// Time:  O(m * n)
// Space: O(min(m, n))

// dp
class Solution {
public:
    int minLargest(vector<int>& nums1, vector<int>& nums2) {
        static const int INF = numeric_limits<int>::max();

        if (size(nums1) < size(nums2)) {
            swap(nums1, nums2);
        }
        vector<int> dp(size(nums2) + 1, INF);
        dp[0] = 0;
        for (int i = 0; i <= size(nums1); ++i) {
            for (int j = 0; j <= size(nums2); ++j) {
                if (!i && !j) {
                    continue;
                }
                int curr = INF;
                if (i - 1 >= 0) {
                    if (dp[j] != INF) {
                        curr = min(curr, dp[j] + (dp[j] % 2 == nums1[i - 1] % 2 ? 2 : 1));
                    }
                }
                if (j - 1 >= 0) {
                    if (dp[j - 1] != INF) {
                        curr = min(curr, dp[j - 1] + (dp[j - 1] % 2 == nums2[j - 1] % 2 ? 2 : 1));
                    }
                }
                dp[j] = curr;
            }
        }
        return dp.back();
    }
};