One of the classical problems of dynamic programming is the 0/1 knapsack problem. The thief has a knapsack of given capacity, and he wishes to maximize the profit with the available valuables. What should he do?

Though it would be weird if he starts implementing DP at the crime scene. But anyways.

Keep track of the maximum profit he can achieve up to a certain weight, and decide whether to pick the next object or not. Here is a simple code for the same.

*Capacity of Knapsack:* W (given)

*Objective:* Maximize profit.

```
int dp[W + 1] = {0};
for (int i=0; i<n; ++i)
for (int j=W; j>=weight[i]; --j)
dp[j] = max(dp[j], value[i]+ dp[j - weight[i]]);
return dp[W];
```

*dp[j]* if he will not pick the ith object, otherwise *value[i]+ dp[j - weight[i]]*

#### Now, let’s see how our thief with his knapsack will commit a series of crimes at LeetCode.

LC416. Partition equal subset sum

Can an array of positive integers be partitioned into two subsets, such that their sums are equal?

Can our thief pick a subset such that the sum of the subset is half the sum of the whole array?

*Capacity of Knapsack:* Half the sum of the given array.

*Objective:* Achieve a total sum exact as the capacity.

```
vector<bool> dp(sum+1, false);
dp[0] = true;
for (auto num : nums)
for (int i=sum; i>0; --i)
if (i >= num) dp[i] = dp[i-num] or dp[i];
return dp[sum];
```

*dp[i]* if the number is not included in the subset, else *dp[i-num]*.

Handle one edge case

Given an array of binary strings strs and two integers m and n. Return the size of the largest subset of strs such that there are at most m 0’s and n 1’s in the subset.

Can our thief pick up at most m 0’s and n 1’s?

*Capacity of Knapsack:* Limitless

*Objective:* Pick up at most m 0’s and 1 n’s

```
vector<vector<int>> dp(M+1, vector<int>(N+1, 0));
for (auto s : strs) {
int zero = zeros(s); // a function to count zeroes in a string
int one = ones(s); // a function to count ones in a string
for (int i=M; i>=zero; --i)
for (int j=N; j>=one; --j)
dp[i][j] = max(dp[i][j], 1 + dp[i-zero][j-one]);
}
return dp[M][N];
```

*dp[i][j]* if the string is not chosen, else *dp[i-zero][j-one]*

Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers. Return the maximum product you can get.

Can our thief pick k numbers from 1 to n, such that their product is maximized and they sum to n?

*Capacity of Knapsack:* n

*Objective:* Maximize the product of chosen smaller integers

```
vector<int> dp(n+1, 0);
dp[0] = 1;
for (int i=1; i<n; ++i)
for (int j=i; j<=n; ++j)
dp[j] = max(dp[j], dp[j-i]*i);
return dp[n];
```

*dp[j]* if the number is not included in the product. dp[j-i]*i if the number is included, multiply with the previous suitable state.

#### Unbound knapsack

In this class of problems, every available item type is present in infinite quantity. Iterate over the choices, and put it any number of combinations you wish.

LC518. Coin Change 2

Given an array of coins and an integer representing a total amount. Return the number of combinations that make up that amount. If impossible, return 0. Each coin is in present infinite quantity.

The order will not matter in this problem. e.g.

```
To achieve a sum of 5 using [1,2,5]
5 = 5
5 = 2+2+1 // other permutations like (2,1,2) are not included
5 = 2+1+1+1
5 = 1+1+1+1+1
result is 4 ways
```

Can our thief pick a total sum equal to the target?

*Capacity of Knapsack:* target

*Objective:* Count the number of ways to achieve the target

```
vector<int> dp(amount+1, 0);
dp[0] = 1;
for (auto coin : coins)
for (int i=1; i<=amount; i++)
if (coin <= i) dp[i] += dp[i-coin];
return dp[amount];
```

Since every coin can be put in any combination, iterate over coins, and fill every achievable target. This will prevent the recounting of the permutations. Why? Because the definition of our state dp[i] is the number of ways to achieve a sum i using all coins. If we’ll first iterate over the sum and next on the coins, we’ll include all three permutations viz. (1,2,2), (2,1,2), and (2,2,1) in the solution. Because we are adding dp[i-coin] every time we find a suitable candidate. Alternatively, you can use a 2D dp, where the state dp[i][j], will be the number of ways to get sum j using first i coins. This approach will be independent of the priority of the for-loops.

LC377. Combination Sum IV

Given an array of distinct integers and a target, return the number of possible combinations that add up to the target.

In this problem, the order matters, unlike Coin Change 2. So, iterate over the target, then the coins (in this case numbers). Else use a 2D dp.

```
vector<unsigned int> dp(target+1, 0);
dp[0] = 1;
for (int i=1; i<= target; i++)
for (auto n : nums)
if (i >= n) dp[i] += dp[i - n];
return dp[target];
```

For every problem, pick one example and dry run the code to fill the states.

### Thank you.

If you witness any other crimes, please report them in the comments.