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2263. Make Array Non-decreasing or Non-increasing

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

  • 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);
        }
    };
    
  • 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]))
    
    
  • 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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