977. Squares of a Sorted Array [Leetcode][C++]

class Solution { public: vector<int> sortedSquares(vector<int>& nums) { // Brute Force Solution Time O(NlogN) & Auxiliary Space O(1) int len=nums.size(); for(int i=0;i<len;i++) nums[i]=nums[i]*nums[i]; sort(nums.begin(),nums.end()); return nums; } }; 

class Solution { public: vector<int> sortedSquares(vector<int>& nums) { // Efficient Solution Time O(N) & Auxiliary Space O(1) // Two Pointer Method int len=nums.size(), lt=0, rt=len; vector<int> res; // Find index of first positive vector element & position right pointer for(int i=0;i<len;i++){ if(nums[i]>=0){ rt=i; break; } } // Position left pointer at smallest negative vector element lt=rt-1; while(lt>=0 && rt<=len-1){ // Compare elements at left & right index and push the square of  // the smaller element in the resultant vector in ascending order // and reposition the left(decrement) & right(increment) pointers if(abs(nums[lt])<nums[rt]){ res.push_back(nums[lt]*nums[lt]); lt--; } else if(abs(nums[lt])>nums[rt]){ res.push_back(nums[rt]*nums[rt]); rt++; } else{ res.push_back(nums[lt]*nums[lt]); lt--; res.push_back(nums[rt]*nums[rt]); rt++; } } // Push the square of the element in the resultant vector for the // case of unvisited elements while(lt>-1){ res.push_back(nums[lt]*nums[lt]); lt--; } while(rt<len){ res.push_back(nums[rt]*nums[rt]); rt++; } return res; } };