## Knapsack: Top-down, Memoization & Bottom-up

### Introduction

Given the weights and profits of ‘N’ items, we are asked to put these items in a knapsack which has a capacity ‘C’. The goal is to get the maximum profit out of the items in the knapsack. Each item can only be selected once, as we don’t have multiple quantities of any item.

Let’s take the example of Merry, who wants to carry some fruits in the knapsack to get maximum profit. Here are the weights and profits of the fruits:

Items: { Apple, Orange, Banana, Melon }
Weights: { 2, 3, 1, 4 }
Profits: { 4, 5, 3, 7 }
Knapsack capacity: 5

Let’s try to put various combinations of fruits in the knapsack, such that their total weight is not more than 5:

Apple + Orange (total weight 5) => 9 profit
Apple + Banana (total weight 3) => 7 profit
Orange + Banana (total weight 4) => 8 profit
Banana + Melon (total weight 5) => 10 profit

This shows that Banana + Melon is the best combination as it gives us the maximum profit and the total weight does not exceed the capacity.

### Problem Statement

Given two integer arrays to represent weights and profits of ‘N’ items, we need to find a subset of these items which will give us maximum profit such that their cumulative weight is not more than a given number ‘C’. Each item can only be selected once, which means either we put an item in the knapsack or we skip it.

### Basic Solution

A basic brute-force solution could be to try all combinations of the given items (as we did above), allowing us to choose the one with maximum profit and a weight that doesn’t exceed ‘C’. To try all the combinations, our algorithm will look like:

  public int solveKnapsack(int[] profits, int[] weights, int capacity) {
return knapsackRecursive(profits, weights, capacity, 0);
}

private int knapsackRecursive(int[] profits, int[] weights, int capacity, int currentIndex) {
// base checks
if (capacity <= 0 || currentIndex >= profits.length)
return 0;

// recursive call after choosing the element at the currentIndex
// if the weight of the element at currentIndex exceeds the capacity, we shouldn't process this
int profit1 = 0;
if( weights[currentIndex] <= capacity )
profit1 = profits[currentIndex] + knapsackRecursive(profits, weights,
capacity - weights[currentIndex], currentIndex + 1);

// recursive call after excluding the element at the currentIndex
int profit2 = knapsackRecursive(profits, weights, capacity, currentIndex + 1);

return Math.max(profit1, profit2);
}


Complexity:

For recursive function with two separate calls of itself, the time complexity is O(2​^n). Since recursive calls work the same way as DF search, space complexity is O(n).

### Top-down Dynamic Programming with Memoization

Memoization is when we store the results of all the previously solved sub-problems and return the results from memory if we encounter a problem that has already been solved.

Since we have two changing values (capacity and currentIndex) in our recursive function knapsackRecursive(), we can use a two-dimensional array to store the results of all the solved sub-problems. As mentioned above, we need to store results for every sub-array (i.e. for every possible index ‘i’) and for every possible capacity ‘c’.

  public int solveKnapsack(int[] profits, int[] weights, int capacity) {
Integer[][] dp = new Integer[profits.length][capacity + 1];
return knapsackRecursive(dp, profits, weights, capacity, 0);
}

private int knapsackRecursive(Integer[][] dp, int[] profits, int[] weights, int capacity,
int currentIndex) {

// base checks
if (capacity <= 0 || currentIndex >= profits.length)
return 0;

// if we have already solved a similar problem, return the result from memory
if(dp[currentIndex][capacity] != null)
return dp[currentIndex][capacity];

// recursive call after choosing the element at the currentIndex
// if the weight of the element at currentIndex exceeds the capacity, we shouldn't process this
int profit1 = 0;
if( weights[currentIndex] <= capacity )
profit1 = profits[currentIndex] + knapsackRecursive(dp, profits, weights,
capacity - weights[currentIndex], currentIndex + 1);

// recursive call after excluding the element at the currentIndex
int profit2 = knapsackRecursive(dp, profits, weights, capacity, currentIndex + 1);

dp[currentIndex][capacity] = Math.max(profit1, profit2);
return dp[currentIndex][capacity];
}


#### Time and Space complexity

Since our memoization array dp[profits.length][capacity+1] stores the results for all subproblems, we can conclude that we will not have more than  N*C subproblems (where ‘N’ is the number of items and ‘C’ is the knapsack capacity). This means that our time complexity will be O(N*C).

The above algorithm will use O(N*C) space for the memoization array. Other than that we will use O(N) space for the recursion call-stack. So the total space complexity will be O(N*C + N), which is asymptotically equivalent to O(N*C)

### Bottom-up Dynamic Programming

Let’s try to populate our dp[][] array from the above solution by working in a bottom-up fashion. Essentially, we want to find the maximum profit for every sub-array and for every possible capacity. This means that dp[i][c] will represent the maximum knapsack profit for capacity ‘c’ calculated for the first ‘i’ items.

So, for each item at index ‘i’ (0 <= i < items.length) and capacity ‘c’ (0 <= c <= capacity), we have two options:

1. Exclude the item at index ‘i’. In this case, we will take whatever profit we get from the sub-array excluding this item => dp[i-1][c]
2. Include the item at index ‘i’ if its weight is not more than the capacity. In this case, we include its profit plus whatever profit we get from the remaining capacity and from previous items => profit[i] + dp[i-1][c-weight[i]]

Finally, our optimal solution will be maximum of the above two values:

    dp[i][c] = max (dp[i-1][c], profit[i] + dp[i-1][c-weight[i]]) 
  public int solveKnapsack(int[] profits, int[] weights, int capacity) {
// basic checks
if (capacity &lt;= 0 || profits.length == 0 || weights.length != profits.length)
return 0;

int n = profits.length;
int[][] dp = new int[n][capacity + 1];

// populate the capacity=0 columns, with '0' capacity we have '0' profit
for(int i=0; i &lt; n; i++)
dp[i][0] = 0;

// if we have only one weight, we will take it if it is not more than the capacity
for(int c=0; c &lt;= capacity; c++) {
if(weights[0] &lt;= c)
dp[0][c] = profits[0];
}

// process all sub-arrays for all the capacities
for(int i=1; i &lt; n; i++) {
for(int c=1; c &lt;= capacity; c++) {
int profit1= 0, profit2 = 0;
// include the item, if it is not more than the capacity
if(weights[i] &lt;= c)
profit1 = profits[i] + dp[i-1][c-weights[i]];
// exclude the item
profit2 = dp[i-1][c];
// take maximum
dp[i][c] = Math.max(profit1, profit2);
}
}

// maximum profit will be at the bottom-right corner.
return dp[n-1][capacity];
}


#### Time and Space complexity

The above solution has the time and space complexity of O(N*C), where ‘N’ represents total items and ‘C’ is the maximum capacity.