# First string from the given array whose reverse is also present in the same array

Last Updated : 19 Sep, 2023

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 the 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 the 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`

## Javascript

 ``

## PHP

 ``

Output

```geeks

```

Time Complexity: O(n3), as nested loops are used
Auxiliary Space: O(1), as no extra space is used

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 reversed 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[0]);` `    ``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 reversed 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 reversed 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 reversed 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

```

Time Complexity: O(nlogn)

Auxiliary Space: O(n)

## Approach: “Iterative Reverse String Comparison”

One approach is to iterate through the array, and for each string, check if its reverse exists in the array. If it does, return that string as the answer. If none of the strings have a reverse that is also present in the array, return -1.

Steps:

1. Iterate through the array using a for loop and a variable i that goes from 0 to n-1, where n is the length of the array.
2. For each string str[i], find its reverse by using the reverse() function.
3. Iterate through the array again using another for loop and a variable j that goes from 0 to n-1.
4. Check if the reverse of str[i] is equal to any of the other strings in the array.
5. If it is, return str[i] as the answer.
6. If none of the strings have a reverse that is also present in the array, return -1.

## C++

 `#include ` `#include ` `#include ` `#include ` `using` `namespace` `std;`   `string findReverse(string s) {` `    ``reverse(s.begin(), s.end());` `    ``return` `s;` `}`   `string firstReverse(string str[], ``int` `n) {` `    ``string rev;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``rev = findReverse(str[i]);` `        ``for` `(``int` `j = 0; j < n; j++) {` `            ``if` `(i != j && str[j] == rev) {` `                ``return` `str[i];` `            ``}` `        ``}` `    ``}` `    ``return` `"-1"``;` `}`   `int` `main() {` `    ``string str[] = {``"geeks"``, ``"for"``, ``"skeeg"``};` `    ``int` `n = ``sizeof``(str)/``sizeof``(str[0]);` `    ``cout << firstReverse(str, n);` `    ``return` `0;` `}`

## Java

 `import` `java.util.Arrays;`   `public` `class` `Main {` `    ``public` `static` `String findReverse(String s) {` `        ``return` `new` `StringBuilder(s).reverse().toString();` `    ``}`   `    ``public` `static` `String firstReverse(String[] str) {` `        ``String rev;` `        ``int` `n = str.length;` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``rev = findReverse(str[i]);` `            ``for` `(``int` `j = ``0``; j < n; j++) {` `                ``if` `(i != j && str[j].equals(rev)) {` `                    ``return` `str[i];` `                ``}` `            ``}` `        ``}` `        ``return` `"-1"``;` `    ``}`   `    ``// Example Usage` `    ``public` `static` `void` `main(String[] args) {` `        ``String[] str = {``"geeks"``, ``"for"``, ``"skeeg"``};` `        ``System.out.println(firstReverse(str)); ``// Output: geeks` `    ``}` `}`

## Python3

 `def` `find_reverse(s):` `    ``return` `s[::``-``1``]`   `def` `first_reverse(str_list):` `    ``n ``=` `len``(str_list)` `    ``for` `i ``in` `range``(n):` `        ``rev ``=` `find_reverse(str_list[i])` `        ``for` `j ``in` `range``(n):` `            ``if` `i !``=` `j ``and` `str_list[j] ``=``=` `rev:` `                ``return` `str_list[i]` `    ``return` `"-1"`   `str_list ``=` `[``"geeks"``, ``"for"``, ``"skeeg"``]` `print``(first_reverse(str_list))`

## C#

 `using` `System;`   `namespace` `GFG` `{` `    ``class` `GFG` `    ``{` `        ``static` `string` `FindReverse(``string` `s)` `        ``{` `            ``char``[] charArray = s.ToCharArray();` `            ``Array.Reverse(charArray);` `            ``return` `new` `string``(charArray);` `        ``}` `        ``static` `string` `FirstReverse(``string``[] str, ``int` `n)` `        ``{` `            ``string` `rev;` `            ``for` `(``int` `i = 0; i < n; i++)` `            ``{` `                ``rev = FindReverse(str[i]);` `                ``for` `(``int` `j = 0; j < n; j++)` `                ``{` `                    ``if` `(i != j && str[j] == rev)` `                    ``{` `                        ``return` `str[i];` `                    ``}` `                ``}` `            ``}` `            ``return` `"-1"``;` `        ``}` `        ``static` `void` `Main(``string``[] args)` `        ``{` `            ``string``[] str = { ``"geeks"``, ``"for"``, ``"skeeg"` `};` `            ``int` `n = str.Length;` `            ``string` `result = FirstReverse(str, n);` `            ``Console.WriteLine(result);` `        ``}` `    ``}` `}`

## Javascript

 `// JavaScript code for above approach`   `// function to find reverse` `function` `findReverse(s) {` `    ``return` `s.split(``''``).reverse().join(``''``);` `}`   `// function to find first reverse` `function` `firstReverse(strList) {` `    ``const n = strList.length;` `    ``for` `(let i = 0; i < n; i++) {` `        ``let rev = findReverse(strList[i]);` `        ``for` `(let j = 0; j < n; j++) {` `            ``if` `(i !== j && strList[j] === rev) {` `                ``return` `strList[i];` `            ``}` `        ``}` `    ``}` `    ``return` `"-1"``;` `}`   `// Driver code` `const strList = [``"geeks"``, ``"for"``, ``"skeeg"``];` `console.log(firstReverse(strList));`

Output

`geeks`

Time complexity: O(n^2) where n is the length of the array,

Auxiliary space: O(m) where m is the length of the longest string in the array.

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