Given a list of strings, the task is to find the sum of all LCP (Longest Common Prefix) of maximum length by selecting any two strings at a time.
Examples:
Input: str[] = {babab, ababb, abbab, aaaaa, babaa, babbb}
Output: 6
Explanation:
Choose 1st and 5th string => length of LCP = 4,
Choose 2nd and 3rd string => length of LCP = 2
Sum of LCP = 4 + 2 = 6
Input: str = [“aa”, “aaaa”, “aaaaaaaa”, “aaaabaaaa”, “aaabaaa”]
Output: 7
Explanation:
Choose 3rd (aaaaaaaa) and 4th string (aaaabaaaa) => length of LCP (aaaa) = 4,
Choose 2nd (aaaa) and 5th (aaabaaa) string => length of LCP (aaa) = 3
Sum of LCP = 4 + 3 = 7
Naive Approach:
- Sort the list of strings in decreasing order of their length
- Then take the first string from the list and find the Longest Common Prefix with all other remaining string in the list and store it in the array
- Choose the maximum value from the array and add it to variable answer and remove the pair of string from the list corresponding to that sum
- Repeat the above procedures for all the next strings till the list is empty or you reach the last string
- The variable answer has the required sum of all LCP of maximum length
Time Complexity: O(M*N2), where M = maximum string length and N = number of strings.
Efficient Approach:
An efficient solution can be obtained using a Trie Data Structure. To find the number of characters common between the strings we will use the variable ‘visited’ to keep track of how many times one character is visited.
Following are the steps:
- Insert list of string in trie such that every string in the list is inserted as an individual trie node.
- For all prefixes of maximum length, count the pairs from deepest node in the trie.
- Use depth-first search (DFS) traversal on trie to count the pairs from deepest node.
- If the value of visited node is more than one, it means that there two or more strings that have common prefix up till that node.
- Add the value of that visited node to a variable count.
- Decrease the value of that visited node from current and previous nodes such that the pair of words chosen for calculation must be removed.
- Repeat the above steps for all nodes and return the value of count.
Below is the implementation of the above approach:
C++
// C++ program to find Sum of all LCP// of maximum length by selecting// any two Strings at a time#include <bits/stdc++.h>using namespace std;class TrieNode {public: char val; // Using map to store the pointers // of children nodes for dynamic // implementation, for making the // program space efficient map<char, TrieNode*> children; // Counts the number of times the node // is visited while making the trie int visited; // Initially visited value for all // nodes is zero TrieNode(char x) { val = x; visited = 0; }};class Trie {public: TrieNode* head; // Head node of the trie is initialize // as '\0', after this all strings add Trie() { head = new TrieNode('\0'); } // Function to insert the strings in // the trie void addWord(string s) { TrieNode* temp = head; const unsigned int n = s.size(); for (int i = 0; i < n; i++) { // Inserting character-by-character char ch = s[i]; // If the node of ch is not present in // map make a new node and add in map if (!temp->children[ch]) { temp->children[ch] = new TrieNode(ch); } temp = temp->children[ch]; temp->visited++; } } // Recursive function to calculate the // answer argument is passed by reference int dfs(TrieNode* node, int& ans, int depth) { // To store changed visited values from // children of this node i.e. number of // nodes visited by its children int vis = 0; for (auto child : node->children) { vis += dfs(child.second, ans, depth + 1); } // Updating the visited variable, telling // number of nodes that have // already been visited by its children node->visited -= vis; int string_pair = 0; // If node->visited > 1, means more than // one string has prefix up till this node // common in them if (node->visited > 1) { // Number of string pair with current // node common in them string_pair = (node->visited / 2); ans += (depth * string_pair); // Updating visited variable of current node node->visited -= (2 * string_pair); } // Returning the total number of nodes // already visited that needs to be // updated to previous node return (2 * string_pair + vis); } // Function to run the dfs function for the // first time and give the answer variable int dfshelper() { // Stores the final answer // as sum of all depths int ans = 0; dfs(head, ans, 0); return ans; }};// Driver Functionint main(){ Trie T; string str[] = { "babab", "ababb", "abbab", "aaaaa", "babaa", "babbb" }; int n = 6; for (int i = 0; i < n; i++) { T.addWord(str[i]); } int ans = T.dfshelper(); cout << ans << endl; return 0;} |
Java
