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 <= 10000 <= nums[i] <= 1000
Follow up: Can you solve it in O(n*log(n)) time complexity?
Solutions
Solution 1
PythonJavaC++Go
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]))(code-box)
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;
}
}(code-box)
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);
}
};(code-box)
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
}(code-box)
