//this is the exceptional case 
  if(stones[1]!=1) return false;


  int n=stones.size();

  vector<vector<bool>> dp(n,{false});
  //in the process of recursion there may be several instances where
  // we will have same combination of lastIndex and currentIndex
  // we will use dp to check whether the same combination of lastIndex and currentIndex has occured or not
 // this will minimize repeated further computation

  return func(stones,0,1,dp);


}

bool func(vector<int>& stones, int lastIndex, int currentIndex, vector<vector<bool>> &dp)
{
  //Reached the last index => we have made it through 
  //hence return true
  if(currentIndex== stones.size()-1)
    return true;
 
  if(dp[lastIndex][currentIndex]) return false;


  int lastJump = stones[currentIndex] - stones[lastIndex];

  int nextIndex = currentIndex + 1;

  //at max we are allowed to move through only the one step greater than the previous step
  //this is what the second step is
  while(nextIndex < stones.size() && stones[nextIndex] <= stones[currentIndex] + lastJump + 1)
  {
      int nextJump= stones[nextIndex] - stones[currentIndex];
      int jump = nextJump - lastJump;

      if(jump>=-1 && jump<=1)
      {  
       if( func(stones, currentIndex, nextIndex, dp))
          return true;
      }
      
      nextIndex ++;


  }

  //we have checked for given lastIndex and currentIndex So
  //tick for this given combination 
  dp[lastIndex][currentIndex]= true;

  return false;
  
}