0977. Squares of a Sorted Array¶
Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.
Example 1:
Input: nums = [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Explanation: After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].
Example 2:
Input: nums = [-7,-3,2,3,11]
Output: [4,9,9,49,121]
Constraints:
1 <= nums.length <= 104-104 <= nums[i] <= 104numsis sorted in non-decreasing order.
Follow up: Squaring each element and sorting the new array is very trivial, could you find an O(n) solution using a different approach?
Analysis¶
This question can be solved using two pointers method to achieve linear time complexity. Since the given array has negative values, there is a possible case when nums[left] ^ 2 is greater than nums[right] ^ 2. However, it is also true that from any points from left or from right, the square of each element is montonic increasing. The only exception to the previous statement is when left == right– when the two pointers met, and we can use that as the termination condition.
- Time: O(n)
- Space: O(n) for the solution array
Code¶
#define abs(x) (x > 0 ? x : -x)
class Solution {
public:
vector<int> sortedSquares(vector<int>& nums) {
int sz = nums.size();
vector<int> res(sz);
for (int l = 0, r = sz - 1, p = sz - 1; l <= r; --p) {
res[p] = (max(nums[l] * nums[l], nums[r] * nums[r]));
// choose the next greatest one by eliminating the current largest one
if (abs(nums[l]) > abs(nums[r])) l ++;
else r --;
}
return res;
}
};