0/1 Knapsack Problem to print all possible solutions

Given weights and profits of N items, put these items in a knapsack of capacity W. The task is to print all possible solutions to the problem in such a way that there are no remaining items left whose weight is less than the remaining capacity of the knapsack. Also, compute the maximum profit.
Examples:

Input: Profits[] = {60, 100, 120, 50} 
Weights[] = {10, 20, 30, 40}, W = 40 
Output:
10: 60, 20: 100, 
10: 60, 30: 120, 
Maximum Profit = 180 
Explanation:
Maximum profit from all the possible solutions is 180
Input: Profits[] = {60, 100, 120, 50} 
Weights[] = {10, 20, 30, 40}, W = 50 
Output:
10: 60, 20: 100, 
10: 60, 30: 120, 
20: 100, 30: 120, 
Maximum Profit = 220 
Explanation:
Maximum profit from all the possible solutions is 220 

Approach: The idea is to make pairs for the weight and the profits of the items and then try out all permutations of the array and including the weights until their is no such item whose weight is less than the remaining capacity of the knapsack. Meanwhile after including an item increment the profit for that solution by the profit of that item. 
Below is the implementation of the above approach:

C++

// C++ implementation to print all
// the possible solutions of the
// 0/1 Knapsack problem

#include <bits/stdc++.h>

using namespace std;

// Utility function to find the
// maximum of the two elements
int max(int a, int b) {
    return (a > b) ? a : b;
}

// Function to find the all the
// possible solutions of the
// 0/1 knapSack problem
int knapSack(int W, vector<int> wt,
            vector<int> val, int n)
{
    // Mapping weights with Profits
    map<int, int> umap;

    set<vector<pair<int, int>>> set_sol;
    // Making Pairs and inserting
    // into the map
    for (int i = 0; i < n; i++) {
        umap.insert({ wt[i], val[i] });
    }

    int result = INT_MIN;
    int remaining_weight;
    int sum = 0;

    // Loop to iterate over all the
    // possible permutations of array
    do {
        sum = 0;

        // Initially bag will be empty
        remaining_weight = W;
        vector<pair<int, int>> possible;

        // Loop to fill up the bag
        // until there is no weight
        // such which is less than
        // remaining weight of the
        // 0-1 knapSack
        for (int i = 0; i < n; i++) {
            if (wt[i] <= remaining_weight) {

                remaining_weight -= wt[i];
                auto itr = umap.find(wt[i]);
                sum += (itr->second);
                possible.push_back({itr->first,
                     itr->second
                });
            }
        }
        sort(possible.begin(), possible.end());
        if (sum > result) {
            result = sum;
        }
        if (set_sol.find(possible) ==
                        set_sol.end()){
            for (auto sol: possible){
                cout << sol.first << ": "
                     << sol.second << ", ";
            }
            cout << endl;
            set_sol.insert(possible);
        }

    } while (
        next_permutation(wt.begin(),
                           wt.end()));
    return result;
}

// Driver Code
int main()
{
    vector<int> val{ 60, 100, 120 };
    vector<int> wt{ 10, 20, 30 };
    int W = 50;
    int n = val.size();
    int maximum = knapSack(W, wt, val, n);
    cout << "Maximum Profit = ";
    cout << maximum;
    return 0;
}

Python3

# Python3 implementation to print all
# the possible solutions of the
# 0/1 Knapsack problem


INT_MIN=-2147483648
def nextPermutation(nums: list) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        if sorted(nums,reverse=True)==nums:
            return None
        n=len(nums)
        brk_point=-1
        for pos in range(n-1,0,-1):
            if nums[pos]>nums[pos-1]:
                brk_point=pos
                break
        else:
            nums.sort()
            return
        replace_with=-1
        for j in range(brk_point,n):
            if nums[j]>nums[brk_point-1]:
                replace_with=j
            else:
                break
        nums[replace_with],nums[brk_point-1]=nums[brk_point-1],nums[replace_with]
        nums[brk_point:]=sorted(nums[brk_point:])
        return nums

# Function to find the all the
# possible solutions of the
# 0/1 knapSack problem
def knapSack(W, wt, val, n):
    # Mapping weights with Profits
    umap=dict()

    set_sol=set()
    # Making Pairs and inserting
    # o the map
    for i in range(n) :
        umap[wt[i]]=val[i]


    result = INT_MIN
    remaining_weight=0
    sum = 0

    # Loop to iterate over all the
    # possible permutations of array
    while True:
        sum = 0

        # Initially bag will be empty
        remaining_weight = W
        possible=[]

        # Loop to fill up the bag
        # until there is no weight
        # such which is less than
        # remaining weight of the
        # 0-1 knapSack
        for i in range(n) :
            if (wt[i] <= remaining_weight) :

                remaining_weight -= wt[i]
                sum += (umap[wt[i]])
                possible.append((wt[i],
                     umap[wt[i]])
                )


        possible.sort()
        if (sum > result) :
            result = sum

        if (tuple(possible) not in set_sol):
            for sol in possible:
                print(sol[0], ": ", sol[1], ", ",end='')

            print()
            set_sol.add(tuple(possible))


        if not nextPermutation(wt):
            break
    return result


# Driver Code
if __name__ == '__main__':
    val=[60, 100, 120]
    wt=[10, 20, 30]
    W = 50
    n = len(val)
    maximum = knapSack(W, wt, val, n)
    print("Maximum Profit =",maximum)

#This code was contributed by Amartya Ghosh

Output:

10: 60, 20: 100, 
10: 60, 30: 120, 
20: 100, 30: 120, 
Maximum Profit = 220