Create minimum-operations-to-maximize-last-elements-in-arrays.cpp

Author: kamyu104

C++/minimum-operations-to-maximize-last-elements-in-arrays.cpp

@@ -0,0 +1,50 @@
+// Time:  O(n)
+// Space: O(1)
+
+// simulation
+class Solution {
+public:
+    int minOperations(vector<int>& nums1, vector<int>& nums2) {
+        int cnt1 = 0, cnt2 = 0;
+        for (int i = 0; i < size(nums1); ++i) {
+            if (!(min(nums1[i], nums2[i]) <= min(nums1.back(), nums2.back()) &&
+                  max(nums1[i], nums2[i]) <= max(nums1.back(), nums2.back()))) {
+                return -1;
+            }
+            if (!(nums1[i] <= nums1.back() && nums2[i] <= nums2.back())) {
+                ++cnt1;
+            }
+            if (!(nums1[i] <= nums2.back() && nums2[i] <= nums1.back())) {
+                ++cnt2;
+            }
+        }
+        return min(cnt1, cnt2);
+    }
+};
+
+// Time:  O(n)
+// Space: O(1)
+// simulation
+class Solution2 {
+public:
+    int minOperations(vector<int>& nums1, vector<int>& nums2) {
+        static const int INF = numeric_limits<int>::max();
+
+        const auto& count = [&](int mx1, int mx2) {
+            int result = 0;
+            for (int i = 0; i < size(nums1); ++i) {
+                if (nums1[i] <= mx1 && nums2[i] <= mx2) {
+                    continue;
+                }
+                if (!(nums2[i] <= mx1 && nums1[i] <= mx2)) {
+                    return INF;
+                }
+                ++result;
+            }
+            return result;
+        };
+
+        const int result = min(count(nums1.back(), nums2.back()), count(nums2.back(), nums1.back()));
+        return result != INF ? result : -1;
+    }
+};