ninjaAndHisFriends.cpp created

Dynamic Programming/ninjaAndHisFriends.cpp

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+// Problem(3-D DP) : We are given an ‘N *M’ matrix.Every cell of the matrix has some chocolates on it, mat[i][j] gives us the
+// number of chocolates.We have two friends ‘Alice’ and ‘Bob’.initially, Alice is standing on the cell(0, 0) and Bob is standing
+// on the cell(0, M - 1).Both of them can move only to the cells below them in these three directions : to the bottom cell(↓),
+// to the bottom - right cell(↘), or to the bottom - left cell(↙). When Alica and Bob visit a cell, they take all the chocolates
+// from that cell with them. It can happen that they visit the same cell, in that case, the chocolates need to be considered only
+// once. They cannot go out of the boundary of the given matrix, we need to return the maximum number of chocolates that Bob and
+// Alice can together collect.
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int f(int i, int j1, int j2, vector<vector<int>> &grid, int n, int m, vector<vector<vector<int>>> &dp)
+{
+    if (i == n)
+    {
+        return 0;
+    }
+    if (dp[i][j1][j2] != -1)
+    {
+        return dp[i][j1][j2];
+    }
+    int mx = INT_MIN;
+    // if Alice can move to the left column in the row below
+    if (((j1 - 1) >= 0))
+    {
+        // if Bob can move to the same column in the row below
+        mx = max(mx, f(i + 1, j1 - 1, j2, grid, n, m, dp));
+        // if Bob can move to the left column in the row below
+        if ((j2 - 1) >= 0)
+        {
+            mx = max(mx, f(i + 1, j1 - 1, j2 - 1, grid, n, m, dp));
+        }
+        // if Bob can move to the right column in the row below
+        if ((j2 + 1) < m)
+        {
+            mx = max(mx, f(i + 1, j1 - 1, j2 + 1, grid, n, m, dp));
+        }
+    }
+    // if Alice can move to the same column in the row below
+    // if Bob can move to the same column in the row below
+    mx = max(mx, f(i + 1, j1, j2, grid, n, m, dp));
+    // if Bob can move to the left column in the row below
+    if ((j2 - 1) >= 0)
+    {
+        mx = max(mx, f(i + 1, j1, j2 - 1, grid, n, m, dp));
+    }
+    // if Bob can move to the right column in the row below
+    if ((j2 + 1) < m)
+    {
+        mx = max(mx, f(i + 1, j1, j2 + 1, grid, n, m, dp));
+    }
+    // if Alice can move to the right column in the row below
+    if (((j1 + 1) < m))
+    {
+        // if Bob can move to the same column in the row below
+        mx = max(mx, f(i + 1, j1 + 1, j2, grid, n, m, dp));
+        // if Bob can move to the left column in the row below
+        if ((j2 - 1) >= 0)
+        {
+            mx = max(mx, f(i + 1, j1 + 1, j2 - 1, grid, n, m, dp));
+        }
+        // if Bob can move to the right column in the row below
+        if ((j2 + 1) < m)
+        {
+            mx = max(mx, f(i + 1, j1 + 1, j2 + 1, grid, n, m, dp));
+        }
+    }
+    if (j1 == j2)
+    {
+        // if same cell for both alice and bob, count chocolates only once
+        return dp[i][j1][j2] = (mx + grid[i][j1]);
+    }
+    else
+    {
+        return dp[i][j1][j2] = (mx + grid[i][j1] + grid[i][j2]);
+    }
+}
+
+int maximumChocolates(int r, int c, vector<vector<int>> &grid)
+{
+    vector<vector<vector<int>>> dp(r, vector<vector<int>>(c, vector<int>(c)));
+    for (int i = 0; i < r; i++)
+    {
+        for (int j = 0; j < c; j++)
+        {
+            for (int k = 0; k < c; k++)
+            {
+                dp[i][j][k] = -1;
+            }
+        }
+    }
+    return f(0, 0, c - 1, grid, r, c, dp);
+}
+
+int main()
+{
+    int r, c;
+    cin >> r >> c;
+    vector<vector<int>> grid(r, vector<int>(c));
+    for (int i = 0; i < r; i++)
+    {
+        for (int j = 0; j < c; j++)
+        {
+            cin >> grid[i][j];
+        }
+    }
+    cout << maximumChocolates(r, c, grid);
+}
+
+// Sample Inputs
+
+// 3 4
+// 2 3 1 2
+// 3 4 2 2
+// 5 6 3 5
+
+// 2 2
+// 1 1
+// 1 2
+
+// Corresponding Outputs
+
+// 21
+
+// 5