# Pair of strings having longest common prefix of maximum length in given array

Given an array of strings arr[], the task is to find the pair of strings from the given array whose length of the longest common prefix between them is maximum. If multiple solutions exist, then print any one of them.

Examples:

Input: arr[] = {“geeksforgeeks”, “geeks”, “geeksforcse”, }
Output: (geeksforgeeks, geeksforcse)
Explanation:
All possible pairs and their longest common prefix are:
(“geeksforgeeks”, “geeks”) has the longest common prefix = “geeks”
(“geeksforgeeks”, “geeksforcse”) has the longest common prefix = “geeksfor”
(“geeks”, “geeksforcse”) has the longest common prefix = “geeks”
Therefore, a pair having maximum length of the longest common prefix is (“geeksforgeeks”, “geeksforcse”)

Input: arr[] = {“abbcbgfh”, “bcdee”, “bcde”, “abbcbde”}
Output: (abbcbgfh, abbcbde)

Naive Approach: The simplest approach to solve this problem is to generate all possible pairs of the given array and calculate the length of the longest common prefix of each pair. Finally, print the pair having a maximum length of the longest common prefix.
Time Complexity: O(N2 * M), Where M denotes the length of the longest string
Auxiliary Space: O(1)

Efficient Approach: The problem can be solved using Trie. The idea is to traverse the given array and for each array element, find the maximum length of the longest prefix present in Trie, and insert the current element into the Trie. Finally, print the pair having a maximum length of the longest common prefix. Follow the steps below to solve the problem:

• Create a Trie having root node, say root to store each element of the given array.
• Traverse the given array and for each array element, find the maximum length of the longest prefix present in Trie and insert the current element into Trie.
• Finally, print the pair having a maximum length of the longest common prefix.

Below is the implementation of the above approach:

 // C++ program to implement // the above approach    #include using namespace std;    // Structure of Trie struct TrieNode {     // Stores characters of     // each string     TrieNode* child[256];        TrieNode()     {         child[0] = child[1] = NULL;     } };    // Function to insert a string into Trie void insertTrie(TrieNode* root,                 string str) {        // Stores length of the string     int M = str.length();        // Traverse the string str     for (int i = 0; i < M;          i++) {            // If str[i] is not present         // in current path of Trie         if (!root->child[str[i]]) {                // Create a new node             // of Trie             root->child[str[i]]                 = new TrieNode();         }            // Update root         root = root->child[str[i]];     } }    // Function to find the maximum length of // longest common prefix in Trie with str int findStrLen(TrieNode* root, string str) {        // Stores length of str     int M = str.length();        // Stores length of longest     // common prefix in Trie with str     int len = 0;        // Traverse the string str     for (int i = 0; i < M; i++) {            // If str[i] is present in         // the current path of Trie         if (root->child[str[i]]) {                // Update len             len++;                // Update root             root                 = root->child[str[i]];         }         else {             return len;         }     }     return len; }    // Function to print the pair having maximum // length of the longest common prefix void findMaxLenPair(vector& arr,                     int N) {     // Stores index of the string having     // maximum length of longest common prefix     int idx = -1;        // Stores maximum length of longest     // common prefix.     int len = 0;        // Create root node of Trie     TrieNode* root = new TrieNode();        // Insert arr[0] into Trie     insertTrie(root, arr[0]);        // Traverse the array.     for (int i = 1; i < N; i++) {            // Stores maximum length of longest         // common prefix in Trie with arr[i]         int temp = findStrLen(root, arr[i]);            // If temp is greater than len         if (temp > len) {                // Update len             len = temp;                // Update idx             idx = i;         }            insertTrie(root, arr[i]);     }        // Traverse array arr[]     for (int i = 0; i < N; i++) {            // Stores length of arr[i]         int M = arr[i].length();            // Check if maximum length of         // longest common prefix > M         if (i != idx && M >= len) {                // Traverse string arr[i]             // and arr[j]             for (int j = 0; j < len;                  j++) {                    // If current character of both                 // string does not match.                 if (arr[i][j]                     != arr[idx][j]) {                     break;                 }             }                // Print pairs having maximum length             // of the longest common prefix             cout << "(" << arr[i] << ", "                  << arr[idx] << ")";             return;         }     } }    // Driver Code int main() {     vector arr         = { "geeksforgeeks",             "geeks", "geeksforcse" };        int N = arr.size();     findMaxLenPair(arr, N); }

Output:
(geeksforgeeks, geeksforcse)

Time Complexity: O(N * M), where M denotes the length of the longest string
Auxiliary Space: 0(N * 256)

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