Skip to content
Related Articles
First string from the given array whose reverse is also present in the same array
• Difficulty Level : Basic
• Last Updated : 27 May, 2021

Given a string array str[], the task is to find the first string from the given array whose reverse is also present in the same array. If there is no such string then print -1.
Examples:

Input: str[] = {“geeks”, “for”, “skeeg”}
Output: geeks
“geeks” is the first string from the array whose reverse is also present in the array i.e. “skeeg”.
Input: str[] = {“there”, “you”, “are”}
Output: -1

Approach: Traverse the array element by element and for every string, check whether there is any string that appears after the current string in the array and is equal to the reverse of it. If yes then print the current string else print -1 in the end.
Below is the implementation of the above approach:

## C++

 `// CPP implementation of the approach``#include``using` `namespace` `std;` `    ``// Function that returns true if s1``    ``// is equal to reverse of s2``    ``bool` `isReverseEqual(string s1, string s2)``    ``{` `        ``// If both the strings differ in length``        ``if` `(s1.length() != s2.length())``            ``return` `false``;` `        ``int` `len = s1.length();``        ``for` `(``int` `i = 0; i < len; i++)` `            ``// In case of any character mismatch``            ``if` `(s1[i] != s2[len - i - 1])``                ``return` `false``;` `        ``return` `true``;``    ``}` `    ``// Function to return the first word whose``    ``// reverse is also present in the array``    ``string getWord(string str[], ``int` `n)``    ``{` `        ``// Check every string``        ``for` `(``int` `i = 0; i < n - 1; i++)` `            ``// Pair with every other string``            ``// appearing after the current string``            ``for` `(``int` `j = i + 1; j < n; j++)` `                ``// If first string is equal to the``                ``// reverse of the second string``                ``if` `(isReverseEqual(str[i], str[j]))``                    ``return` `str[i];` `        ``// No such string exists``        ``return` `"-1"``;``    ``}` `    ``// Driver code``    ``int` `main()``    ``{``        ``string str[] = { ``"geeks"``, ``"for"``, ``"skeeg"` `};``        ` `        ``cout<<(getWord(str, 3));``    ``}``    ` `// This code is contributed by``// Surendra_Gangwar`

## Java

 `// Java implementation of the approach``class` `GFG {` `    ``// Function that returns true if s1``    ``// is equal to reverse of s2``    ``static` `boolean` `isReverseEqual(String s1, String s2)``    ``{` `        ``// If both the strings differ in length``        ``if` `(s1.length() != s2.length())``            ``return` `false``;` `        ``int` `len = s1.length();``        ``for` `(``int` `i = ``0``; i < len; i++)` `            ``// In case of any character mismatch``            ``if` `(s1.charAt(i) != s2.charAt(len - i - ``1``))``                ``return` `false``;` `        ``return` `true``;``    ``}` `    ``// Function to return teh first word whose``    ``// reverse is also present in the array``    ``static` `String getWord(String str[], ``int` `n)``    ``{` `        ``// Check every string``        ``for` `(``int` `i = ``0``; i < n - ``1``; i++)` `            ``// Pair with every other string``            ``// appearing after the current string``            ``for` `(``int` `j = i + ``1``; j < n; j++)` `                ``// If first string is equal to the``                ``// reverse of the second string``                ``if` `(isReverseEqual(str[i], str[j]))``                    ``return` `str[i];` `        ``// No such string exists``        ``return` `"-1"``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str[] = { ``"geeks"``, ``"for"``, ``"skeeg"` `};``        ``int` `n = str.length;` `        ``System.out.print(getWord(str, n));``    ``}``}`

## Python3

 `# Python implementation of the approach` `# Function that returns true if s1``# is equal to reverse of s2``def` `isReverseEqual(s1, s2):` `    ``# If both the strings differ in length``    ``if` `len``(s1) !``=` `len``(s2):``        ``return` `False``    ` `    ``l ``=` `len``(s1)` `    ``for` `i ``in` `range``(l):` `        ``# In case of any character mismatch``        ``if` `s1[i] !``=` `s2[l``-``i``-``1``]:``            ``return` `False``    ``return` `True` `# Function to return the first word whose``# reverse is also present in the array``def` `getWord(``str``, n):` `    ``# Check every string``    ``for` `i ``in` `range``(n``-``1``):` `        ``# Pair with every other string``        ``# appearing after the current string``        ``for` `j ``in` `range``(i``+``1``, n):` `            ``# If first string is equal to the``            ``# reverse of the second string``            ``if` `(isReverseEqual(``str``[i], ``str``[j])):``                ``return` `str``[i]``    ` `    ``# No such string exists``    ``return` `"-1"`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``str` `=` `[``"geeks"``, ``"for"``, ``"skeeg"``]``    ``print``(getWord(``str``, ``3``))` `# This code is contributed by``# sanjeev2552`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `    ``// Function that returns true if s1``    ``// is equal to reverse of s2``    ``static` `bool` `isReverseEqual(String s1, String s2)``    ``{` `        ``// If both the strings differ in length``        ``if` `(s1.Length != s2.Length)``            ``return` `false``;` `        ``int` `len = s1.Length;``        ``for` `(``int` `i = 0; i < len; i++)` `            ``// In case of any character mismatch``            ``if` `(s1[i] != s2[len - i - 1])``                ``return` `false``;` `        ``return` `true``;``    ``}` `    ``// Function to return teh first word whose``    ``// reverse is also present in the array``    ``static` `String getWord(String []str, ``int` `n)``    ``{` `        ``// Check every string``        ``for` `(``int` `i = 0; i < n - 1; i++)` `            ``// Pair with every other string``            ``// appearing after the current string``            ``for` `(``int` `j = i + 1; j < n; j++)` `                ``// If first string is equal to the``                ``// reverse of the second string``                ``if` `(isReverseEqual(str[i], str[j]))``                    ``return` `str[i];` `        ``// No such string exists``        ``return` `"-1"``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String []str = { ``"geeks"``, ``"for"``, ``"skeeg"` `};``        ``int` `n = str.Length;` `        ``Console.Write(getWord(str, n));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## PHP

