Find K-th Smallest Pair Distance | Multiple Approaches | Leetcode 719 | codestorywithMIK
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This is the 31st Video of our Playlist "Binary Search : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a very good Problem on Binary Search on Answer topic : Find K-th Smallest Pair Distance | Multiple Approaches | Leetcode 719 | codestorywithMIK
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Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Find K-th Smallest Pair Distance | Multiple Approaches | Leetcode 719 | codestorywithMIK
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Summary :
Approach-1 (Storing the Distances and Finding the k-th Smallest)
Time Complexity: O(n²)
Space Complexity: O(maxEl)
Explanation:
This approach first calculates all possible pairwise distances between elements in the array and stores the frequency of each distance in a vec array.
It then iterates through the vec array to find the k-th smallest distance by subtracting the frequencies from k until k is less than or equal to zero.
This method is efficient in terms of space when the range of distances (maxEl) is small.
Approach-2 (Using nth_element)
Time Complexity: O(n²)
Space Complexity: O(n²)
Explanation:
This approach calculates all pairwise distances and stores them in a vector.
It then uses nth_element to efficiently find the k-th smallest element in the unsorted vector.
While this method is simple and leverages efficient partitioning, it requires O(n²) space to store all distances and may perform less efficiently in the worst-case scenario.
Approach-3 (Binary Search + Sliding Window)
Time Complexity: O(n log n + n log M), where n log n is for sorting the array and n log M is for the binary search and sliding window, with M being the maximum possible distance.
Space Complexity: O(1)
Explanation:
This approach uses binary search to find the k-th smallest distance.
It sorts the array and then performs binary search on the possible distances. For each mid-distance, it uses a sliding window technique to count the number of pairs with a distance less than or equal to mid.
This method is space-efficient and generally faster due to the logarithmic nature of the binary search, making it more suitable for large inputs.
✨ Timelines✨
00:00 - Introduction
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