# 2460. Apply Operations to an Array

## Description

You are given a 0-indexed array nums of size n consisting of non-negative integers.

You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums:

• If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.

After performing all the operations, shift all the 0's to the end of the array.

• For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0].

Return the resulting array.

Note that the operations are applied sequentially, not all at once.

Example 1:

Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]
Explanation: We do the following operations:
- i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
- i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
- i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
- i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
- i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].


Example 2:

Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.


Constraints:

• 2 <= nums.length <= 2000
• 0 <= nums[i] <= 1000

## Solutions

Solution 1: Simulation

We can directly simulate according to the problem description.

First, we traverse the array $nums$. For any two adjacent elements $nums[i]$ and $nums[i+1]$, if $nums[i] = nums[i+1]$, then we double the value of $nums[i]$ and change the value of $nums[i+1]$ to $0$.

Then, we create an answer array $ans$ of length $n$, and put all non-zero elements of $nums$ into $ans$ in order.

Finally, we return the answer array $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

• class Solution {
public int[] applyOperations(int[] nums) {
int n = nums.length;
for (int i = 0; i < n - 1; ++i) {
if (nums[i] == nums[i + 1]) {
nums[i] <<= 1;
nums[i + 1] = 0;
}
}
int[] ans = new int[n];
int i = 0;
for (int x : nums) {
if (x > 0) {
ans[i++] = x;
}
}
return ans;
}
}

• class Solution {
public:
vector<int> applyOperations(vector<int>& nums) {
int n = nums.size();
for (int i = 0; i < n - 1; ++i) {
if (nums[i] == nums[i + 1]) {
nums[i] <<= 1;
nums[i + 1] = 0;
}
}
vector<int> ans(n);
int i = 0;
for (int& x : nums) {
if (x) {
ans[i++] = x;
}
}
return ans;
}
};

• class Solution:
def applyOperations(self, nums: List[int]) -> List[int]:
n = len(nums)
for i in range(n - 1):
if nums[i] == nums[i + 1]:
nums[i] <<= 1
nums[i + 1] = 0
ans = [0] * n
i = 0
for x in nums:
if x:
ans[i] = x
i += 1
return ans


• func applyOperations(nums []int) []int {
n := len(nums)
for i := 0; i < n-1; i++ {
if nums[i] == nums[i+1] {
nums[i] <<= 1
nums[i+1] = 0
}
}
ans := make([]int, n)
i := 0
for _, x := range nums {
if x > 0 {
ans[i] = x
i++
}
}
return ans
}

• function applyOperations(nums: number[]): number[] {
const n = nums.length;
for (let i = 0; i < n - 1; ++i) {
if (nums[i] === nums[i + 1]) {
nums[i] <<= 1;
nums[i + 1] = 0;
}
}
const ans: number[] = Array(n).fill(0);
let i = 0;
for (const x of nums) {
if (x !== 0) {
ans[i++] = x;
}
}
return ans;
}


• impl Solution {
pub fn apply_operations(nums: Vec<i32>) -> Vec<i32> {
let mut nums = nums;

for i in 0..nums.len() - 1 {
if nums[i] == nums[i + 1] {
nums[i] <<= 1;
nums[i + 1] = 0;
}
}

let mut cur = 0;
for i in 0..nums.len() {
if nums[i] != 0 {
nums.swap(i, cur);
cur += 1;
}
}

nums
}
}