JavaScript Program to Find Second Most Repeated Word in a Sequence
Last Updated :
29 Aug, 2024
Finding the second most repeated word in a sequence involves analyzing a given text or sequence of words to determine which word appears with the second-highest frequency. This task often arises in natural language processing and text analysis. To solve it, we need to parse the input sequence, count word frequencies, and identify the word with the second-highest count.
Examples:
Input : {"1", "2", "1", "1", "2", "3"}
Output : "2"
Input : {"a", "b", "c", "a", "a", "b"}
Output : "b"Approach 1: Using Map() in JavaScript
This approach involves iterating through the words in the sequence and using a hash map to count their frequencies. By maintaining two variables to track the most repeated and second most repeated words, we can efficiently identify the second most repeated word.
Syntax:
let frequencyMap = new Map();
frequency.sort()
Example: Below is the implementation of the above approach.
JavaScript
function findSecondMostFrequentElement(arr) {
let frequencyMap = new Map();
// Counting frequency of each element
for (let i = 0; i < arr.length; i++) {
if (frequencyMap.has(arr[i])) {
frequencyMap.set(arr[i],
frequencyMap.get(arr[i]) + 1);
} else {
frequencyMap.set(arr[i], 1);
}
}
let maxFrequency = Number.MIN_SAFE_INTEGER;
let frequencies = [];
// Finding the maximum frequency
for (let [key, value] of frequencyMap) {
if (value > maxFrequency) {
maxFrequency = value;
}
}
for (let [key, value] of frequencyMap) {
if (value !== maxFrequency) {
frequencies.push(value);
}
}
frequencies.sort((a, b) => a - b);
// Returning the second most frequent element
for (let [key, value] of frequencyMap) {
if (value ===
frequencies[frequencies.length - 1]) {
return key;
}
}
return "-1";
}
let arr = ["1", "2", "3", "1", "1", "2"];
let ans = findSecondMostFrequentElement(arr);
console.log(ans);
Time Complexity : O(Nlog(N))
Approach 2: Using Set() in JavaScript
Set() approach is to count the occurrences of each word in the array using a Map, then find the second most repeated word by comparing the occurrence counts. It iterates through the array to count word occurrences and efficiently identifies the second most repeated word using a Map data structure.
Syntax:
let occurrences = new Map();
occurrences.set()
Example: Below is the implementation of the above approach.
JavaScript
// Function to find the second most repeated word
function findSecondMostRepeatedWordInArray(words) {
// Store all the words with their occurrences
let occurrences = new Map();
for (let i = 0; i < words.length; i++) {
if (occurrences.has(words[i])) {
occurrences.set(words[i],
occurrences.get(words[i]) + 1);
} else {
occurrences.set(words[i], 1);
}
}
// Find the second largest occurrence
let firstMax =
Number.MIN_VALUE,
secondMax = Number.MIN_VALUE;
for (let [key, value] of occurrences) {
if (value > firstMax) {
secondMax = firstMax;
firstMax = value;
} else if (value > secondMax
&& value !== firstMax) {
secondMax = value;
}
}
// Return the word with
// an occurrence equal to secondMax
for (let [key, value] of occurrences) {
if (value === secondMax) {
return key;
}
}
}
// Driver program
let wordsArray = ["a", "b", "c", "a", "a", "b"];
console.log(findSecondMostRepeatedWordInArray(wordsArray));
Approach 3: Using Object.entries() and Reduce()
Using Object.entries() and reduce() to find the second most repeated word involves counting word frequencies with reduce(), converting the object into an array of entries, sorting it by frequency, and returning the second entry.
Example:
JavaScript
function findSecondMostRepeatedWord(sequence) {
const frequency = sequence.reduce((acc, word) => {
acc[word] = (acc[word] || 0) + 1;
return acc;
}, {});
const sorted = Object.entries(frequency).sort((a, b) => b[1] - a[1]);
return sorted[1] ? sorted[1][0] : null;
}
const sequence = ["apple", "banana", "banana", "apple", "cherry", "banana", "cherry"];
console.log(findSecondMostRepeatedWord(sequence)); // Output: "apple"
Approach 4: Using Sorting and Iteration
In this approach, we will first count the occurrences of each word using an object. Then, we will convert the object into an array of [word, count] pairs, sort this array by count in descending order, and finally, retrieve the second element from the sorted array to find the second most repeated word.
Example:
JavaScript
const findSecondMostRepeatedWord = (words) => {
const wordCounts = {};
words.forEach(word => {
wordCounts[word] = (wordCounts[word] || 0) + 1;
});
const wordCountPairs = Object.entries(wordCounts);
wordCountPairs.sort((a, b) => b[1] - a[1]);
return wordCountPairs[1] ? wordCountPairs[1][0] : null;
};
console.log(findSecondMostRepeatedWord(["1", "2", "1", "1", "2", "3"]));
console.log(findSecondMostRepeatedWord(["a", "b", "c", "a", "a", "b"]));
console.log(findSecondMostRepeatedWord(["apple", "banana", "apple", "orange", "banana", "banana"]));
console.log(findSecondMostRepeatedWord(["cat", "dog", "cat", "mouse", "dog", "dog", "cat"]));
Approach 5: Using Min-Heap (Priority Queue)
Use a priority queue to maintain the top two frequencies while iterating through the words. The heap data structure helps to keep track of the top elements efficiently.
Steps:
- Count Frequencies: Count the occurrences of each word using a hash map.
- Use Min-Heap: Maintain a min-heap to keep track of the top two frequencies. The heap ensures that you can easily access the smallest frequency among the top frequencies.
- Extract Second Most Frequent: After processing all words, the heap will contain the two highest frequencies. The second element in the heap represents the second most frequent word.
Example:
JavaScript
class PriorityQueue {
constructor() {
this.heap = [];
}
push(item) {
this.heap.push(item);
this.heap.sort((a, b) => a[1] - b[1]);
if (this.heap.length > 2) {
this.heap.shift(); // Keep only top 2 elements
}
}
top() {
return this.heap;
}
}
function findSecondMostRepeatedWord(words) {
const frequencyMap = new Map();
words.forEach(word => {
frequencyMap.set(word, (frequencyMap.get(word) || 0) + 1);
});
// Use priority queue to track top two frequencies
const pq = new PriorityQueue();
frequencyMap.forEach((count, word) => {
pq.push([word, count]);
});
const topTwo = pq.top();
return topTwo.length < 2 ? null : topTwo[0][0];
}
console.log(findSecondMostRepeatedWord(["1", "2", "1", "1", "2", "3"]));
console.log(findSecondMostRepeatedWord(["a", "b", "c", "a", "a", "b"]));
console.log(findSecondMostRepeatedWord(["apple", "banana", "apple", "orange", "banana", "banana"]));
console.log(findSecondMostRepeatedWord(["cat", "dog", "cat", "mouse", "dog", "dog", "cat"]));
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