input: Write a function to solve the 0-1 Knapsack problem using dynamic programming and return the maximum value that can be obtained with the given capacity of the knapsack. 
  def knapSack(W, wt, val, n): 
    
      # Base Case 
      if n == 0 or W == 0 : 
          return 0
    
      # If weight of the nth item is more 
      # than Knapsack capacity W, then 
      # this item cannot be included in the optimal solution 
      if (wt[n-1] > W): 
          return knapSack(W, wt, val, n-1) 
    
      # return the maximum of two cases: 
      # (1) nth item included 
      # (2) not included 
      else: 
          return max(val[n-1] + knapSack(W-wt[n-1], wt, val, n-1), 
                      knapSack(W, wt, val, n-1)) 
  
  # Test cases
  items = [(60, 10), (100, 20), (120, 30)]
  values = [item[0] for item in items]
  weights = [item[1] for item in items]
  assert knapSack(50, weights, values, len(values)) == 220
  assert knapSack(0, weights, values, len(values)) == 0
  assert knapSack(50, [], [], 0) == 0
  assert knapSack(30, weights, values, len(values)) == 160
  assert knapSack(10, weights, values, len(values)) == 60
output: def knapSack(W, wt, val, n): 
      dp = [[0 for _ in range(W + 1)] for _ in range(n + 1)]
      
      for i in range(1, n + 1):
          for w in range(1, W + 1):
              if wt[i-1] <= w:
                  dp[i][w] = max(val[i-1] + dp[i-1][w-wt[i-1]], dp[i-1][w])
              else:
                  dp[i][w] = dp[i-1][w]
                  
      return dp[n][W]