2934. Minimum Operations to Maximize Last Elements in Arrays

You are given two 0-indexed integer arrays, nums1 and nums2, both having length n.

You are allowed to perform a series of operations (possibly none).

In an operation, you select an index i in the range [0, n - 1] and swap the values of nums1[i] and nums2[i].

Your task is to find the minimum number of operations required to satisfy the following conditions:

• nums1[n - 1] is equal to the maximum value among all elements of nums1, i.e., nums1[n - 1] = max(nums1[0], nums1[1], ..., nums1[n - 1]).
• nums2[n - 1] is equal to the maximum value among all elements of nums2, i.e., nums2[n - 1] = max(nums2[0], nums2[1], ..., nums2[n - 1]).

Return an integer denoting the minimum number of operations needed to meet both conditions, or -1 if it is impossible to satisfy both conditions.

class Solution {
    int helper(vector<int>& A, vector<int>& B) {
        int n = A.size(), ma1 = A.back(), ma2 = B.back();
        int res = 0;
        for(int i = 0; i < n-1; i++) {
            if(A[i] <= ma1 and B[i] <= ma2) continue;
            if(A[i] <= ma2 and B[i] <= ma1) res += 1;
            else return 3 * n;
        }
        return res;
    }
public:
    int minOperations(vector<int>& nums1, vector<int>& nums2) {
        int res = helper(nums1, nums2), n = nums1.size();
        swap(nums1[n-1], nums2[n-1]);
        res = min(res, 1 + helper(nums1,nums2));
        if(res > 2 * n) return -1;
        return res;
    }
};