Leetcode 1144: Decrease Elements To Make Array Zigzag
You are given an array nums of integers. A move consists of selecting any element and decreasing it by 1. A zigzag array is an array where either every even-indexed element is greater than its adjacent elements or every odd-indexed element is greater than its adjacent elements. Your task is to return the minimum number of moves required to transform the given array into a zigzag array.
Problem
Approach
Steps
Complexity
Input: The input consists of an array nums of integers.
Example: Input: nums = [3, 1, 4]
Constraints:
• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000
Output: Return the minimum number of moves required to transform the given array into a zigzag array.
Example: Output: 1
Constraints:
• The output will be a single integer representing the minimum number of moves.
Goal: The goal is to transform the array into a zigzag array by minimizing the number of moves.
Steps:
• 1. Iterate over the array and compare each element with its adjacent elements.
• 2. For each element, check if it violates the zigzag condition. If it does, calculate the number of moves to fix it.
• 3. Track the total number of moves for both the even-indexed and odd-indexed conditions.
• 4. Return the minimum number of moves required.
Goal: The algorithm must work efficiently for arrays with a length up to 1000.
Steps:
• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000
Assumptions:
• The array may contain both small and large numbers.
• The length of the array will be between 1 and 1000.
• Input: Input: nums = [3, 1, 4]
• Explanation: In this case, we can decrease 4 to 2, resulting in the zigzag array [3, 1, 2], and the number of moves is 1.
• Input: Input: nums = [6, 4, 2, 8, 10]
• Explanation: In this case, we decrease 6 to 4 and 8 to 6, resulting in the zigzag array [4, 4, 2, 6, 10], and the number of moves is 6.
Approach: We can solve this problem by calculating the minimum number of moves required to fix the zigzag condition for both the even-indexed and odd-indexed conditions. For each condition, we will check each element and adjust it if necessary.
Observations:
• The problem can be solved by iterating through the array and comparing each element with its neighbors.
• A greedy approach that minimizes the number of moves for each element will work well for this problem.
Steps:
• 1. Initialize two variables to track the number of moves for even-indexed and odd-indexed conditions.
• 2. Loop through the array and check for violations of the zigzag condition for both even and odd indexed elements.
• 3. For each violation, calculate how many moves are needed to fix it.
• 4. Return the minimum of the two calculated move counts.
Empty Inputs:
• If the array is empty, the result should be 0.
Large Inputs:
• The algorithm should be optimized to handle arrays of length 1000 efficiently.
Special Values:
• If all elements are the same, no moves are required.
Constraints:
• The solution should be efficient in terms of both time and space complexity.
int movesToMakeZigzag(vector<int>& nums) {
int n = nums.size(), left, right;
vector<int> res(2, 0);
for(int i = 0; i < n; i++) {
left = ( i > 0 ) ? nums[i - 1] : 1001;
right = ( i + 1 < n ) ? nums[i + 1] : 1001;
res[i % 2] += max(0, nums[i] - min(left, right) + 1);
}
return min(res[0], res[1]);
}
1 : Function Definition
int movesToMakeZigzag(vector<int>& nums) {
This defines the function `movesToMakeZigzag`, which takes a vector of integers `nums` and returns the minimum number of moves to convert the array into a zigzag pattern.
2 : Variable Declaration
int n = nums.size(), left, right;
Here, the variable `n` stores the size of the array `nums`, while `left` and `right` are used to store the adjacent elements for comparison when calculating the required changes for the zigzag pattern.
3 : Vector Initialization
vector<int> res(2, 0);
This initializes a vector `res` with two elements, both set to 0. The vector will be used to store the number of changes required to convert the array into a zigzag pattern for even and odd indexed elements.
4 : Loop Initialization
for(int i = 0; i < n; i++) {
This initiates a loop that iterates through each element of the array `nums`.
5 : Left Neighbor Calculation
left = ( i > 0 ) ? nums[i - 1] : 1001;
This calculates the value of the left neighbor of the current element. If `i` is greater than 0, it assigns `left` to the previous element (`nums[i - 1]`), otherwise, it assigns a large value (1001) to `left` as a default.
6 : Right Neighbor Calculation
right = ( i + 1 < n ) ? nums[i + 1] : 1001;
This calculates the value of the right neighbor of the current element. If `i + 1` is less than `n` (the size of the array), it assigns `right` to the next element (`nums[i + 1]`), otherwise, it assigns a large value (1001) to `right` as a default.
7 : Zigzag Move Calculation
res[i % 2] += max(0, nums[i] - min(left, right) + 1);
This line calculates the number of moves required to make the current element a part of a zigzag sequence. It updates the corresponding index in the `res` vector based on whether `i` is even or odd, and adds the minimum number of moves calculated between `left` and `right`.
8 : Final Calculation
return min(res[0], res[1]);
This returns the minimum value between the two possible results stored in `res[0]` and `res[1]`, representing the minimum number of moves required for the zigzag pattern starting with even and odd indexed elements.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n), where n is the length of the input array, as we only iterate through the array once.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) since we only need a constant amount of space to track the number of moves.
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