Skip to content

Instantly share code, notes, and snippets.

@rohit012
Created October 22, 2017 03:40
Show Gist options
  • Star 0 You must be signed in to star a gist
  • Fork 0 You must be signed in to fork a gist
  • Save rohit012/289665de0c9470b9b9fbf9f4e72e8996 to your computer and use it in GitHub Desktop.
Save rohit012/289665de0c9470b9b9fbf9f4e72e8996 to your computer and use it in GitHub Desktop.
/*Given a non-empty list of words, return the k most frequent elements.
Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.
Example 1:
Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
Note that "i" comes before "love" due to a lower alphabetical order.
Example 2:
Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
with the number of occurrence being 4, 3, 2 and 1 respectively.
Note:
You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Input words contain only lowercase letters.
Follow up:
Try to solve it in O(n log k) time and O(n) extra space.
*/
var topKFrequent = function(words, k) {
var map = {};
words.forEach(function(item){
if(map[item]){
map[item] = map[item]+1;
} else {
map[item] = 1;
}
});
var arr = [];
for(key in map) {
arr.push({key: key, val: map[key]});
}
return arr.sort(function(a, b){
if(a.val===b.val) {
return a.key.localeCompare(b.key);
}
return b.val-a.val;
}).slice(0,k).map(function(item){
return item.key
});
};
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment