Problem Description

In this problem, we're given an array of integers called nums. Our task is to make the array into a zigzag array with the least number of moves. A zigzag array is defined in one of the following two ways:

1. Even-index Zigzag: For every even index i, the element at i is greater than its adjacent elements (nums[i] > nums[i - 1] and nums[i] > nums[i + 1]), like A[0] > A[1] < A[2] > A[3] < A[4] > ....

2. Odd-index Zigzag: For every odd index i, the element at i is greater than its adjacent elements (nums[i] > nums[i - 1] and nums[i] > nums[i + 1]), like A[0] < A[1] > A[2] < A[3] > A[4] < ....

A move consists of choosing any element of the array and decreasing its value by 1. The goal is to find out the minimum number of such moves required to achieve a zigzag pattern.

Intuition

To solve this problem, we can approach it by considering the two possible patterns of zigzag separately (even-index zigzag and odd-index zigzag), then choose the one which requires the fewest moves.

1. We can iterate through every even index i and check if nums[i] is already greater than its adjacent elements. If not, we calculate the difference by how much nums[i] needs to be decreased to satisfy the zigzag condition. We add this difference to our total moves for even-index zigzag pattern.

2. Similarly, we iterate through every odd index i doing the same check and calculation for the moves required to transform array nums into an odd-index zigzag pattern.

An important point to note is that when checking for the zigzag condition, we only need to ensure that each chosen element (nums[i] for even or odd i) is greater than its adjacent elements if they exist. This implies that for the first and last elements, we only consider one adjacent element.

After calculating moves for both patterns, we return the minimum of the two totals as the answer. This way, we ensure that we find the zigzag array that requires the least amount of modification (moves).

Learn more about Greedy patterns.

Solution Approach

The solution approach for converting an array nums into a zigzag array is as follows:

• We utilize a list ans with two elements initialized to 0. These two elements represent the minimum number of moves required for turning nums into an even-index zigzag array (first element of ans) and an odd-index zigzag array (second element of ans), respectively.

• We iterate through the array twice, first starting at index 0 for even-index zigzag array, and then starting at index 1 for odd-index zigzag array. This aligns with the two possible zigzag patterns.

• For each iteration (for even and odd patterns), we loop through the appropriate indices (i for even and i+1 for odd) with step 2, to only look at the elements that should be larger than their neighbors.

• At each selected index j, we calculate the amount to decrease the current element nums[j] to make it lower than both of its neighbors if they exist. This is done with two if conditions:

  • If j is not the first element (j > 0), we calculate the required decrease to ensure nums[j] is greater than nums[j - 1], which is nums[j] - nums[j - 1] + 1.

  • If j is not the last element (j < n - 1), we calculate the required decrease to ensure nums[j] is greater than nums[j + 1], which is nums[j] - nums[j + 1] + 1.

• The amount d is the maximum of the decreases required for both sides (left and right) of the current element nums[j]. Only one side is considered if j is at the boundary of the array.

• We accumulate the total decreases in ans[i] for each iteration pattern (0 for even-indexed and 1 for odd-indexed zigzag).

• After the loops, ans contains two numbers: the total moves to make the nums array a zigzag array by promoting even-indexed elements and the total moves to make it a zigzag array by promoting odd-indexed elements.

• The final result is the minimum of the values in ans, which represents the minimum number of moves to convert nums into a zigzag array, regardless of whether we are following the even-indexed or odd-indexed pattern.

This approach efficiently computes the desired outcome using array manipulation and condition checking without the need for complex data structures or advanced algorithms. The simplicity of this solution lies in the clever use of iteration, conditionals, and element comparison to meet the zigzag pattern requirements with a minimal number of moves.

Example Walkthrough

Let's consider a small example array nums:

1nums = [4, 3, 7, 6, 2]

We want to convert this array into either an even-index zigzag or an odd-index zigzag array in the least number of moves possible, following the solution approach described earlier.

Transforming into Even-Index Zigzag

Starting with index 0 (even), the selected indices will be 0, 2, 4, hence we check and modify only these indices.

• At index 0, nums[0] is 4. Since there is no element at -1, we only need to compare it to nums[1]. No moves are required here because 4 > 3.
• At index 2, nums[2] is 7. It is already greater than both its neighbors nums[1] and nums[3]. No moves required.
• At index 4, nums[4] is 2. Since there is no element at index 5, we only compare with nums[3]. We need to decrease nums[3] (6) to at least 1 for it to be smaller than nums[4]. Therefore, 6 - 2 + 1 is 5 moves required.

Total moves required for the even-index zigzag is 0 + 0 + 5 = 5.

Transforming into Odd-Index Zigzag

Now, starting with index 1 (odd), the selected indices to check will be 1, 3.

• At index 1, nums[1] is 3. We need to reduce it to 3 to less than nums[0] (4). Hence no moves required as 4 > 3.
• At index 3, nums[3] is 6. It should be greater than nums[2] (7) but it's not. So, we need to decrease nums[2] to 5 (nums[3] - nums[2] + 1), which is 6 - 7 + 1 = 0 moves. Additionally, nums[3] should also be greater than nums[4] (2), but it already is, so no further moves required.

Total moves required for the odd-index zigzag is 0 + 0 = 0.

Conclusion

Comparing both patterns, odd-index zigzag requires 0 moves and even-index zigzag requires 5 moves. Hence, the minimum number of moves required to make the given nums a zigzag array is 0.

Using the solution approach, we can thus determine that the original array is already an odd-index zigzag array and no moves are needed.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(n), where n is the length of the input list nums. This is because the code contains two nested loops. The outer loop runs twice (once for each of the two possible zigzag configurations), and the inner loop iterates over every other element of the list. Since at most, each element is considered once per outer loop iteration, the total number of operations is proportional to the size of the input list, resulting in a linear time complexity.

Space Complexity

The space complexity is O(1), meaning it requires constant additional space regardless of the input size. The array ans stores only two elements corresponding to the two possible zigzag configurations, making its size fixed. The variables d and n also do not depend on the input size, so the memory usage does not grow with the size of nums.

Learn more about how to find time and space complexity quickly using problem constraints.