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2263. Make Array Non-decreasing or Non-increasing πŸ”’

DifficultyHard

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

Thinking

We may increment or decrement an element by one and want a non-decreasing or non-increasing array at minimum total cost. \(n\) and the value range are both \(10^3\), so we can DP on the final value of position \(i\). Non-increasing is non-decreasing on the reversed array.

\(f[i][j]\) is the cost of the first \(i\) positions with the \(i\)-th equal to \(j\). Then \(f[i][j] = \min_{k\le j} f[i-1][k] + |j-nums[i-1]|\), and the running min is maintained as \(j\) grows. Take the better of the array and its reverse.

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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]))

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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;
    }
}

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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);
    }
};

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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
}

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