Sort an array of strings based on the frequency of good words in them
Given a set of product reviews (R) by different customers and a string S containing good words separated by a _, the task is to sort the reviews in decreasing order of their goodness value.
Goodness Value is defined by the number of good words present in that review.
Examples:
Input: S = “geeks_for_geeks_is_great”,
R = {“geeks_are_geeks”, “geeks_dont_lose”, “geeks_for_geeks_is_love”}
Output:
geeks for geeks is love
geeks are geeks
geeks dont lose
Input: S = “cool_wifi_ice”,
R = {“water_is_cool”, “cold_ice_drink”, “cool_wifi_speed”}
Output:
cool wifi speed
water is cool
cold ice drink
Naive approach: Insert all the good words in an unordered_set and then iterate through each word of each sentence of the reviews array and keep a count of the good words by checking if this word is present in that set of good words. We then use a stable sorting algorithm and sort the array R according to the count of good words in each review present in R. It’s clear that the time complexity of this method is greater than O(N * NlogN) .
Efficient approach: Make a Trie of all the good words and check the goodness of each word in a review using the trie.
- Insert all the good words in a trie.
- For each review, calculate the number of good words in it by checking whether a given word exists in the trie or not.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define F first #define S second #define MAX 26 // Comparator function for sorting bool cmp(const pair<int, int>& a, const pair<int, int>& b) { // Compare the number of good words if (a.F == b.F) return a.S < b.S; return a.F > b.F; } // Structure of s Trie node struct node { bool exist; node* arr[MAX]; node(bool bul = false) { exist = bul; for (int i = 0; i < MAX; i++) arr[i] = NULL; } }; // Function to add a string to the trie void add(string s, node* trie) { // Add a node to the trie int n = s.size(); for (int i = 0; i < n; i++) { // If trie doesn't already contain // the current node then create one if (trie->arr[s[i] - 'a'] == NULL) trie->arr[s[i] - 'a'] = new node(); trie = trie->arr[s[i] - 'a']; } trie->exist = true; return; } // Function that returns true if // the trie contains the string s bool search(string s, node* trie) { // Search for a node in the trie for (int i = 0; i < s.size(); i++) { if (trie->arr[s[i] - 'a'] == NULL) return false; trie = trie->arr[s[i] - 'a']; } return trie->exist; } // Function to replace every '_' with a // white space in the given string void convert(string& str) { // Convert '_' to spaces for (int i = 0; i < str.size(); i++) if (str[i] == '_') str[i] = ' '; return; } // Function to sort the array based on good words void sortArr(string good, vector<string>& review) { // Extract all the good words which // are '_' separated convert(good); node* trie = new node(); string word; stringstream ss; ss << good; // Building the entire trie by stringstreaming // the 'good words' string while (ss >> word) add(word, trie); int k, n = review.size(); // To store the number of good words // and the string index pairs vector<pair<int, int> > rating(n); for (int i = 0; i < n; i++) { convert(review[i]); ss.clear(); ss << review[i]; k = 0; while (ss >> word) { // If this word is present in the trie // then increment its count if (search(word, trie)) k++; } // Store the number of good words in the // current string paired with its // index in the original array rating[i].F = k; rating[i].S = i; } // Using comparator function to // sort the array as required sort(rating.begin(), rating.end(), cmp); // Print the sorted array for (int i = 0; i < n; i++) cout << review[rating[i].S] << "\n"; } // Driver code int main() { // String containing good words string S = "geeks_for_geeks_is_great"; // Vector of strings to be sorted vector<string> R = { "geeks_are_geeks", "geeks_dont_lose", "geeks_for_geeks_is_love" }; // Sort the array based on the given conditions sortArr(S, R); } |
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Output:
geeks for geeks is love geeks are geeks geeks dont lose
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