Problem Description

In this problem, you are given an array nums with n non-negative integers. The focus is on the array's integers' transformation through a series of operations to minimize the value of the maximum integer in the array. Each operation includes three steps:

1. Select an integer i where i is between 1 and n-1 (inclusive), and nums[i] is greater than 0.
2. Decrease nums[i] by 1.
3. Increase nums[i - 1] by 1.

The objective is to find out the smallest possible value of the largest number in the array after performing any number of these operations.

Intuition

The intuition behind the solution is to use binary search to find the minimum possible value of the maximum integer in the array after all the operations. Since we are asked for the minimum possible maximum value, we can guess that there is some maximum value that we cannot go below no matter how many operations we perform.

We perform binary search between the lowest possible maximum value, which is 0, and the current maximum value in the array. For each guess (possible maximum value) in the binary search, we check if we can achieve that maximum value after performing the operations. Here's the step-by-step reasoning:

1. Start with a binary search between 0 and the maximum value present in the array nums.
2. Calculate the middle value between the current range as our guess for the possible maximum value.
3. Define a check function that will determine if it is possible to achieve the guessed maximum value with the array.
4. In the check function, we work backwards from the end of the array towards the beginning (right to left). For each element, we calculate the surplus or deficit compared to our guessed maximum and carry this difference to the previous element.
5. The conditions that tell us if we can achieve the guessed maximum are:
   • If at any point the required decrease is so great we cannot make up the difference with the carried surplus, the check fails.
   • If we can carry the difference through to the first element, we then check if the total sum of increases needed for the first element does not exceed our guessed maximum.
6. If the check function returns true, it means we can achieve the guessed maximum or even lower, so we continue searching towards the lower half of the binary search range.
7. If the check function returns false, the guessed maximum is too low, and we have to adjust our range towards the higher values.
8. Repeat this process until the binary search narrows down to the smallest possible maximum integer value that satisfies the condition (the left and right pointers converge).

The solution uses the fact that after all possible optimizations, the array nums will become non-decreasing from the beginning to the end. This is because the maximum possible number in an optimal solution cannot be less than the average of the numbers in the array. Thus, binary search optimizes the process by eliminating the need to test each value sequentially.

Learn more about Greedy, Binary Search, Dynamic Programming and Prefix Sum patterns.

Solution Approach

The solution utilizes a binary search algorithm to narrow down the possible maximum value that can be achieved in the array after performing the operations. Here's an in-depth walkthrough of the algorithm, elaborating on the reference solution provided:

1. Binary Search Initialization: Define two pointers, left and right, which represent the range within which the final answer could be. Initialize left to 0, assuming the minimum possible value in the array could be 0 after all operations, and initialize right to the maximum value in nums, which is the upper limit of the answer.

2. Binary Search Loop: While left is less than right, calculate the middle of the current left and right boundary, mid = (left + right) >> 1. (>> 1 is a bitwise operation equivalent to dividing by 2, shifting the bits to the right once.)

3. Check Function: The check function is defined to test whether a certain maximum value mx can be attained after performing the allowed operations. It works as follows:

   • Initialize a d variable to 0, which represents the cumulative difference needed to reach the guessed maximum mx from the end of the array.
   • Iterate in reverse (from the last element to the second, hence nums[:0:-1]) to calculate d. At each step, update d by adding the difference between the current element x and mx only if that difference is positive. In other words, if the element x is larger than mx, we need to 'push' some of its value to its left neighbor.
   • Finally, add d to the first element nums[0] and check if it still does not exceed mx. If it does not, return True; otherwise, False.

4. Binary Search Logic: Use the result of the check function inside the binary search loop. If check(mid) returns True, the current guess can be achieved, and perhaps a lower maximum value is possible, so we update right to mid. Conversely, if check(mid) returns False, we must increase our guess and move the lower boundary up, setting left to mid + 1.

5. Binary Search Conclusion: The binary search concludes when left equals right, indicating that the search range has been narrowed down to a single element—this element is the minimum possible maximal value that can be achieved in the nums array after performing any number of operations.

This solution effectively combines binary search with a greedy strategy in the check function to ensure that all potential operations are taken into account from right to left, simulating the effect that the operations would have on an array and determining the viability of a selected maximum value. The use of binary search drastically reduces the number of checks needed compared to a linear search and finds the optimal solution efficiently.

Example Walkthrough

Assume we have an array nums = [3, 1, 5, 6, 8, 7] with n = 6 non-negative integers. We need to minimize the value of the maximum integer in the array after performing our operations.

1. Initialize our binary search range with left = 0 and right = max(nums) = 8.

2. Conduct a binary search. The first mid value will be (0 + 8) / 2 = 4.

3. Run the check function with mx = 4:

   • Start iterating from the end:
     • nums[5] = 7, d += 7 - 4 = 3 (since 7 is greater than 4)
     • nums[4] = 8, d += 8 - 4 = 7 (total d becomes 10)
     • nums[3] = 6, d += 6 - 4 = 12 (total d becomes 12)
     • nums[2] = 5, d += 5 - 4 = 13 (total d becomes 13)
     • nums[1] does not contribute since 1 is not greater than 4
     Now, adding the total d (13) to nums[0] which is 3, we have 3 + 13 = 16, which is greater than mx = 4, so our check function returns False.

4. Since the check function returned False, we can't attain a maximum with 4. We adjust our binary search range, setting left to mid + 1, which is 5.

5. With a new search range left = 5 and right = 8, we calculate a new mid value of (5 + 8) / 2 = 6 and repeat the check function. We keep iterating this process until we find the mid value for which the check function returns True.

6. When we finally find the smallest mid for which the check function returns True, we have found the smallest possible value of the largest number in the array after performing the operations. For this example, let's assume through the binary search process, we find that the smallest maximum we can achieve is 5.

7. The solution approach successfully minimizes the maximum value in the array with the optimal use of operations detailed in the given problem.

Solution Implementation

Time and Space Complexity

The given code implements a binary search algorithm to minimize the maximum value in the array after applying the specified operation. Here's the analysis of its time and space complexity:

Time Complexity:

The while loop in the code is running a binary search over the range 0 to max(nums), which has at most log(max(nums)) iterations since we're halving the search space in every step. The check function is called inside the loop and it iterates through the array in reverse, which takes O(n) time. Therefore, the time complexity of the code is O(n * log(max(nums))).

Space Complexity:

The space complexity of the code is O(1). The algorithm uses a constant number of variables (d, mx, left, right, mid) and the space occupied by these does not depend on the size of the input array nums. The function check also does not use any additional space that scales with the size of the input, as it works in place.

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