Given an array nums of integers, a move consists of choosing any element and decreasing it by 1.
An array A is a zigzag array if either:
- Every even-indexed element is greater than adjacent elements, ie.
A[0] > A[1] < A[2] > A[3] < A[4] > ... - OR, every odd-indexed element is greater than adjacent elements, ie.
A[0] < A[1] > A[2] < A[3] > A[4] < ...
Return the minimum number of moves to transform the given array nums into a zigzag array.
Example 1:
Input: nums = [1,2,3] Output: 2 Explanation: We can decrease 2 to 0 or 3 to 1.
Example 2:
Input: nums = [9,6,1,6,2] Output: 4
Constraints:
1 <= nums.length <= 10001 <= nums[i] <= 1000
Solution: Greedy
One pass, making each element local minimum.
[9,6,1,6,2]
i = 0, [inf, 9, 6], 9 => 5, even cost 4
i = 1, [9, 6, 1], 6 => 0, odd cost 6
i = 2, [6, 1, 6], 1 => 1, even cost 0
i = 3, [1, 6, 2], 6 => 0, odd cost 12
i = 4, [6, 2, inf], 2 => 2, even cost 0
total even cost 4, new array => [5, 6, 1, 6, 2]
total odd cost 18, new array => [9, 0, 1, 0, 2]
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int movesToMakeZigzag(vector<int>& nums) { const int n = nums.size(); vector<int> moves(2); for (int i = 0; i < n; i++) { int l = i == 0 ? INT_MAX : nums[i - 1]; int r = i == n - 1 ? INT_MAX : nums[i + 1]; moves[i % 2] += max(0, nums[i] - min(l, r) + 1); } return min(moves[0], moves[1]); } }; |