Add solution and approach of hard daily problem

Author: SartHak-0-Sach

042-11th_feb-Cherry_pickup_II/approach-explanation.txt

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+Intuition:
+1st try uses recursive DP solution which has large time complexity.
+
+2nd approach is a bottom-up design, an iterative DP version, converted from the 1st one with space optimization by using & tick
+
+Think how to reduce the TC.
+
+Approach:
+This is a 3D array named dp used for memoization to store intermediate results of the dynamic programming solution.
+
+int f(int r, int c0, int c1, vector<vector<int>>& grid): This is a recursive function f that calculates the maximum number of cherries that can be picked starting from cell (r, c0) and (r, c1).
+
+ans is initialized based on whether the two pickers are on the same or different cells.
+The nested loops iterate over the possible next positions for both pickers and calculate the maximum cherries that can be picked from the next row.
+
+ans += next;
+
+Complexity
+Time complexity:
+O(row×col2)
+
+Space complexity:
+O(70^3)→O(70^2)

042-11th_feb-Cherry_pickup_II/code.cpp

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+#include <iostream>
+#include <vector>
+using namespace std;
+#pragma GCC optimize("O3", "unroll-loops")
+class Solution
+{
+public:
+    int cherryPickup(vector<vector<int>> &grid)
+    {
+        int row = grid.size(), col = grid[0].size();
+        int dp[2][70][70] = {0};
+
+        // base case for the last row
+        for (int c0 = 0; c0 < col; c0++)
+            for (int c1 = 0; c1 < col; c1++)
+            {
+                dp[(row - 1) & 1][c0][c1] = (c0 != c1) ? grid[row - 1][c0] + grid[row - 1][c1] : grid[row - 1][c0];
+            }
+        // DP from r=row-2 to 0
+        for (int r = row - 2; r >= 0; r--)
+        {
+            for (int c0 = 0; c0 < col; c0++)
+                for (int c1 = 0; c1 < col; c1++)
+                {
+                    int cherry = (c0 != c1) ? grid[r][c0] + grid[r][c1] : grid[r][c0];
+                    int next = 0;
+
+                    for (int d0 : {c0 - 1, c0, c0 + 1})
+                        for (int d1 : {c1 - 1, c1, c1 + 1})
+                        {
+                            if (d0 < 0 || d0 >= col || d1 < 0 || d1 >= col)
+                                continue;
+                            next = max(next, dp[(r + 1) & 1][d0][d1]);
+                        }
+                    dp[r & 1][c0][c1] = next + cherry;
+                }
+        }
+
+        return dp[0][0][col - 1];
+    }
+};
+
+auto init = []()
+{
+    ios::sync_with_stdio(0);
+    cin.tie(0);
+    cout.tie(0);
+    return 'c';
+}();