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LeetCode Solutions

360. Sort Transformed Array

Time: $O(n)$

Space: $O(n)$

			


class Solution {
 public:
  vector<int> sortTransformedArray(vector<int>& nums, int a, int b, int c) {
    const int n = nums.size();
    const bool upward = a > 0;
    vector<int> ans(n);
    vector<int> quad;

    for (const int num : nums)
      quad.push_back(f(num, a, b, c));

    int i = upward ? n - 1 : 0;
    for (int l = 0, r = n - 1; l <= r;)
      if (upward)  // Maximum in both ends
        ans[i--] = quad[l] > quad[r] ? quad[l++] : quad[r--];
      else  // Minimum in both ends
        ans[i++] = quad[l] < quad[r] ? quad[l++] : quad[r--];

    return ans;
  }

 private:
  // The concavity of f only depends on a's sign
  int f(int x, int a, int b, int c) {
    return (a * x + b) * x + c;
  }
};




			


class Solution {
  public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
    final int n = nums.length;
    final boolean upward = a > 0;
    int[] ans = new int[n];
    int[] quad = new int[n];

    for (int i = 0; i < nums.length; ++i)
      quad[i] = f(nums[i], a, b, c);

    int i = upward ? n - 1 : 0;
    for (int l = 0, r = n - 1; l <= r;)
      if (upward) // Maximum in both ends
        ans[i--] = quad[l] > quad[r] ? quad[l++] : quad[r--];
      else // Minimum in both ends
        ans[i++] = quad[l] < quad[r] ? quad[l++] : quad[r--];

    return ans;
  }

  // The concavity of f only depends on a's sign
  private int f(int x, int a, int b, int c) {
    return (a * x + b) * x + c;
  }
}




			


class Solution:
  def sortTransformedArray(self, nums: List[int], a: int, b: int, c: int) -> List[int]:
    n = len(nums)
    upward = a > 0
    ans = [0] * n

    # The concavity of f only depends on a's sign
    def f(x: int, a: int, b: int, c: int) -> int:
      return (a * x + b) * x + c

    quad = [f(num, a, b, c) for num in nums]

    i = n - 1 if upward else 0
    l = 0
    r = n - 1
    while l <= r:
      if upward:  # Maximum in both ends
        if quad[l] > quad[r]:
          ans[i] = quad[l]
          l += 1
        else:
          ans[i] = quad[r]
          r -= 1
        i -= 1
      else:  # Minimum in both ends
        if quad[l] < quad[r]:
          ans[i] = quad[l]
          l += 1
        else:
          ans[i] = quad[r]
          r -= 1
        i += 1

    return ans