Square Each Element in Array

function squareArray(nums) {
  return nums.map(num => num * num);
}
                  

Explanation: We iterate over the array and replace each element with its square.

⏱ O(n) | 💾 O(n)

class Solution {
  public int[] squareArray(int[] nums) {
    int[] result = new int[nums.length];
    for (int i = 0; i < nums.length; i++) {
      result[i] = nums[i] * nums[i];
    }
    return result;
  }
}
                  

Explanation: We create a new array and store the square of each element.

⏱ O(n) | 💾 O(n)

def square_array(nums):
    return [num * num for num in nums]
                  

Explanation: List comprehension allows us to square each element efficiently.

⏱ O(n) | 💾 O(n)