Top k frequent elements

Contains Duplicate.js → ContainsDuplicate.js

@@ -13,3 +13,5 @@ const containsDuplicate = function (nums) {
 
   return newArr.some((x) => x >= 2);
 };
+
+// https://leetcode.com/problems/contains-duplicate/description/

LongestConsecutiveSequence.js

@@ -0,0 +1,29 @@
+// Given an unsorted array of integers nums,
+// return the length of the longest consecutive elements sequence.
+// You must write an algorithm that runs in O(n) time.
+
+const longestConsecutive = (nums) => {
+  if (!nums.length) return 0;
+
+  const x = [...new Set(nums)].sort((a, b) => a - b);
+  const answer = [];
+
+  let cnt = 1;
+
+  x.forEach((v, i) => {
+    if (v + 1 === x[i + 1]) {
+      cnt += 1;
+    } else {
+      answer.push(cnt);
+      cnt = 1;
+    }
+  });
+
+  return Math.max(...answer);
+};
+
+const nums = [1, 2, 0, 1];
+const x = longestConsecutive(nums);
+console.log(x);
+
+// https://leetcode.com/problems/longest-consecutive-sequence/description/

TwoSum.js

@@ -20,3 +20,5 @@ const nums = [3, 2, 3];
 const target = 6;
 const a = twoSum(nums, target);
 console.log(a);
+
+// https://leetcode.com/problems/two-sum/description/

topKFrequent.js

@@ -0,0 +1,27 @@
+// Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.
+
+const topKFrequent = (nums, k) => {
+  const obj = {};
+
+  for (let i = 0; i < nums.length; i++) {
+    const v = nums[i];
+    if (obj[v]) {
+      obj[v] += 1;
+    } else {
+      obj[v] = 1;
+    }
+  }
+
+  const sortedArray = Object.entries(obj).sort((a, b) => b[1] - a[1]);
+  const answer = sortedArray.slice(0, k).map(([num, _]) => Number(num));
+  return answer;
+
+  //[ [ '1', 3 ], [ '2', 2 ], [ '3', 3 ] ]
+  // 0:숫자  1:몇개가 존재하는지
+  // [1] 번째 인덱스 () 순서대로 정렬.
+};
+
+const nums = [1, 1, 1, 2, 2, 3, 3, 3];
+
+const k = 2;
+const a = topKFrequent(nums, k);