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360. Sort Transformed Array πŸ”’

Description

Given a sorted integer array nums and three integers a, b and c, apply a quadratic function of the form f(x) = ax2 + bx + c to each element nums[i] in the array, and return the array in a sorted order.

 

Example 1:

Input: nums = [-4,-2,2,4], a = 1, b = 3, c = 5
Output: [3,9,15,33]

Example 2:

Input: nums = [-4,-2,2,4], a = -1, b = 3, c = 5
Output: [-23,-5,1,7]

 

Constraints:

  • 1 <= nums.length <= 200
  • -100 <= nums[i], a, b, c <= 100
  • nums is sorted in ascending order.

 

Follow up: Could you solve it in O(n) time?

Solutions

Solution 1

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class Solution:
    def sortTransformedArray(
        self, nums: List[int], a: int, b: int, c: int
    ) -> List[int]:
        def f(x):
            return a * x * x + b * x + c

        n = len(nums)
        i, j, k = 0, n - 1, 0 if a < 0 else n - 1
        res = [0] * n
        while i <= j:
            v1, v2 = f(nums[i]), f(nums[j])
            if a < 0:
                if v1 <= v2:
                    res[k] = v1
                    i += 1
                else:
                    res[k] = v2
                    j -= 1
                k += 1
            else:
                if v1 >= v2:
                    res[k] = v1
                    i += 1
                else:
                    res[k] = v2
                    j -= 1
                k -= 1
        return res

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class Solution {
    public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
        int n = nums.length;
        int i = 0, j = n - 1, k = a < 0 ? 0 : n - 1;
        int[] res = new int[n];
        while (i <= j) {
            int v1 = f(a, b, c, nums[i]), v2 = f(a, b, c, nums[j]);
            if (a < 0) {
                if (v1 <= v2) {
                    res[k] = v1;
                    ++i;
                } else {
                    res[k] = v2;
                    --j;
                }
                ++k;
            } else {
                if (v1 >= v2) {
                    res[k] = v1;
                    ++i;
                } else {
                    res[k] = v2;
                    --j;
                }
                --k;
            }
        }
        return res;
    }

    private int f(int a, int b, int c, int x) {
        return a * x * x + b * x + c;
    }
}

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class Solution {
public:
    vector<int> sortTransformedArray(vector<int>& nums, int a, int b, int c) {
        int n = nums.size();
        int i = 0, j = n - 1, k = a < 0 ? 0 : n - 1;
        vector<int> res(n);
        while (i <= j) {
            int v1 = f(a, b, c, nums[i]), v2 = f(a, b, c, nums[j]);
            if (a < 0) {
                if (v1 <= v2) {
                    res[k] = v1;
                    ++i;
                } else {
                    res[k] = v2;
                    --j;
                }
                ++k;
            } else {
                if (v1 >= v2) {
                    res[k] = v1;
                    ++i;
                } else {
                    res[k] = v2;
                    --j;
                }
                --k;
            }
        }
        return res;
    }

    int f(int a, int b, int c, int x) {
        return a * x * x + b * x + c;
    }
};

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func sortTransformedArray(nums []int, a int, b int, c int) []int {
    n := len(nums)
    i, j, k := 0, n-1, 0
    if a >= 0 {
        k = n - 1
    }
    res := make([]int, n)
    for i <= j {
        v1, v2 := f(a, b, c, nums[i]), f(a, b, c, nums[j])
        if a < 0 {
            if v1 <= v2 {
                res[k] = v1
                i++
            } else {
                res[k] = v2
                j--
            }
            k++
        } else {
            if v1 >= v2 {
                res[k] = v1
                i++
            } else {
                res[k] = v2
                j--
            }
            k--
        }
    }
    return res
}

func f(a, b, c, x int) int {
    return a*x*x + b*x + c
}

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