Minimize Maximum of Array
<p>You are given a <strong>0-indexed</strong> array <code>nums</code> comprising of <code>n</code> non-negative integers.</p>

<p>In one operation, you must:</p>

<ul>
	<li>Choose an integer <code>i</code> such that <code>1 &lt;= i &lt; n</code> and <code>nums[i] &gt; 0</code>.</li>
	<li>Decrease <code>nums[i]</code> by 1.</li>
	<li>Increase <code>nums[i - 1]</code> by 1.</li>
</ul>

<p>Return<em> the <strong>minimum</strong> possible value of the <strong>maximum</strong> integer of </em><code>nums</code><em> after performing <strong>any</strong> number of operations</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,7,1,6]
<strong>Output:</strong> 5
<strong>Explanation:</strong>
One set of optimal operations is as follows:
1. Choose i = 1, and nums becomes [4,6,1,6].
2. Choose i = 3, and nums becomes [4,6,2,5].
3. Choose i = 1, and nums becomes [5,5,2,5].
The maximum integer of nums is 5. It can be shown that the maximum number cannot be less than 5.
Therefore, we return 5.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,1]
<strong>Output:</strong> 10
<strong>Explanation:</strong>
It is optimal to leave nums as is, and since 10 is the maximum value, we return 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Greedy; Array; Binary Search; Dynamic Programming; Prefix Sum
class Solution {
public:
    int minimizeArrayValue(vector<int>& nums) {
        int left = 0, right = *max_element(nums.begin(), nums.end());
        auto check = [&](int mx) {
            long d = 0;
            for (int i = nums.size() - 1; i; --i) {
                d = max(0l, d + nums[i] - mx);
            }
            return nums[0] + d <= mx;
        };
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(mid))
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }
};