All Divisions With the Highest Score of a Binary Array
<p>You are given a <strong>0-indexed</strong> binary array <code>nums</code> of length <code>n</code>. <code>nums</code> can be divided at index <code>i</code> (where <code>0 &lt;= i &lt;= n)</code> into two arrays (possibly empty) <code>nums<sub>left</sub></code> and <code>nums<sub>right</sub></code>:</p>

<ul>
	<li><code>nums<sub>left</sub></code> has all the elements of <code>nums</code> between index <code>0</code> and <code>i - 1</code> <strong>(inclusive)</strong>, while <code>nums<sub>right</sub></code> has all the elements of nums between index <code>i</code> and <code>n - 1</code> <strong>(inclusive)</strong>.</li>
	<li>If <code>i == 0</code>, <code>nums<sub>left</sub></code> is <strong>empty</strong>, while <code>nums<sub>right</sub></code> has all the elements of <code>nums</code>.</li>
	<li>If <code>i == n</code>, <code>nums<sub>left</sub></code> has all the elements of nums, while <code>nums<sub>right</sub></code> is <strong>empty</strong>.</li>
</ul>

<p>The <strong>division score</strong> of an index <code>i</code> is the <strong>sum</strong> of the number of <code>0</code>&#39;s in <code>nums<sub>left</sub></code> and the number of <code>1</code>&#39;s in <code>nums<sub>right</sub></code>.</p>

<p>Return <em><strong>all distinct indices</strong> that have the <strong>highest</strong> possible <strong>division score</strong></em>. You may return the answer in <strong>any order</strong>.</p>

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

<pre>
<strong>Input:</strong> nums = [0,0,1,0]
<strong>Output:</strong> [2,4]
<strong>Explanation:</strong> Division at index
- 0: nums<sub>left</sub> is []. nums<sub>right</sub> is [0,0,<u><strong>1</strong></u>,0]. The score is 0 + 1 = 1.
- 1: nums<sub>left</sub> is [<u><strong>0</strong></u>]. nums<sub>right</sub> is [0,<u><strong>1</strong></u>,0]. The score is 1 + 1 = 2.
- 2: nums<sub>left</sub> is [<u><strong>0</strong></u>,<u><strong>0</strong></u>]. nums<sub>right</sub> is [<u><strong>1</strong></u>,0]. The score is 2 + 1 = 3.
- 3: nums<sub>left</sub> is [<u><strong>0</strong></u>,<u><strong>0</strong></u>,1]. nums<sub>right</sub> is [0]. The score is 2 + 0 = 2.
- 4: nums<sub>left</sub> is [<u><strong>0</strong></u>,<u><strong>0</strong></u>,1,<u><strong>0</strong></u>]. nums<sub>right</sub> is []. The score is 3 + 0 = 3.
Indices 2 and 4 both have the highest possible division score 3.
Note the answer [4,2] would also be accepted.</pre>

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

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [3]
<strong>Explanation:</strong> Division at index
- 0: nums<sub>left</sub> is []. nums<sub>right</sub> is [0,0,0]. The score is 0 + 0 = 0.
- 1: nums<sub>left</sub> is [<u><strong>0</strong></u>]. nums<sub>right</sub> is [0,0]. The score is 1 + 0 = 1.
- 2: nums<sub>left</sub> is [<u><strong>0</strong></u>,<u><strong>0</strong></u>]. nums<sub>right</sub> is [0]. The score is 2 + 0 = 2.
- 3: nums<sub>left</sub> is [<u><strong>0</strong></u>,<u><strong>0</strong></u>,<u><strong>0</strong></u>]. nums<sub>right</sub> is []. The score is 3 + 0 = 3.
Only index 3 has the highest possible division score 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1]
<strong>Output:</strong> [0]
<strong>Explanation:</strong> Division at index
- 0: nums<sub>left</sub> is []. nums<sub>right</sub> is [<u><strong>1</strong></u>,<u><strong>1</strong></u>]. The score is 0 + 2 = 2.
- 1: nums<sub>left</sub> is [1]. nums<sub>right</sub> is [<u><strong>1</strong></u>]. The score is 0 + 1 = 1.
- 2: nums<sub>left</sub> is [1,1]. nums<sub>right</sub> is []. The score is 0 + 0 = 0.
Only index 0 has the highest possible division score 2.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Array
impl Solution {
    pub fn max_score_indices(nums: Vec<i32>) -> Vec<i32> {
        let mut l0 = 0;
        let mut r1: i32 = nums.iter().sum();
        let mut mx = r1;
        let mut ans = vec![0];

        for i in 1..=nums.len() {
            let x = nums[i - 1];
            l0 += x ^ 1;
            r1 -= x;
            let t = l0 + r1;
            if mx == t {
                ans.push(i as i32);
            } else if mx < t {
                mx = t;
                ans = vec![i as i32];
            }
        }

        ans
    }
}