 ``

## Javascript

 ``
Output:

`geeks`

Efficient Approach: O(n) approach. This approach requires a Hashmap to store words as traversed. As we traverse, if reverse of current word found in the map, then reversed word is the first occurrence that is the answer. If not found at the end of traversal, return -1.
Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Method that returns first occurrence of revered word.``string getReversed(string words[], ``int` `length)``{``  ` `  ``// Hashmap to store word as we traverse``  ``map reversedWordMap;``  ``for``(``int` `i = 0; i < length; i++)``  ``{``    ` `    ``string reversedString = words[i];``    ``reverse(reversedString.begin(),reversedString.end());``     ` `    ``// check if reversed word exists in map.``    ``if` `(reversedWordMap.find(reversedString) !=``        ``reversedWordMap.end() and reversedWordMap[reversedString])``      ``return` `reversedString;``    ``else``      ` `      ``// else put the word in map``      ``reversedWordMap[words[i]] = ``true``;``  ``}``  ``return` `"-1"``;``}``  ` `// Driver code``int` `main()``{``    ``string words[] = {``"some"``, ``"geeks"``, ``"emos"``, ``"for"``, ``"skeeg"``};``    ``int` `length = ``sizeof``(words) / ``sizeof``(words);``    ``cout << getReversed(words, length);``    ``return` `0;``}` `// This code is contributed by divyesh072019`

## Java

 `import` `java.util.HashMap;``import` `java.util.Map;` `public` `class` `ReverseExist {` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args) {``        ``String[] words = {``"some"``, ``"geeks"``, ``"emos"``, ``"for"``, ``"skeeg"``};``        ``System.out.println(getReversed(words, words.length));``    ``}` `    ``// Method that returns first occurrence of revered word.``    ``private` `static` `String getReversed(String[] words, ``int` `length) {``        ` `        ``// Hashmap to store word as we traverse``        ``Map reversedWordMap = ``new` `HashMap<>();` `        ``for` `(String word : words) {``            ``StringBuilder reverse = ``new` `StringBuilder(word);``            ``String reversed = reverse.reverse().toString();``            ` `            ``// check if reversed word exists in map.``            ``Boolean exists = reversedWordMap.get(reversed);``            ``if` `(exists != ``null` `&& exists.booleanValue()) {``                ``return` `reversed;``            ``} ``else` `{``                ``// else put the word in map``                ``reversedWordMap.put(word, ``true``);``            ``}` `        ``}``        ``return` `"-1"``;``    ``}``}``// Contributed by srika21m`

## Python3

 `# Method that returns first occurrence of revered word.``def` `getReversed(words, length):``  ` `  ``# Hashmap to store word as we traverse``  ``reversedWordMap ``=` `{}``  ``for` `word ``in` `words:``    ``reversedString ``=` `word[::``-``1``]``    ` `    ``# check if reversed word exists in map.``    ``if` `(reversedString ``in` `reversedWordMap ``and` `reversedWordMap[reversedString]):``      ``return` `reversedString``    ``else``:``      ` `      ``# else put the word in map``      ``reversedWordMap[word] ``=` `True``  ``return` `"-1"` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``  ``words ``=` `[``"some"``, ``"geeks"``, ``"emos"``, ``"for"``, ``"skeeg"``]``  ``print``(getReversed(words, ``len``(words)))` `  ``# This code is contributed by chitranayal`

## C#

 `using` `System;``using` `System.Collections.Generic;``class` `GFG``{``    ` `    ``// Method that returns first occurrence of revered word.``    ``static` `string` `getReversed(``string``[] words, ``int` `length)``    ``{``       ` `      ``// Hashmap to store word as we traverse``      ``Dictionary<``string``, ``bool``> reversedWordMap =``        ``new` `Dictionary<``string``, ``bool``>();``      ``for``(``int` `i = 0; i < length; i++)``      ``{``         ` `        ``char``[] reversedString = words[i].ToCharArray();``        ``Array.Reverse(reversedString);``          ` `        ``// check if reversed word exists in map.``        ``if` `(reversedWordMap.ContainsKey(``new` `string``(reversedString)) &&``            ``reversedWordMap[``new` `string``(reversedString)])``          ``return` `new` `string``(reversedString);``        ``else``           ` `          ``// else put the word in map``          ``reversedWordMap[words[i]] = ``true``;``      ``}``      ``return` `"-1"``;``    ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``string``[] words = {``"some"``, ``"geeks"``, ``"emos"``, ``"for"``, ``"skeeg"``};``    ``int` `length = words.Length;``    ``Console.Write(getReversed(words, length));``  ``}``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``
Output:
`some`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up