Assign Cookies With Matching Parity

You are given two integer arrays g and s.
g[i] is the greed factor of child i. s[j] is the size of cookie j.
Assign at most one cookie to each child. Cookie j may be assigned to child i only when s[j] >= g[i] and s[j] and g[i] have the same parity: both even or both odd.
Return the maximum number of children who can be assigned a cookie.

Function
findContentChildren(g: int[], s: int[]) → int

Examples
Example 1
g = [1,2,3]
s = [1,2,3]
return = 3
Each child can take the cookie of equal size. The pairs 1, 2, and 3 all have matching parity.
Example 2
g = [1,3]
s = [2,4]
return = 0
Both children have odd greed, and both cookies are even, so no assignment is legal.

Constraints
1 <= g.length <= 3 * 10^4.
0 <= s.length <= 3 * 10^4.
1 <= g[i], s[j] <= 2^31 - 1.

The trick: a cookie can only go to a child with the same parity, so the two groups never interact. Split g and s into odd and even lists. Sort each list. For each group, run the classic two-pointer greedy: walk through cookies in ascending order, and if the current cookie is at least the current child's greed, assign it and advance the child pointer. Add the two counts. That's O(n log n) from sorting, which fits the 3 * 10^4 limit easily. The common pitfall is sorting everything together and checking parity on the fly, which wastes cookies on children they can't feed. Another miss is forgetting s can be empty. Values go up to 2^31 - 1, so watch overflow if you add anything in a typed language. If you freeze on the split, StealthCoder is the hedge during the live OA.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.