Minimum Number of Moves to Seat Everyone | Counting Sort | Sort | Leetcode 2037 | codestorywithMIK
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This is the 43rd Video of our Playlist "Leetcode Easy : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good practice Array problem : Minimum Number of Moves to Seat Everyone | Counting Sort | Normal Sort | Leetcode 2037 | 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 : Minimum Number of Moves to Seat Everyone | Counting Sort | Normal Sort | Leetcode 2037 | codestorywithMIK
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Summary :
Approach 1: Frequency Counting
This approach uses frequency arrays to keep track of the number of students and seats at each position. Here is a summary:
Initialization: Create two vectors position_seat and position_stud of size 101 (to cover seat positions from 0 to 100) initialized to 0.
Frequency Counting: Iterate over the seats and students arrays to populate position_seat and position_stud respectively.
Matching Positions: Use two pointers i and j to traverse the frequency arrays. For each non-zero entry in position_seat and position_stud, compute the moves as the absolute difference between the indices and decrement the counts. Continue this until all students are seated.
Result: The total moves are accumulated in the variable result.
Time Complexity: O(n + max_position)
Space Complexity: O(max_position)
Approach 2: Sorting
This approach involves sorting both the seats and students arrays and then directly pairing them based on their sorted order. Here is a summary:
Sorting: Sort both the seats and students arrays.
Computing Moves: Iterate through the sorted arrays and calculate the absolute difference between corresponding elements to get the total number of moves.
Result: The total moves are accumulated in the variable moves.
Time Complexity: O(n log n) due to the sorting step
Space Complexity: O(1) as it uses a constant amount of extra space
✨ Timelines✨
00:00 - Introduction
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