grid47 Exploring patterns and algorithms
Sep 29, 2024
5 min read
Solution to LeetCode 390: Elimination Game Problem
Given an integer n, simulate a process on a list of integers from 1 to n where every other number is removed, alternating the direction from left to right and right to left until only one number remains. Return the last remaining number.
Problem
Approach
Steps
Complexity
Input: The input is an integer n representing the length of a list starting from 1 to n.
Example: input = 10
Constraints:
• 1 <= n <= 10^9
Output: The output is the last remaining number in the list after applying the alternating removal process.
Example: Output: 4
Constraints:
• The last number should be a positive integer.
Goal: The goal is to determine the last remaining number after alternating removal of numbers from the list.
Steps:
• Simulate the process of removing every other number starting from the left, then from the right, and alternate until only one number remains.
• Keep track of the position of the remaining number after each step.
Goal: The algorithm should handle large inputs up to 10^9 efficiently.
Steps:
• The solution should work within time and space limits for inputs up to 10^9.
Assumptions:
• The input is guaranteed to be a positive integer n where 1 <= n <= 10^9.
• Input: Input: 10
• Explanation: Start with a list of numbers from 1 to 10, and simulate the alternating removal process. The last remaining number is 4.
Approach: To solve this problem efficiently, we can use a mathematical approach that exploits the alternating removal pattern without having to simulate the entire process for large inputs.
Observations:
• Simulating the alternating process for large values of n would be computationally expensive, so we need to optimize.
• The problem follows a predictable pattern, and the answer can be determined using mathematical reasoning rather than directly simulating the process.
Steps:
• Start by simulating the removal from left to right, then from right to left, keeping track of the last number after each step.
• For large n, observe that the pattern of removals follows a predictable cycle that can be exploited mathematically.
Empty Inputs:
• There will always be a valid input where n >= 1.
Large Inputs:
• Ensure the solution can handle values of n up to 10^9 efficiently.
Special Values:
• When n = 1, the result is trivially 1.
Constraints:
• The algorithm must handle large values of n without directly simulating the process.
intlastRemaining(int n) {
bool left =true;
int rm = n;
int step =1;
int head =1;
while(rm >1) {
if(left || rm %2==1) {
head += step;
}
rm /=2;
step *=2;
left =!left;
}
return head;
}
1 : Function Definition
intlastRemaining(int n) {
Define the function 'lastRemaining' which takes an integer 'n' representing the number of elements in the sequence, and returns the last remaining element.
2 : Variable Initialization
bool left =true;
Initialize 'left' as true to track whether numbers should be removed from the left or right in each step.
3 : Variable Initialization
int rm = n;
Initialize 'rm' as the total number of elements 'n'. This will keep track of the remaining elements in each step.
4 : Variable Initialization
int step =1;
Initialize 'step' to 1. This will track the step size by which elements are removed in each iteration.
5 : Variable Initialization
int head =1;
Initialize 'head' to 1. This variable represents the first element in the sequence that will be returned as the last remaining element.
6 : While Loop
while(rm >1) {
Start a while loop that runs until only one element remains in the sequence (i.e., rm > 1).
7 : Conditional Check
if(left || rm %2==1) {
Check if we are removing elements from the left side or if the remaining number of elements is odd, in which case the element at the 'head' should be included in the remaining sequence.
8 : Path Update
head += step;
If removing from the left or the number of remaining elements is odd, update 'head' to include the next element to be kept in the sequence.
9 : Division
rm /=2;
Divide 'rm' by 2 to halve the remaining elements in the sequence after each round of removals.
10 : Step Increase
step *=2;
Double the 'step' after each round of removals to account for the increasing gap between remaining elements.
11 : Switch Direction
left =!left;
Switch the direction of removal (from left to right or right to left) by toggling the 'left' flag.
12 : Return Statement
return head;
Return the value of 'head', which represents the last remaining number in the sequence.
Best Case: O(log n)
Average Case: O(log n)
Worst Case: O(log n)
Description: The time complexity is O(log n) as we reduce the number of elements by half in each step.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) since we do not need extra space other than a few variables to track the current state.
