Add top-k-frequent-elements solution in TypeScript

Author: Jeehay28

top-k-frequent-elements/Jeehay28.ts

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+// Approach 2: Min-Heap
+// ⏳ Time Complexity: O(n log(k))
+// 💾 Space Complexity: O(n)
+
+function topKFrequent(nums: number[], k: number): number[] {
+  const map = new Map<number, number>();
+  // step 1: count frequency of each number
+  for (const num of nums) {
+    map.set(num, (map.get(num) || 0) + 1);
+  }
+
+  // step 2: use a min-heap to keep only top k frequent elements
+  // A Min-Heap is a special type of binary heap (tree-based data structure) where:
+  // The smallest element is always at the root (top).
+
+  const minHeap: [number, number][] = [];
+
+  // Time Complexity: O(n log k)
+  // Sorting is within a fixed size k which is much smaller than O(n log n) sorting all elements.
+  for (const [num, freq] of map.entries()) {
+    if (minHeap.length < k) {
+      minHeap.push([num, freq]);
+      minHeap.sort((a, b) => a[1] - b[1]);
+    } else if (freq > minHeap[0][1]) {
+      minHeap.shift(); // Remove smallest frequency
+      minHeap.push([num, freq]);
+      minHeap.sort((a, b) => a[1] - b[1]);
+    }
+  }
+
+  // step 3: extract the top k freqeuent elements from the heap
+  return minHeap.map(([num, freq]) => num);
+}
+
+
+// Approach 1: Map & Sorting (Using Map in TypeScript)
+// ⏳ Time Complexity: O(n log(n))
+// 💾 Space Complexity: O(n)
+// function topKFrequent(nums: number[], k: number): number[] {
+
+//     const map = new Map<number, number>()
+
+//     for (const num of nums) {
+//         map.set(num, (map.get(num) || 0) + 1);
+//     }
+
+//     // const sorted = [...map.entries()].sort((a, b) => b[1] - a[1]);
+//     const sorted = Array.from(map).sort((a, b) => b[1] - a[1]);
+
+//     return sorted.slice(0, k).map(item => item[0]);
+// };
+