Techieindoor

Leetcode 2460: Apply Operations to an Array Solution

techieindoor

Here, we will see how to solve Apply Operations to an Array Solution of leet code 2460 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:

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

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.

Apply Operations to an Array Solution code in C++

Code 1:

class Solution {
public:
    vector<int> applyOperations(vector<int>& nums) {
        for(int i = 0; i < nums.size() - 1; i++) {
            if(nums[i] == nums[i+1]) {
                nums[i] *= 2;
                nums[i+1] = 0;
            }
        }
        int i = 0;
        for(int j = 0; j < nums.size();) {
            if(nums[j] == 0) {
                j++;
            } else {
                nums[i] = nums[j];
                i++;
                j++;
            }
        }
        while(i < nums.size()) {
            nums[i++] = 0;
        }
        return nums;
    }
};

Code 2:

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] *= 2;
                nums[i+1] = 0;
            }
        }
        vector<int> ret;
        for (int i: nums) {
            if (i != 0) {
                ret.push_back(i);
            }
        }
        while (ret.size() < n) {
            ret.push_back(0);
        }
        return ret;
    }
};

Code 3:

class Solution {
public:
    vector<int> applyOperations(vector<int>& nums) {
        for(int i=1;i<nums.size();i++)
        {
            if(nums[i]==nums[i-1])
            {
                nums[i] = 0;
                nums[i - 1] *= 2;
            }
        }
        int cnt=0;
        vector<int> ans;
        for(int x:nums)
        {
            if(x==0)
                cnt++;
            else
                ans.push_back(x);
        }
        while(cnt>0)
        {
            ans.push_back(0);
            cnt--;
        }
        return ans;
    }
};

Code in Go

Code 1:

func applyOperations(nums []int) []int {
    var i, j = 0, 0
    
    for ;i < len(nums) - 1; i++ {
        if nums[i] == nums[i+1] {
            nums[i] *= 2
            nums[i+1] = 0
        }
    }
    i = 0
    
    for ;j < len(nums); j++ {
        if nums[j] != 0 {
            nums[i] = nums[j]
            i += 1
        }
    }
    for i < len(nums) {
        nums[i] = 0
        i += 1
    }
    return nums
}

Output:

Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]

To check more leetcode problem’s solution. Pls click given below link:

Alivecor Interview Go questions | set-1

https://www.techieindoor.com/category/leetcode/

Exit mobile version