Leetcode 2439: Minimize Maximum of Array
You are given a 0-indexed array
nums of length n, where each element represents the number of items in a container. In one operation, you can pick an index i such that 1 <= i < n and nums[i] > 0, and move one item from nums[i] to nums[i-1]. Your goal is to minimize the maximum value in the array nums after performing any number of operations.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums` where `n` is the length of the array and `nums[i]` is the value at index `i`.
Example: nums = [5, 10, 3, 8]
Constraints:
• 2 <= n <= 10^5
• 0 <= nums[i] <= 10^9
Output: Return the minimum possible value of the maximum integer in `nums` after performing any number of operations.
Example: Output: 6
Constraints:
• The final result must be an integer.
Goal: The goal is to minimize the maximum value in the `nums` array after redistributing the elements. This can be done by calculating the sum of the array and spreading the items evenly.
Steps:
• 1. Calculate the sum of the entire array `nums`.
• 2. For each index `i`, calculate the maximum possible value that can be achieved in the array after redistributing the items.
• 3. Return the minimized maximum value.
Goal: The solution should handle large inputs and work efficiently for arrays of size up to 100,000.
Steps:
• The input array `nums` will always contain at least two elements.
• The values of the array elements can be very large, up to 10^9.
Assumptions:
• The input array will always contain at least two elements.
• You are allowed to perform any number of operations to achieve the optimal result.
• Input: nums = [3, 7, 1, 6]
• Explanation: In this example, we can perform operations to redistribute the values in `nums`. Here’s one set of operations: 1. Move one item from `nums[1]` to `nums[0]`, resulting in [4, 6, 1, 6]; 2. Move one item from `nums[3]` to `nums[2]`, resulting in [4, 6, 2, 5]; 3. Move one item from `nums[1]` to `nums[0]`, resulting in [5, 5, 2, 5]; The maximum value in the array is now 5, and this is the minimal possible maximum value.
• Input: nums = [10, 1]
• Explanation: Here, the array is already optimal since 10 is the maximum value, and no operations are needed. The result is 10.
Approach: The key idea is to minimize the maximum value by redistributing items from higher-valued elements to lower-valued ones. This can be done by calculating the cumulative sum and determining the point at which the values start to stabilize.
Observations:
• To minimize the maximum value, it's helpful to consider redistributing the values across the array rather than focusing on individual operations.
• One approach is to calculate the cumulative sum while iterating through the array, and at each step, track the maximum value that could result from the redistribution of items.
Steps:
• 1. Calculate the sum of all elements in the array `nums`.
• 2. For each index `i`, calculate the potential maximum value of the array after redistributing the values.
• 3. Return the minimized maximum value after processing all elements.
Empty Inputs:
• There will always be at least two elements in `nums`, so empty input is not a concern.
Large Inputs:
• The solution must be able to handle up to 100,000 elements in the array, so time efficiency is important.
Special Values:
• If the array is already sorted with all values equal, no operations are needed.
Constraints:
• The array can have very large numbers, up to 10^9, so avoid overflow and ensure the solution handles such cases efficiently.
int minimizeArrayValue(vector<int>& nums) {
long sum = 0, res = 0;
for(int i = 0; i < nums.size(); i++) {
sum += nums[i];
res = max(res, (sum + i) / (i + 1));
}
return res;
}
1 : Function Definition
int minimizeArrayValue(vector<int>& nums) {
Defining the function `minimizeArrayValue`, which takes a vector `nums` and returns an integer. The function aims to minimize the maximum value in the array after adjusting the values.
2 : Variable Initialization
long sum = 0, res = 0;
Initializing two variables: `sum`, which stores the cumulative sum of the array, and `res`, which tracks the result—the minimum maximum value of the array.
3 : Loop
for(int i = 0; i < nums.size(); i++) {
A for loop that iterates over each element in the `nums` array. It updates the sum and checks for the maximum value after adjusting the array.
4 : Mathematical Operation
sum += nums[i];
Adding the current element of the array `nums[i]` to the `sum` variable.
5 : Mathematical Operation
res = max(res, (sum + i) / (i + 1));
Calculating the average of the first `i+1` elements (considering the sum up to index `i`) and updating `res` to store the maximum average encountered so far.
6 : Return Statement
return res;
Returning the result stored in `res`, which is the minimized maximum value of the array after adjustments.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) because we only need to iterate over the array once to calculate the sum and the optimal maximum value.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we are only using a constant amount of extra space, aside from the input array.
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