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 $O(1)$.

Python3JavaC++GoTypeScriptRust

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
    }
}

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