// Java program to find Sum of all LCP// of maximum length by selecting// any two Strings at a timeimport java.util.*;class GFG{static class TrieNode { char val; // Using map to store the pointers // of children nodes for dynamic // implementation, for making the // program space efficient HashMap<Character, TrieNode> children; // Counts the number of times the node // is visited while making the trie int visited; // Initially visited value for all // nodes is zero TrieNode(char x) { val = x; visited = 0; children = new HashMap<>(); }}static class Trie { TrieNode head; int ans; // Head node of the trie is initialize // as '\0', after this all Strings add Trie() { head = new TrieNode('\0'); ans = 0; } // Function to insert the Strings in // the trie void addWord(String s) { TrieNode temp = head; int n = s.length(); for (int i = 0; i < n; i++) { // Inserting character-by-character char ch = s.charAt(i); // If the node of ch is not present in // map make a new node and add in map if (temp.children.get(ch) == null) { temp.children.put(ch, new TrieNode(ch)); } temp = temp.children.get(ch); temp.visited++; } } // Recursive function to calculate the // answer argument is passed by reference int dfs(TrieNode node, int depth) { // To store changed visited values from // children of this node i.e. number of // nodes visited by its children int vis = 0; Iterator hmIterator = node.children.entrySet().iterator(); while (hmIterator.hasNext()) { Map.Entry child = (Map.Entry)hmIterator.next(); vis += dfs((TrieNode)child.getValue(), depth + 1); } // Updating the visited variable, telling // number of nodes that have // already been visited by its children node.visited -= vis; int String_pair = 0; // If node.visited > 1, means more than // one String has prefix up till this node // common in them if (node.visited > 1) { // Number of String pair with current // node common in them String_pair = (node.visited / 2); ans += (depth * String_pair); // Updating visited variable of current node node.visited -= (2 * String_pair); } // Returning the total number of nodes // already visited that needs to be // updated to previous node return (2 * String_pair + vis); } // Function to run the dfs function for the // first time and give the answer variable int dfshelper() { // Stores the final answer // as sum of all depths ans = 0; dfs(head, 0); return ans; }}// Driver codepublic static void main(String args[]){ Trie T = new Trie(); String str[] = { "babab", "ababb", "abbab", "aaaaa", "babaa", "babbb" }; int n = 6; for (int i = 0; i < n; i++) { T.addWord(str[i]); } int ans = T.dfshelper(); System.out.println( ans );}}// This code is contributed by Arnab Kundu |
Python
# python program to find Sum of all LCP# of maximum length by selecting# any two Strings at a timeclass TrieNode: # Using map to store the pointers # of children nodes for dynamic # implementation, for making the # program space efficient def __init__(self, x): self.val = x self.children = {} # Counts the number of times the node # is visited while making the trie self.visited = 0class Trie: def __init__(self): # Head node of the trie is initialize # as '\0', after this all strings add self.head = TrieNode('') # Function to insert the strings in # the trie def addWord(self, s): temp = self.head n = len(s) for i in range(n): ch = s[i] if ch not in temp.children: # If the node of ch is not present in # map make a new node and add in map temp.children[ch] = TrieNode(ch) temp = temp.children[ch] temp.visited += 1 # Recursive function to calculate the # answer argument is passed by reference def dfs(self, node, ans, depth): # To store changed visited values from # children of this node i.e. number of # nodes visited by its children vis = 0 for child in node.children.values(): vis += self.dfs(child, ans, depth + 1) # Updating the visited variable, telling # number of nodes that have # already been visited by its children node.visited -= vis string_pair = 0 # If node->visited > 1, means more than # one string has prefix up till this node # common in them if node.visited > 1: # Number of string pair with current # node common in them string_pair = node.visited // 2 ans[0] += (depth * string_pair) node.visited -= (2 * string_pair) # Returning the total number of nodes # already visited that needs to be # updated to previous node return 2 * string_pair + vis # Function to run the dfs function for the # first time and give the answer variable def dfshelper(self): # Stores the final answer # as sum of all depths ans = [0] self.dfs(self.head, ans, 0) return ans[0]# Driver codeT = Trie()str_list = ["babab", "ababb", "abbab", "aaaaa", "babaa", "babbb"]for s in str_list: T.addWord(s)ans = T.dfshelper()print(ans)# This code is contributed by bhardwajji |
C#
