Question

2460. Apply Operations to an Array

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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:
  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

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;
  }
};

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
  }
}