Minimum Equal Sum of Two Arrays After Replacing Zeros
<p>You are given two arrays <code>nums1</code> and <code>nums2</code> consisting of positive integers.</p>

<p>You have to replace <strong>all</strong> the <code>0</code>&#39;s in both arrays with <strong>strictly</strong> positive integers such that the sum of elements of both arrays becomes <strong>equal</strong>.</p>

<p>Return <em>the <strong>minimum</strong> equal sum you can obtain, or </em><code>-1</code><em> if it is impossible</em>.</p>

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

<pre>
<strong>Input:</strong> nums1 = [3,2,0,1,0], nums2 = [6,5,0]
<strong>Output:</strong> 12
<strong>Explanation:</strong> We can replace 0&#39;s in the following way:
- Replace the two 0&#39;s in nums1 with the values 2 and 4. The resulting array is nums1 = [3,2,2,1,4].
- Replace the 0 in nums2 with the value 1. The resulting array is nums2 = [6,5,1].
Both arrays have an equal sum of 12. It can be shown that it is the minimum sum we can obtain.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [2,0,2,0], nums2 = [1,4]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is impossible to make the sum of both arrays equal.
</pre>

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

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Greedy; Array
function minSum(nums1: number[], nums2: number[]): number {
    let [s1, s2] = [0, 0];
    let hasZero = false;
    for (const x of nums1) {
        if (x === 0) {
            hasZero = true;
        }
        s1 += Math.max(x, 1);
    }
    for (const x of nums2) {
        s2 += Math.max(x, 1);
    }
    if (s1 > s2) {
        return minSum(nums2, nums1);
    }
    if (s1 === s2) {
        return s1;
    }
    return hasZero ? s2 : -1;
}
