Special Array With X Elements Greater Than or Equal X | 3 Approaches |Leetcode 1608|codestorywithMIK
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This is the 27th Video of our Playlist "Binary Search : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a very good Binary Search Practice Problem : Special Array With X Elements Greater Than or Equal X | 3 Approaches | Binary Search | Count Sort | Prefix Sum | Leetcode 1608 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Special Array With X Elements Greater Than or Equal X | 3 Approaches | Binary Search | Count Sort | Leetcode 1608 | codestorywithMIK
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Summary :
Approach 1: Using Binary Search
Time Complexity: O(nlogn)
Space Complexity: O(1)
Method:
Sort the array.
Iterate through possible values of x from 0 to n (length of the array).
Use lower_bound to find the first position where elements are greater than or equal to x.
Check if the count of elements greater than or equal to x equals x. If true, return x.
If no such x is found, return -1.
Approach 2: Binary Search on Answer
Time Complexity: O(nlogn)
Space Complexity: O(1)
Method:
Sort the array.
Use binary search to find the special value x.
Set initial search range between 0 and n.
Calculate the midpoint mid
x
and use lower_bound to find the count of elements greater than or equal to mid
x
.
Adjust the search range based on the comparison between mid
x
and the count of elements.
Return mid
x
if it matches the condition, otherwise return -1 if no such x is found.
Approach 3: Using Counting Sort + Prefix Sum
Time Complexity: O(n)
Space Complexity: O(n)
Method:
Create a frequency array of size n+1 to count occurrences of each element up to n, treating elements greater than n as n.
Populate the frequency array.
Iterate from n down to 0, maintaining a cumulative sum of frequencies.
Check if the current index equals the cumulative sum. If true, return the index.
If no such index is found, return -1.
Each approach efficiently finds the special number x such that there are exactly x elements in the array that are greater than or equal to x, using different strategies to balance between sorting, binary search, and counting techniques.
✨ Timelines✨
00:00 - Introduction
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