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Formatted question description: https://leetcode.ca/all/2460.html

2460. Apply Operations to an Array

  • Difficulty: Easy.
  • Related Topics: Array, Simulation.
  • Similar Questions: Remove Duplicates from Sorted Array, Move Zeroes.

Problem

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

Solution (Java, C++, Python)

  • 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 j = 0;
        for (int i = 0; i < n; ++i) {
            if (nums[i] != 0) {
                ans[j++] = nums[i];
            }
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                ans[j++] = nums[i];
            }
        }
        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 j = 0;
        for (int i = 0; i < n; ++i) if (nums[i]) ans[j++] = nums[i];
        for (int i = 0; i < n; ++i) if (!nums[i]) ans[j++] = nums[i];
        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 = [v for v in nums if v]
        ans += [0] * (n - len(ans))
        return ans


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

  • 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
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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