How do you solve Find the Kth Largest Integer in the Array on LeetCode?

Find the Kth Largest Integer in the Array (LeetCode #1985) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log k) time complexity and O(k) space complexity. Key insight: When comparing strings that represent numbers, the length of the string is the primary factor in determining which is larger.

What is the time complexity of Find the Kth Largest Integer in the Array?

The optimal solution for Find the Kth Largest Integer in the Array runs in O(n log k) time and uses O(k) space. We maintain a heap of size k, which takes O(log k) time for each of the n elements, resulting in O(n log k) overall.

What is the brute force solution for Find the Kth Largest Integer in the Array?

The brute force approach for Find the Kth Largest Integer in the Array is Brute Force, with O(n log n) time complexity and O(1) space complexity. The brute force approach involves sorting the array of strings and then simply accessing the k-th largest element. This is straightforward but not the most efficient. The optimal Optimal Solution approach improves this to O(n log k).

What are the interview tips for Find the Kth Largest Integer in the Array?

Always clarify whether duplicates should be counted distinctly. Consider edge cases, such as all elements being the same or very large numbers. Think about the efficiency of your solution; can you reduce the time complexity?

What are common mistakes when solving Find the Kth Largest Integer in the Array?

Not considering the string comparison properly, leading to incorrect sorting. Confusing the indices when accessing the k-th largest element after sorting.

#1985

Find the Kth Largest Integer in the Array

Medium

ArrayStringDivide and ConquerSortingHeap (Priority Queue)QuickselectHeap (Priority Queue)Sorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n log n)	O(n log k)
Space	O(1)	O(k)

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Intuition

Time O(n log k)Space O(k)

Using a min-heap allows us to efficiently track the k largest elements without sorting the entire array, making this approach faster.

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Algorithm

4 steps

1Step 1: Create a min-heap to store the k largest numbers.
2Step 2: Iterate through each number in the array and add it to the heap.
3Step 3: If the heap exceeds size k, remove the smallest element (the root of the min-heap).
4Step 4: After processing all numbers, the root of the heap will be the k-th largest number.

solution.py9 lines

1import heapq
2
3def kthLargestNumber(nums, k):
4    min_heap = []
5    for num in nums:
6        heapq.heappush(min_heap, num)
7        if len(min_heap) > k:
8            heapq.heappop(min_heap)
9    return min_heap[0]

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Complexity note: We maintain a heap of size k, which takes O(log k) time for each of the n elements, resulting in O(n log k) overall.

1When comparing strings that represent numbers, the length of the string is the primary factor in determining which is larger.
2Using a min-heap allows us to efficiently keep track of the k largest elements without sorting the entire array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.