Solution to problem 390 of leetcode

Leetcode/Elimination-game.cpp

@@ -0,0 +1,50 @@
+/*
+Time Complexity : O(log N)
+Space Complexity : O(log N)
+*/
+
+//Author : Deepak Mehrotra (StealthX2k20)
+
+class Solution {
+public:
+
+/*
+  if direction is 0 that means we are going from left to right
+  otherwise we are going from right to left
+*/
+
+int find_ans(int first, int last, int direction, int diff)
+{
+    if(first == last) return first; // we have a single element remaining
+
+    if(first + diff == last) // we have 2 elements remaining
+    {
+        return (direction == 0 ? last : first); // return the element which will not get removed
+    }
+
+    int len = 1 + ((last - first) / diff); // find length of remaining segment
+
+    if(len & 1) // check if len is odd
+    {
+        return find_ans(first + diff, last - diff, direction ^ 1, diff * 2);
+    }
+
+    else // len is even
+    {
+        if(!direction) // if direction is 0 then we are going from left to right -> -> ->
+        {
+            return find_ans(first + diff, last, direction ^ 1, diff * 2);
+        }
+
+        else // if direction is 1 then we are going from right to left <- <- <-
+        {
+            return find_ans(first, last - diff, direction ^ 1, diff * 2);
+        }
+    }
+}
+
+    int lastRemaining(int n)
+    {
+        return find_ans(1, n, 0, 1); /* initially we have segment from 1, 2, 3 .. n and we have to go from left to right and our initial difference is 1                                      */
+    }
+};

Leetcode/Leetcode Problem 1465.cpp

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+//Problem link : https://leetcode.com/problems/maximum-area-of-a-piece-of-cake-after-horizontal-and-vertical-cuts/
+
+class Solution {
+public:
+    int maxArea(int h, int w, vector<int>& horizontalCuts, vector<int>& verticalCuts) {
+       sort(horizontalCuts.begin(), horizontalCuts.end());
+       sort(verticalCuts.begin(), verticalCuts.end());
+
+       int mxh = -1, mxv = -1;
+       int prevh = 0, prevv = 0; 
+       for(int i = 0; i < horizontalCuts.size(); i++)
+       {
+           mxh = max(mxh, horizontalCuts[i] - prevh);
+           prevh = horizontalCuts[i];
+       }
+
+       mxh = max(mxh, h - prevh);
+
+       for(int i = 0; i < verticalCuts.size(); i++)
+       {
+           mxv = max(mxv, verticalCuts[i] - prevv);
+           prevv = verticalCuts[i];
+       }
+
+       mxv = max(mxv, w - prevv);
+
+       // mxv++, mxh++;
+
+       long long res, mod = 1000000007;
+
+       res = (((mxv * 1ll) % mod) * ((mxh * 1ll) % mod)) % mod;
+
+       return res;
+    }
+};