grid47 Exploring patterns and algorithms
Oct 16, 2024
5 min read
Solution to LeetCode 215: Kth Largest Element in an Array Problem
Given an array of integers, find the kth largest element without sorting the entire array.
Problem
Approach
Steps
Complexity
Input: The input consists of an integer array nums and an integer k. The array contains n integers, and k is an integer specifying the rank of the largest element to find.
Example: [7, 2, 5, 8, 3, 9], k = 3
Constraints:
• 1 <= k <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4
Output: The output should be the kth largest element from the array.
Example: For input [7, 2, 5, 8, 3, 9] and k = 3, the output is 7.
Constraints:
Goal: Efficiently find the kth largest element in the array.
Steps:
• Use a priority queue (min-heap) to keep track of the k largest elements encountered so far.
• Iterate through the array, pushing each element into the heap. If the heap exceeds size k, remove the smallest element.
• After processing the array, the root of the heap will contain the kth largest element.
Goal: The algorithm should handle up to 10^5 elements efficiently.
Steps:
• The array size will be at most 10^5 elements.
• The element values range from -10^4 to 10^4.
Assumptions:
• k will always be a valid number, such that 1 <= k <= nums.length.
• The array will not be empty.
• Input: Example 1
• Explanation: In this example, the 3rd largest element is 7. After sorting the array in descending order, the array becomes [9, 8, 7, 5, 3, 2]. The 3rd largest element is 7.
• Input: Example 2
• Explanation: In this example, the 2nd largest element is 8. The sorted array is [10, 8, 6, 4, 2, 1]. The 2nd largest element is 8.
Approach: Use a heap (priority queue) to maintain the k largest elements efficiently without sorting the entire array.
Observations:
• Sorting the entire array would be inefficient for large arrays. A heap can help achieve this in O(n log k) time.
• The heap's size will be kept at k, and we will only insert new elements into the heap when necessary.
Steps:
• Initialize a min-heap.
• Iterate through the array, adding each element to the heap.
• If the heap exceeds size k, remove the smallest element.
• After processing all elements, the root of the heap will be the kth largest element.
Empty Inputs:
• The array will not be empty.
Large Inputs:
• Ensure that the solution works for arrays up to size 10^5.
Special Values:
• If the array contains duplicate elements, the algorithm will still correctly identify the kth largest element.
Constraints:
• Handle cases where all elements in the array are the same.
intfindKthLargest(vector<int>& nums, int k) {
priority_queue<int> pq;
for(intx: nums)
pq.push(x);
int x;
while(k-->0) {
x = pq.top();
pq.pop();
}
return x;
}
1 : Method Definition
intfindKthLargest(vector<int>& nums, int k) {
The method definition for the 'findKthLargest' function, which takes a vector of integers 'nums' and an integer 'k' as input.
2 : Priority Queue Initialization
priority_queue<int> pq;
Initializes a priority queue 'pq' (a max heap by default in C++) to store the elements of the array 'nums'.
3 : Loop Iteration
for(intx: nums)
Iterates over each element 'x' in the 'nums' vector.
4 : Priority Queue Insertion
pq.push(x);
Pushes each element 'x' from the 'nums' vector into the priority queue 'pq'.
5 : Variable Declaration
int x;
Declares an integer variable 'x' to store the current top element from the priority queue.
6 : While Loop
while(k-->0) {
A loop that runs 'k' times, each time removing the largest element from the priority queue.
7 : Priority Queue Access
x = pq.top();
Accesses the top (largest) element of the priority queue and assigns it to the variable 'x'.
8 : Priority Queue Removal
pq.pop();
Removes the top element from the priority queue.
9 : Return Statement
return x;
Returns the 'kth' largest element (the last element 'x' retrieved from the priority queue).
Best Case: O(n log k), where n is the length of the array.
Average Case: O(n log k), as we iterate through the array and perform heap operations.
Worst Case: O(n log k), since each heap operation takes O(log k) time, and we do this for each of the n elements.
Description: The time complexity is proportional to the number of elements in the array, with the log k factor accounting for the heap operations.
Best Case: O(k), since the heap size is always k.
Worst Case: O(k), as we store at most k elements in the heap.
Description: The space complexity is dominated by the size of the heap, which is k.
