Question

2263. Make Array Non-decreasing or Non-increasing

中文文档

Description

You are given a 0-indexed integer array nums. In one operation, you can:

• Choose an index i in the range 0 <= i < nums.length
• Set nums[i] to nums[i] + 1 or nums[i] - 1

Return _the minimum number of operations to make _nums_ non-decreasing or non-increasing._

 

Example 1:

Input: nums = [3,2,4,5,0]
Output: 4
Explanation:
One possible way to turn nums into non-increasing order is to:
- Add 1 to nums[1] once so that it becomes 3.
- Subtract 1 from nums[2] once so it becomes 3.
- Subtract 1 from nums[3] twice so it becomes 3.
After doing the 4 operations, nums becomes [3,3,3,3,0] which is in non-increasing order.
Note that it is also possible to turn nums into [4,4,4,4,0] in 4 operations.
It can be proven that 4 is the minimum number of operations needed.

Example 2:

Input: nums = [2,2,3,4]
Output: 0
Explanation: nums is already in non-decreasing order, so no operations are needed and we return 0.

Example 3:

Input: nums = [0]
Output: 0
Explanation: nums is already in non-decreasing order, so no operations are needed and we return 0.

 

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 1000

 

Follow up: Can you solve it in O(n*log(n)) time complexity?

Solutions

Solution 1

class Solution:
  def convertArray(self, nums: List[int]) -> int:
    def solve(nums):
      n = len(nums)
      f = [[0] * 1001 for _ in range(n + 1)]
      for i, x in enumerate(nums, 1):
        mi = inf
        for j in range(1001):
          if mi > f[i - 1][j]:
            mi = f[i - 1][j]
          f[i][j] = mi + abs(x - j)
      return min(f[n])

    return min(solve(nums), solve(nums[::-1]))

class Solution {
  public int convertArray(int[] nums) {
    return Math.min(solve(nums), solve(reverse(nums)));
  }

  private int solve(int[] nums) {
    int n = nums.length;
    int[][] f = new int[n + 1][1001];
    for (int i = 1; i <= n; ++i) {
      int mi = 1 << 30;
      for (int j = 0; j <= 1000; ++j) {
        mi = Math.min(mi, f[i - 1][j]);
        f[i][j] = mi + Math.abs(j - nums[i - 1]);
      }
    }
    int ans = 1 << 30;
    for (int x : f[n]) {
      ans = Math.min(ans, x);
    }
    return ans;
  }

  private int[] reverse(int[] nums) {
    for (int i = 0, j = nums.length - 1; i < j; ++i, --j) {
      int t = nums[i];
      nums[i] = nums[j];
      nums[j] = t;
    }
    return nums;
  }
}

class Solution {
public:
  int convertArray(vector<int>& nums) {
    int a = solve(nums);
    reverse(nums.begin(), nums.end());
    int b = solve(nums);
    return min(a, b);
  }

  int solve(vector<int>& nums) {
    int n = nums.size();
    int f[n + 1][1001];
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n; ++i) {
      int mi = 1 << 30;
      for (int j = 0; j <= 1000; ++j) {
        mi = min(mi, f[i - 1][j]);
        f[i][j] = mi + abs(nums[i - 1] - j);
      }
    }
    return *min_element(f[n], f[n] + 1001);
  }
};

func convertArray(nums []int) int {
  return min(solve(nums), solve(reverse(nums)))
}

func solve(nums []int) int {
  n := len(nums)
  f := make([][1001]int, n+1)
  for i := 1; i <= n; i++ {
    mi := 1 << 30
    for j := 0; j <= 1000; j++ {
      mi = min(mi, f[i-1][j])
      f[i][j] = mi + abs(nums[i-1]-j)
    }
  }
  ans := 1 << 30
  for _, x := range f[n] {
    ans = min(ans, x)
  }
  return ans
}

func reverse(nums []int) []int {
  for i, j := 0, len(nums)-1; i < j; i, j = i+1, j-1 {
    nums[i], nums[j] = nums[j], nums[i]
  }
  return nums
}

func abs(x int) int {
  if x < 0 {
    return -x
  }
  return x
}