Kth Largest Element in an Array

#️⃣ Array, Priority Queue, Quick Select

https://leetcode.com/problems/kth-largest-element-in-an-array/description

🔥 Understand Problem

If the array is [8, 6, 12, 9, 3, 4] and k is 2, you need to find the 2nd largest element in this array. First, you will sort the array: [3, 4, 6, 8, 9, 12] The output will be 9 because it is the second-largest element.

✅ Bruteforce

var findKthLargest = function(nums, k) {
    // Sort the array in ascending order
    nums.sort((a, b) => a - b);

    // Return the kth largest element
    return nums[nums.length - k];
};

Explanation:

1. Sorting the Array: The array is sorted in ascending order using the sort method.
2. Finding the kth Largest Element: The kth largest element is found by accessing the element at the index nums.length - k.

Time Complexity:

• Sorting: The time complexity of sorting an array is (O(nlog n)), where (n) is the length of the array.
• Accessing the Element: Accessing an element in an array is O(1).

So, the overall time complexity is O(n log n).

Space Complexity:

• In-Place Sorting: The sort method sorts the array in place, so the space complexity is O(1) for the sorting operation.
• Overall: Since we are not using any additional data structures, the overall space complexity is O(1).

✅ Better

var findKthLargest = function(nums, k) {
        // Create a min-heap using a priority queue
        let minHeap = new MinPriorityQueue();

        // Add the first k elements to the heap
        for (let i = 0; i < k; i++) {
            //minHeap.enqueue(nums[i]): Adds the element nums[i] to the min-heap.
            minHeap.enqueue(nums[i]);
        }

        // Iterate through the remaining elements
        for (let i = k; i < nums.length; i++) {
            //minHeap.front().element: Retrieves the smallest element in the min-heap without removing it.
            if (minHeap.front().element < nums[i]) {
                // minHeap.dequeue(): Removes the smallest element from the min-heap.
                minHeap.dequeue();
                // Add the current element
                minHeap.enqueue(nums[i]);
            }
        }

        // The root of the heap is the kth largest element
        return minHeap.front().element;
    };

Explanation:

1. Initial Array: [2, 1, 6, 3, 5, 4]
2. k = 3: We need to find the 3rd largest element.

Step 1: Add the first k elements to the min-heap

• Heap after adding 2: [2]
• Heap after adding 1: [1, 2]
• Heap after adding 6: [1, 2, 6]

Step 2: Iterate through the remaining elements

• Current element = 3

  • Heap before comparison: [1, 2, 6]
  • Smallest element in heap (minHeap.front().element): 1
  • Comparison: 3 > 1
  • Action: Remove 1 and add 3
  • Heap after update: [2, 6, 3]

• Current element = 5

  • Heap before comparison: [2, 6, 3]
  • Smallest element in heap (minHeap.front().element): 2
  • Comparison: 5 > 2
  • Action: Remove 2 and add 5
  • Heap after update: [3, 6, 5]

• Current element = 4

  • Heap before comparison: [3, 6, 5]
  • Smallest element in heap (minHeap.front().element): 3
  • Comparison: 4 > 3
  • Action: Remove 3 and add 4
  • Heap after update: [4, 6, 5]

Final Step: Return the root of the heap

• Heap: [4, 6, 5]
• 3rd largest element: 4 (the root of the heap)

Time Complexity:

• Heap Operations: Inserting and removing elements from the heap takes O(log k) time.
• Overall: Since we perform these operations for each of the n elements in the array, the overall time complexity is O(n log k).

Space Complexity:

• Heap Storage: The space complexity is O(k) because the heap stores at most k elements.

✅ Best

Note: Even though Leetcode restricts quick select, you should remember this approach because it passes most test cases

//Quick Select Algo

function quickSelect(list, left, right, k)

   if left = right
      return list[left]

   Select a pivotIndex between left and right

   pivotIndex := partition(list, left, right, 
                                  pivotIndex)
   if k = pivotIndex
      return list[k]
   else if k < pivotIndex
      right := pivotIndex - 1
   else
      left := pivotIndex + 1

var findKthLargest = function(nums, k) {
    // Call the quickSelect function to find the kth largest element
    return quickSelect(nums, 0, nums.length - 1, nums.length - k);
};

function quickSelect(nums, low, high, index) {
    // If the low and high pointers are the same, return the element at low
    if (low === high) return nums[low];

    // Partition the array and get the pivot index
    let pivotIndex = partition(nums, low, high);

    // If the pivot index is the target index, return the element at pivot index
    if (pivotIndex === index) {
        return nums[pivotIndex];
    } else if (pivotIndex > index) {
        // If the pivot index is greater than the target index, search in the left partition
        return quickSelect(nums, low, pivotIndex - 1, index);
    } else {
        // If the pivot index is less than the target index, search in the right partition
        return quickSelect(nums, pivotIndex + 1, high, index);
    }
}

function partition(nums, low, high) {
    // Choose the pivot element
    let pivot = nums[high];
    let pointer = low;

    // Rearrange the elements based on the pivot
    for (let i = low; i < high; i++) {
        if (nums[i] <= pivot) {
            [nums[i], nums[pointer]] = [nums[pointer], nums[i]];
            pointer++;
        }
    }

    // Place the pivot element in its correct position
    [nums[pointer], nums[high]] = [nums[high], nums[pointer]];
    return pointer;
}

Explanation:

1. Initial Array: [3, 2, 1, 5, 6, 4]
2. k = 2: We need to find the 2nd largest element.

Step 1: Partition the array

• Pivot element = 4
• Array after partitioning: [3, 2, 1, 4, 6, 5]
• Pivot index = 3

Step 2: Recursive Selection

• Target index = 4 (since we need the 2nd largest element, which is the 4th index in 0-based indexing)
• Pivot index (3) < Target index (4): Search in the right partition [6, 5]

Step 3: Partition the right partition

• Pivot element = 5
• Array after partitioning: [3, 2, 1, 4, 5, 6]
• Pivot index = 4

Final Step: Return the element at the target index

• Element at index 4: 5

Time Complexity:

• Average Case: The average time complexity of Quickselect is O(n).
• Worst Case: The worst-case time complexity is O(n^2), but this is rare with good pivot selection.

Space Complexity:

• In-Place: The space complexity is O(1) because the algorithm works in place.