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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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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
      }
  }
  
  

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