# Solution For Elimination Game

Problem:

Given an integer array nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and using only constant extra space.

Solution:

Approach 1: Hashset

One of the simplest solutions to this problem is to use a Hashset to store the elements of the given array nums. We can iterate through the array and for every element, we check if it exists in the Hashset. If it doesn’t exist, we add it to the Hashset. If it does exist, we remove it from the Hashset. By the end of the iteration, the Hashset will only contain the element that appears once in the original array.

Time complexity: O(n). We iterate through the entire array once.
Space complexity: O(n). We store at most n elements in the Hashset.

Approach 2: Bit manipulation

Another approach to this problem is to use bit manipulation. The idea is to XOR all the elements in the array. Since every element appears twice except for one, the XOR of all elements will cancel out the duplicates, leaving only the element that appears once.

Time complexity: O(n). We iterate through the entire array once.
Space complexity: O(1). We only need constant extra space to store the result.

Code:

Approach 1: Hashset

class Solution {
public int singleNumber(int[] nums) {
HashSet set = new HashSet<>();
for (int num : nums) {
if (set.contains(num)) {
set.remove(num);
} else {
}
}
return set.iterator().next();
}
}

Approach 2: Bit manipulation

class Solution {
public int singleNumber(int[] nums) {
int ans = 0;
for (int num : nums) {
ans ^= num;
}
return ans;
}
}

## Step by Step Implementation For Elimination Game

```There are several ways to solve this problem. One way is to use a stack. First, we push all the elements onto the stack. Then, we pop off the top element and check if it is divisible by 2. If it is, we divide it by 2 and push it back onto the stack. Otherwise, we simply discard it. We continue doing this until the stack is empty. The final answer will be the last element remaining on the stack.

Another way to solve this problem is to use a queue. First, we enqueue all the elements onto the queue. Then, we dequeue the front element and check if it is divisible by 2. If it is, we divide it by 2 and enqueue it back onto the queue. Otherwise, we simply discard it. We continue doing this until the queue is empty. The final answer will be the last element remaining on the queue.```
```#There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.

#Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.

#We keep repeating the steps again, alternating left to right and right to left, until a single number remains.

#Find the last number that remains starting with a list of length n.

#Example:

#Input:
#n = 9,
#1 2 3 4 5 6 7 8 9
#2 4 6 8
#2 6
#6

#Output:
#6```
```var lastRemaining = function(n) {
// Base case
if (n == 1) {
return 1;
}

// Recursive case
if (n % 2 == 1) {
// If n is odd, the last remaining number is the
// same as the last remaining number in n-1
return lastRemaining(n-1);
} else {
// If n is even, the last remaining number is n/2
// when the head is removed
return 2 * lastRemaining(n/2);
}
};```
```int lastRemaining(int n)
{
return n == 1 ? 1 : 2 * (n / 2 + 1 - lastRemaining(n / 2));
}```
```int lastRemaining(int n) {
return n == 1 ? 1 : (n - 1) * 2 - lastRemaining(n - 1);
}```

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