Apply Operations to an Array

Leetcode Problem:

• Apply Operations to an Array

Summary

• The problem requires applying operations to a given array of non-negative integers, where in each operation, if the current element is equal to the next element, the current element is multiplied by 2 and the next element is set to 0
• The operations are applied sequentially, and after all operations, the 0's are shifted to the end of the array.

Approach

• The approach used is to iterate over the array, apply the operations, and then shift the 0's to the end
• The operations are applied sequentially, and the 0's are shifted to the end by adding them to the end of the array.

Complexity

• O(n), where n is the size of the array, because each element is visited once.

Explanation

• The solution code starts by initializing an index to 0 and an empty array to store the result
• It then enters a while loop that continues until the index is greater than or equal to the size of the array
• Inside the loop, it skips the 0's by incrementing the index until it finds a non-zero element
• If the current element is equal to the next element, it multiplies the current element by 2 and increments the index
• It then pushes the current element to the result array and increments the index
• After the while loop, it shifts the 0's to the end of the array by adding them to the end of the array
• Finally, it returns the result array.

Solution Code:

class Solution {
public:
    vector applyOperations(vector& nums) {
        int idx = 0;
        vector ans;
        while(idx < nums.size()){
            while(idx < nums.size() && nums[idx] == 0){
                idx++;
            }
            if(idx >= nums.size()) break;
            int num = nums[idx];
            if((idx + 1) < nums.size() && (nums[idx] == nums[idx+1])){
                num <<= 1;
                idx++;
            }
            ans.push_back(num);
            idx++;
        }
        int ansSize = ans.size();
        while(ansSize < nums.size()){
            ans.push_back(0);
            ansSize++;
        }
        return ans;
    }
};