0% completed
Problem Statement
Given a positive integer array nums
, return the minimum possible value
of the maximum integer
of nums after performing multiple operations.
In one operation, you must:
- Select any index
i
such that1 <= i < n
andnums[i] > 0
. - Increase
nums[i - 1]
by1
. - Decrease
nums[i]
by1
.
Examples
Example 1:
- Input: nums =
[4, 5, 3, 2, 1, 6]
- Expected Output:
5
- Justification: We can perform operations to decrease the last element (6) to 5 and increase the fifth element to 2. Here, the maximum array element is 5.
Example 2:
- Input: nums =
[1, 2, 3, 4]
- Expected Output:
3
- Justification: We can decrease the last element (4) to 3 and increase the third element to 4, then decrease the third element (now 4) to 3, increasing the second element to 3. The array becomes [1, 3, 3, 3], with the maximum being 3.
Example 3:
- Input: nums =
[8, 2, 2, 3]
- Expected Output:
8
- Justification: We can't perform any operation to minimize the maximum value of the array. So,
8
is the minimum possible value of the maximum integer.
Solution
To solve this problem, we adopt a strategy that leverages the cumulative sum of the array elements to find the minimum possible maximum value after performing the allowed operations. The core idea is to iteratively calculate the cumulative sum up to each index and use this sum to determine the maximum average value that can be achieved by redistributing the numbers.
This approach is grounded in the insight that the minimum possible maximum value, after any number of operations, is essentially the ceiling of the maximum average of the cumulative sums up to each point in the array. By calculating these averages and keeping track of the highest one, we can deduce the minimum maximum value achievable. This method is effective because it directly addresses the problem's goal without needing to explicitly perform the redistributions, thereby offering a more efficient solution.
Step-by-Step Algorithm
- Initialize a variable
sum
to 0 to keep track of the cumulative sum of the array elements. - Initialize another variable
ans
to 0 to store the maximum average found during the iteration. - Iterate through the array using a loop:
- For each element at index
i
, add it tosum
to update the cumulative sum. - Calculate the average up to the current index as
(sum + i) / (i + 1)
. This formula accounts for the zero-based indexing and ensures that the division is done considering the total number of elements counted so far. - Update
ans
with the maximum value between the currentans
and the calculated average. This step ensures thatans
always holds the highest average value found.
- For each element at index
- After completing the iteration,
ans
will hold the minimum possible maximum value of the array after the operations. Convertans
to an integer (if necessary) and return it as the solution.
Algorithm Walkthrough
Input: nums = [4, 5, 3, 2, 1, 6]
.
-
Initial State:
nums = [4, 5, 3, 2, 1, 6]
sum = 0
ans = 0
-
Iteration 1:
i = 0
,nums[i] = 4
sum = 4
average = (4 + 0) / (0 + 1) = 4
ans = max(0, 4) = 4
-
Iteration 2:
i = 1
,nums[i] = 5
sum = 4 + 5 = 9
average = (9 + 1) / (1 + 1) = 5
ans = max(4, 5) = 5
-
Iteration 3:
i = 2
,nums[i] = 3
sum = 9 + 3 = 12
average = (12 + 2) / (2 + 1) = 4.666...
ans = max(5, 4.666...) = 5
-
Iteration 4:
i = 3
,nums[i] = 2
sum = 12 + 2 = 14
average = (14 + 3) / (3 + 1) = 4.25
ans = max(5, 4.25) = 5
-
Iteration 5:
i = 4
,nums[i] = 1
sum = 14 + 1 = 15
average = (15 + 4) / (4 + 1) = 3.8
ans = max(5, 3.8) = 5
-
Iteration 6:
i = 5
,nums[i] = 6
sum = 15 + 6 = 21
average = (21 + 5) / (5 + 1) = 4.333...
ans = max(5, 4.333...) = 5
-
Final State:
ans = 5
Code
Complexity Analysis
- Time Complexity: The algorithm iterates through the array once. Therefore, the time complexity is O(n), where n is the length of the array.
- Space Complexity: The algorithm operates in-place, requiring no additional space proportional to the input size. Thus, the space complexity is O(1), as only a constant amount of extra space is used.
.....
.....
.....
Table of Contents
Contents are not accessible
Contents are not accessible
Contents are not accessible
Contents are not accessible
Contents are not accessible