Back-To-CP Day 1

13-10-2022

Two Sum II

For the array I used two pointers, one for the beginning of the and one for the end of the array. In a loop I checked if sum of those pointer value greater or less than the target value and decrease end pointer or increase start pointer by 1. Otherwise stop the loop.
Returned a vector containing start pointer + 1 and end pointer + 1 as the index in result supposed to start from 1 and not 0.

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
         vector<int> res;
         int l = 0, h = numbers.size() - 1;
         for(int i = 0; i < numbers.size(); i++){
            if(numbers[l] + numbers[h] < target){
                l++;
            }else if(numbers[l] + numbers[h] > target){
                h--;
            }
            else{
                res.push_back(l+1);
                res.push_back(h+1);
                break;              
            }
        }
        return res;
    }
};

Move Zero
The approach is quite easy and solved it with one go. Firstly, I declared a vector. Then checked if the array contains non-zero, append that that vector otherwise counts how many 0s are there. After the counting is done we just append the number of 0s that was connected. At the end I just replaced the given vector with the new vector. Quite Simple and fun.

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        vector<int> res;
        int cc = 0;
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] != 0){
                res.push_back(nums[i]);
            }else{
                cc++;
            }
        }
        for(int i = nums.size() - cc; i < nums.size(); i++){
            res.push_back(0);
        }
        nums = res;
    }
};