string.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to delete pair of adjacent // characters which are equal string removeAdjacent(string s) { // Base Case if (s.length() == 1) return s; // Stores the update string string sb = ""; // Traverse the string s for(int i = 0; i < s.length() - 1; i++) { char c = s[i]; char d = s[i + 1]; // If two unequal pair of // adjacent characters are found if (c != d) { sb = sb + c; if (i == s.length() - 2) { sb = sb + d; } } // If two equal pair of adjacent // characters are found else { for(int j = i + 2; j < s.length(); j++) // Append the remaining string // after removing the pair sb = sb + s[j]; return sb; } } // Return the final String return sb; } // Function to find the shortest string // after repeatedly removing pairs of // adjacent characters which are equal void reduceString(string s) { // Stores the resultant String string result = ""; // Keeps track of previously // iterated string string pre = s; while (true) { // Update the result after // deleting adjacent pair of // characters which are similar result = removeAdjacent(pre); // Termination Conditions if (result == pre) break; // Update pre variable with // the value of result pre = result; } if (result.length() != 0) cout << (result) << endl; // Case for "Empty String" else cout << "Empty String" << endl; } // Driver code int main(){ string S = "aaabccddd"; reduceString(S); return 0;}// This code is contributed by divyesh072019 |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to delete pair of adjacent // characters which are equal static String removeAdjacent(String s) { // Base Case if (s.length() == 1) return s; // Stores the update string StringBuilder sb = new StringBuilder(""); // Traverse the string s for (int i = 0; i < s.length() - 1; i++) { char c = s.charAt(i); char d = s.charAt(i + 1); // If two unequal pair of // adjacent characters are found if (c != d) { sb.append(c); if (i == s.length() - 2) { sb.append(d); } } // If two equal pair of adjacent // characters are found else { for (int j = i + 2; j < s.length(); j++) // Append the remaining string // after removing the pair sb.append(s.charAt(j)); return sb.toString(); } } // Return the final String return sb.toString(); } // Function to find the shortest string // after repeatedly removing pairs of // adjacent characters which are equal public static void reduceString(String s) { // Stores the resultant String String result = ""; // Keeps track of previously // iterated string String pre = s; while (true) { // Update the result after // deleting adjacent pair of // characters which are similar result = removeAdjacent(pre); // Termination Conditions if (result.equals(pre)) break; // Update pre variable with // the value of result pre = result; } if (result.length() != 0) System.out.println(result); // Case for "Empty String" else System.out.println("Empty String"); } // Driver Code public static void main(String[] args) { String S = "aaabccddd"; reduceString(S); }} |
Python3
# Python program for the above approach# Function to delete pair of adjacent# characters which are equaldef removeAdjacent(s): # Base Case if (len(s) == 1): return s # Stores the update string sb = "" # Traverse the string s for i in range(len(s)-1): c = s[i] d = s[i + 1] # If two unequal pair of # adjacent characters are found if (c != d): sb = sb + c if (i == len(s) - 2): sb = sb + d # If two equal pair of adjacent # characters are found else: for j in range(i + 2,len(s)): # Append the remaining string # after removing the pair sb = sb + s[j] return sb # Return the final String return sb# Function to find the shortest string# after repeatedly removing pairs of# adjacent characters which are equaldef reduceString(s): # Stores the resultant String result = "" # Keeps track of previously # iterated string pre = s while True: # Update the result after # deleting adjacent pair of # characters which are similar result = removeAdjacent(pre) # Termination Conditions if (result == pre): break # Update pre variable with # the value of result pre = result if (len(result)!= 0): print(result) # Case for "Empty String" else: print("Empty String")# Driver codeS = "aaabccddd"reduceString(S) # This code is contributed by shinjanpatra |
C#
// C# program for the above approachusing System;class GFG { // Function to delete pair of adjacent // characters which are equal static string removeAdjacent(string s) { // Base Case if (s.Length == 1) return s; // Stores the update string string sb = ""; // Traverse the string s for(int i = 0; i < s.Length - 1; i++) { char c = s[i]; char d = s[i + 1]; // If two unequal pair of // adjacent characters are found if (c != d) { sb = sb + c; if (i == s.Length - 2) { sb = sb + d; } } // If two equal pair of adjacent // characters are found else { for(int j = i + 2; j < s.Length; j++) // Append the remaining string // after removing the pair sb = sb + s[j]; return sb; } } // Return the final String return sb; } // Function to find the shortest string // after repeatedly removing pairs of // adjacent characters which are equal static void reduceString(string s) { // Stores the resultant String string result = ""; // Keeps track of previously // iterated string string pre = s; while (true) { // Update the result after // deleting adjacent pair of // characters which are similar result = removeAdjacent(pre); // Termination Conditions if (result == pre) break; // Update pre variable with // the value of result pre = result; } if (result.Length != 0) Console.WriteLine(result); // Case for "Empty String" else Console.WriteLine("Empty String"); } static void Main() { string S = "aaabccddd"; reduceString(S); }} |
Javascript
<script> // Javascript program for the above approach // Function to delete pair of adjacent // characters which are equal function removeAdjacent(s) { // Base Case if (s.length == 1) return s; // Stores the update string let sb = ""; // Traverse the string s for(let i = 0; i < s.length - 1; i++) { let c = s[i]; let d = s[i + 1]; // If two unequal pair of // adjacent characters are found if (c != d) { sb = sb + c; if (i == s.length - 2) { sb = sb + d; } } // If two equal pair of adjacent // characters are found else { for(let j = i + 2; j < s.length; j++) // Append the remaining string // after removing the pair sb = sb + s[j]; return sb; } } // Return the final String return sb; } // Function to find the shortest string // after repeatedly removing pairs of // adjacent characters which are equal function reduceString(s) { // Stores the resultant String let result = ""; // Keeps track of previously // iterated string let pre = s; while (true) { // Update the result after // deleting adjacent pair of // characters which are similar result = removeAdjacent(pre); // Termination Conditions if (result == pre) break; // Update pre variable with // the value of result pre = result; } if (result.length != 0) document.write(result); // Case for "Empty String" else document.write("Empty String"); } let S = "aaabccddd"; reduceString(S); // This code is contributed by suresh07</script> |
Output:
abd
Time Complexity: O(N2)
Auxiliary Space: O(N)
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