### Algorithms

# A knapsack has max capacity C and there are n items each with weight w[i] and value v[i] each. Maximize the value in the knapsack without exceeding its max capacity.

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# A knapsack has max capacity C and there are n items each with weight w[i] and value v[i] each. Maximize the value in the knapsack without exceeding its max capacity.

**Category:** Algorithms
**Asked at:** Uber
**Difficulty:**

## Question Explanation

This question is asking you to solve a classic problem in computer science known as the Knapsack problem. The purpose is to find the combination of objects with individual values and weights, to maximize the sum of the values of the objects in your knapsack while adhering to the weight limit of the knapsack �C�. Some key points to consider while solving the Knapsack Problem are: It's a combinatorial optimization problem. The objective is to find the optimal solution from a large set of possible solutions. It's a 0/1 problem, because an item canʼt be partially added or duplicated. You either take the whole item or don't take it at all. You will likely need to use dynamic programming to solve this problem efficiently.

## Answer Example 1

# Here's a way to solve the problem using the Python programming language. This approach uses dynamic programming: def def knapSack knapSack ( ( C C , , w w , , v v , , n n ) ) : : if if n n

# == 0 0 or or C C

## == 0 0 : : return return 0 0 if if ( ( w w [ [ n n

1 1 ] ]

## C C ) ) : : return return knapSack knapSack ( ( C C , , w w , , v v , , n n

## 1 1 ) ) else else : : return return max max ( ( v v [ [ n n

## 1 1 ] ] + + knapSack knapSack ( ( C C

## w w [ [ n n

## 1 1 ] ] , , w w , , v v , , n n

## 1 1 ) ) , , knapSack knapSack ( ( C C , , w w , , v v , , n n

1 1 ) ) ) ) In this example, the knapSack function iterates over each item and checks whether the item can be placed in the knapsack. If the item's weight is more than the total capacity �C�, the item is discarded. If the item can be added to the knapsack, the function calculates the maximum value of the current item and the remaining capacity of the knapsack, and returns the maximum of these values.

## Answer Example 2

# This is another example in Python, slightly more optimized using dynamic programming: def def knapSack knapSack ( ( C C , , w w , , v v , , n n ) ) : : K K

# = [ [ [ [ 0 0 for for w w in in range range ( ( C C + + 1 1 ) ) ] ] 0Share for for i i in in range range ( ( n n + + 1 1 ) ) ] ] for for i i in in range range ( ( n n + + 1 1 ) ) : : for for wt wt in in range range ( ( C C + + 1 1 ) ) : : if if i i

# == 0 0 or or wt wt

# == 0 0 : : K K [ [ i i ] ] [ [ wt wt ] ]

## = 0 0 elif elif w w [ [ i i

# 1 1 ] ] <= <= wt wt : : K K [ [ i i ] ] [ [ wt wt ] ]

## = max max ( ( v v [ [ i i

## 1 1 ] ] + + K K [ [ i i

## 1 1 ] ] [ [ wt wt

## w w [ [ i i

## 1 1 ] ] ] ] , , K K [ [ i i

# 1 1 ] ] [ [ wt wt ] ] ) ) else else : : K K [ [ i i ] ] [ [ wt wt ] ]

## = K K [ [ i i

1 1 ] ] [ [ wt wt ] ] return return K K [ [ n n ] ] [ [ C C ] ] In this example, a 2�D matrix K[][] is built in bottom up manner. For each weight from 0 to C, the maximum value that can be put in the knapsack is stored in the matrix. The final answer will be K[n] �C�. This solution avoids recalculations and repetition of subproblems, improving efficiency.