// C# program to find Sum of all LCP// of maximum length by selecting// any two Strings at a timeusing System;using System.Collections.Generic;class Program{ class TrieNode { public char val; // Using dictionary to store the pointers // of children nodes for dynamic // implementation, for making the // program space efficient public Dictionary<char, TrieNode> children; // Counts the number of times the node // is visited while making the trie public int visited; // Initially visited value for all // nodes is zero public TrieNode(char x) { val = x; visited = 0; children = new Dictionary<char, TrieNode>(); } } class Trie { TrieNode head; int ans; // Head node of the trie is initialize // as '\0', after this all Strings add public Trie() { head = new TrieNode('\0'); ans = 0; } // Function to insert the Strings in // the trie public void addWord(string s) { TrieNode temp = head; int n = s.Length; for (int i = 0; i < n; i++) { // Inserting character-by-character char ch = s[i]; // If the node of ch is not present in // dictionary make a new node and add in dictionary if (!temp.children.ContainsKey(ch)) { temp.children.Add(ch, new TrieNode(ch)); } temp = temp.children[ch]; temp.visited++; } } // Recursive function to calculate the // answer argument is passed by reference int dfs(TrieNode node, int depth) { // To store changed visited values from // children of this node i.e. number of // nodes visited by its children int vis = 0; foreach (var child in node.children) { vis += dfs(child.Value, depth + 1); } // Updating the visited variable, telling // number of nodes that have // already been visited by its children node.visited -= vis; int String_pair = 0; // If node.visited > 1, means more than // one String has prefix up till this node // common in them if (node.visited > 1) { // Number of String pair with current // node common in them String_pair = (node.visited / 2); ans += (depth * String_pair); // Updating visited variable of current node node.visited -= (2 * String_pair); } // Returning the total number of nodes // already visited that needs to be // updated to previous node return (2 * String_pair + vis); } // Function to run the dfs function for the // first time and give the answer variable public int dfshelper() { // Stores the final answer // as sum of all depths ans = 0; dfs(head, 0); return ans; }} // Driver codepublic static void Main(){ Trie T = new Trie(); string[] str = { "babab", "ababb", "abbab", "aaaaa", "babaa", "babbb" }; int n = 6; for (int i = 0; i < n; i++) { T.addWord(str[i]); } int ans = T.dfshelper(); Console.WriteLine(ans);}}// This code is contributed by Aman Kumar. |
Javascript
<script>// Javascript program to find Sum of all LCP// of maximum length by selecting// any two Strings at a timeclass TrieNode { // Initially visited value for all // nodes is zero constructor(x) { this.val = x; // Counts the number of times the node // is visited while making the trie this.visited = 0; // Using map to store the pointers // of children nodes for dynamic // implementation, for making the // program space efficient this.children = new Map(); }}class Trie { // Head node of the trie is initialize // as '\0', after this all Strings add constructor() { this.head = new TrieNode('\0'); this.ans = 0; } // Function to insert the Strings in // the trie addWord(s) { let temp = this.head; let n = s.length; for (let i = 0; i < n; i++) { // Inserting character-by-character let ch = s[i]; // If the node of ch is not present in // map make a new node and add in map if (temp.children.get(ch) == null) { temp.children.set(ch, new TrieNode(ch)); } temp = temp.children.get(ch); temp.visited++; } } // Recursive function to calculate the // answer argument is passed by reference dfs(node,depth) { // To store changed visited values from // children of this node i.e. number of // nodes visited by its children let vis = 0; for(let [key, value] of node.children.entries()) { vis += this.dfs(value, depth + 1); } // Updating the visited variable, telling // number of nodes that have // already been visited by its children node.visited -= vis; let String_pair = 0; // If node.visited > 1, means more than // one String has prefix up till this node // common in them if (node.visited > 1) { // Number of String pair with current // node common in them String_pair = (node.visited / 2); this.ans += (depth * String_pair); // Updating visited variable of current node node.visited -= (2 * String_pair); } // Returning the total number of nodes // already visited that needs to be // updated to previous node return (2 * String_pair + vis); } // Function to run the dfs function for the // first time and give the answer variable dfshelper() { // Stores the final answer // as sum of all depths this.ans = 0; this.dfs(this.head, 0); return this.ans; }}// Driver codelet T = new Trie();let str= [ "babab", "ababb", "abbab", "aaaaa", "babaa", "babbb" ];let n = 6;for (let i = 0; i < n; i++) { T.addWord(str[i]);}let ans = T.dfshelper();document.write( ans );// This code is contributed by unknown2108</script> |
Output:
6
Time Complexity:
For inserting all the strings in the trie: O(MN)
For performing trie traversal: O(26*M) ~ O(M)
Therefore, overall Time complexity: O(M*N), where:
N = Number of strings M = Length of the largest string
Auxiliary Space: O(M)
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