Added frog-1 problem code

dynamic_programming/frog1.cpp

@@ -0,0 +1,37 @@
+// There are N stones, numbered 1,2,…,N. For each i (1≤i≤N), the height of Stone i is
+// h[i]
+// There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
+// If the frog is currently on Stone i, jump to Stone i+1 or Stone i+2. Here, a cost of  
+// ∣h(i) −h(j) ∣ is incurred, where j is the stone to land on.
+// Find the minimum possible total cost incurred before the frog reaches Stone N.
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int jump(int n , vector<int>& height , vector<int>& dp)
+{
+  if (n <= 0)
+    return 0;
+  if (dp[n] != -1)
+    return dp[n];
+
+  int op1 = abs(height[n] - height[n - 1]) + jump(n - 1 , height , dp);//jump form i-1 stone ;
+  int op2 = INT_MAX;
+  if (n - 2 >= 0)
+    op2 = abs(height[n] - height[n - 2]) + jump(n - 2 , height , dp);//jump form i-2 stone;
+  return dp[n] = min(op1 , op2);
+}
+
+int main()
+{
+  int n;
+  cin >> n ;
+  vector<int> height(n);
+  for (int i = 0 ; i < n ; i++) {
+    cin >> height[i];
+  }
+  vector<int>dp(n , -1);
+  int ans = jump(n - 1 , height , dp);
+  cout << ans << endl;
+  return 0;
+}