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Top K Frequent Words Heap Solution [WIP]
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var topKFrequent = function(words, k, frequency = {}) { | |
words.forEach(word => frequency[word] = frequency[word] + 1 || 1); | |
return Object.keys(frequency).sort((left, right) => { | |
const freqDiff = frequency[right] - frequency[left]; | |
return freqDiff === 0 ? left.localeCompare(right) : freqDiff; | |
}).slice(0, k); | |
}; |
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/** | |
* @param {string[]} words | |
* @param {number} k | |
* @return {string[]} | |
*/ | |
class FreqHeap { | |
constructor () { | |
this.heap = []; | |
this.counts = {}; | |
this.position = {} | |
} | |
add (word) { | |
if (word in this.counts) { | |
this.counts[word]++; | |
} else { | |
this.counts[word] = 1; | |
this.position[word] = this.heap.length; | |
this.heap.push(word); | |
} | |
this.bubbleUp(this.position[word]); | |
} | |
bubbleUp (index) { | |
index = index == null ? this.heap.length - 1 : index; | |
if (index === 0) return; | |
const parentIndex = Math.floor(index / 2); | |
const currentWord = this.heap[index]; | |
const parentWord = this.heap[parentIndex]; | |
const currentValue = this.counts[currentWord]; | |
const parentValue = this.counts[parentWord]; | |
const valueUnordered = currentValue > parentValue; | |
// To fix this, we need to keep it completely ordered. | |
// Will potentially have to do a neighbouring check, followed with a bubble down | |
const lexicallyUnordered = currentWord.localeCompare(parentWord) === -1; | |
if (valueUnordered || valueUnordered && lexicallyUnordered) { | |
this.heap[index] = parentWord; | |
this.position[parentWord] = index; | |
this.heap[parentIndex] = currentWord; | |
this.position[currentWord] = parentIndex; | |
this.bubbleUp(parentIndex); | |
} | |
} | |
topK (k) { | |
return this.heap.slice(0, k); | |
} | |
} | |
var topKFrequent = function(words, k) { | |
const freqHeap = new FreqHeap(); | |
words.forEach(word => freqHeap.add(word)); | |
console.log(freqHeap); | |
return freqHeap.topK(k); | |
}; |
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