New Problem Solution - "1824. Minimum Sideway Jumps"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1824|[Minimum Sideway Jumps](https://leetcode.com/problems/minimum-sideway-jumps/) | [C++](./algorithms/cpp/minimumSidewayJumps/MinimumSidewayJumps.cpp)|Medium|
 |1823|[Find the Winner of the Circular Game](https://leetcode.com/problems/find-the-winner-of-the-circular-game/) | [C++](./algorithms/cpp/findTheWinnerOfTheCircularGame/FindTheWinnerOfTheCircularGame.cpp)|Medium|
 |1822|[Sign of the Product of an Array](https://leetcode.com/problems/sign-of-the-product-of-an-array/) | [C++](./algorithms/cpp/signOfTheProductOfAnArray/SignOfTheProductOfAnArray.cpp)|Easy|
 |1819|[Number of Different Subsequences GCDs](https://leetcode.com/problems/number-of-different-subsequences-gcds/) | [C++](./algorithms/cpp/numberOfDifferentSubsequencesGcds/NumberOfDifferentSubsequencesGcds.cpp)|Hard|

algorithms/cpp/minimumSidewayJumps/MinimumSidewayJumps.cpp

@@ -0,0 +1,101 @@
+// Source : https://leetcode.com/problems/minimum-sideway-jumps/
+// Author : Hao Chen
+// Date   : 2021-04-11
+
+/***************************************************************************************************** 
+ *
+ * There is a 3 lane road of length n that consists of n + 1 points labeled from 0 to n. A frog starts 
+ * at point 0 in the second lane and wants to jump to point n. However, there could be obstacles along 
+ * the way.
+ * 
+ * You are given an array obstacles of length n + 1 where each obstacles[i] (ranging from 0 to 3) 
+ * describes an obstacle on the lane obstacles[i] at point i. If obstacles[i] == 0, there are no 
+ * obstacles at point i. There will be at most one obstacle in the 3 lanes at each point.
+ * 
+ * 	For example, if obstacles[2] == 1, then there is an obstacle on lane 1 at point 2.
+ * 
+ * The frog can only travel from point i to point i + 1 on the same lane if there is not an obstacle 
+ * on the lane at point i + 1. To avoid obstacles, the frog can also perform a side jump to jump to 
+ * another lane (even if they are not adjacent) at the same point if there is no obstacle on the new 
+ * lane.
+ * 
+ * 	For example, the frog can jump from lane 3 at point 3 to lane 1 at point 3.
+ * 
+ * Return the minimum number of side jumps the frog needs to reach any lane at point n starting from 
+ * lane 2 at point 0.
+ * 
+ * Note: There will be no obstacles on points 0 and n.
+ * 
+ * Example 1:
+ * 
+ * Input: obstacles = [0,1,2,3,0]
+ * Output: 2 
+ * Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps (red arrows).
+ * Note that the frog can jump over obstacles only when making side jumps (as shown at point 2).
+ * 
+ * Example 2:
+ * 
+ * Input: obstacles = [0,1,1,3,3,0]
+ * Output: 0
+ * Explanation: There are no obstacles on lane 2. No side jumps are required.
+ * 
+ * Example 3:
+ * 
+ * Input: obstacles = [0,2,1,0,3,0]
+ * Output: 2
+ * Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps.
+ * 
+ * Constraints:
+ * 
+ * 	obstacles.length == n + 1
+ * 	1 <= n <= 5 * 10^5
+ * 	0 <= obstacles[i] <= 3
+ * 	obstacles[0] == obstacles[n] == 0
+ ******************************************************************************************************/
+
+class Solution {
+private:
+    int min (int x, int y) {
+        return x < y ? x : y;
+    }
+    int min(int x, int y, int z) {
+        return min(x, min(y,z));
+    }
+    void print(vector<vector<int>>& matrix) {
+        int n = matrix.size();
+        int m = matrix[0].size();
+
+        for(int i=0; i<m; i++) {
+            for (int j=0; j<n-1; j++){
+                if (matrix[j][i] == n) {
+                    cout << setw(2) << "X"<<",";
+                } else {
+                    cout << setw(2) <<matrix[j][i] << ",";
+                }
+            }
+            cout << matrix[n-1][i] << endl;
+        }
+    }
+public:
+    int minSideJumps(vector<int>& obstacles) {
+        int n = obstacles.size();
+        vector<vector<int>> dp(n, vector(3,0));
+        dp[0][0] = dp[0][2] = 1;
+
+        for(int i = 1; i < n; i++){
+
+            dp[i][0] = dp[i-1][0];
+            dp[i][1] = dp[i-1][1];
+            dp[i][2] = dp[i-1][2];
+            if (obstacles[i] > 0 ) dp[i][obstacles[i]-1] = n;
+
+            if (obstacles[i]-1 != 0 ) dp[i][0] = min(dp[i-1][0], dp[i][1]+1, dp[i][2]+1);     
+            if (obstacles[i]-1 != 1 ) dp[i][1] = min(dp[i][0]+1, dp[i-1][1], dp[i][2]+1);
+            if (obstacles[i]-1 != 2 ) dp[i][2] = min(dp[i][0]+1, dp[i][1]+1, dp[i-1][2]);
+
+        }
+        //print(dp);
+        //cout << endl;
+        return min(dp[n-1][0], dp[n-1][1], dp[n-1][2]);
+    }
+};