DynamicProgramming/knapsack.CPP at master · PurushottamParakh/DynamicProgramming · GitHub

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#include <iostream>
#include <string>
#include <vector>
using namespace std;


vector<int> knapsackContent(const vector<int> &val, const vector<int> &wt, const int &W){

    int rows = val.size();
    int cols = W + 1;

    vector<vector<int>> T (rows, vector<int>( cols , 0 ));

    for(int i=0; i<rows; ++i){
        for(int j=1; j< cols; ++j){
            // when considering item i of weight wt[i] and val[i] to fill the sack of capacity j
            // we have two option either pick up the item i or do not pick up
            // if you pick up the item, the sack is left with capacity : j-wt[i] and we already know how to fill it using earlier 0..(i-1) items.

            if(i==0){
                if(j>=wt[i])
                    T[0][j] = val[i];
                continue;
            }

            if(j < wt[i])
                T[i][j] = T[i-1][j];
            else
                T[i][j] = max(val[i] + T[i-1][j-wt[i]] , T[i-1][j]);
        }
    }

    //-----------till now same as knapsack function

    int MaxVal = T[rows-1][cols-1];
    int totalwt = W;
    vector<int> result;
    int i=rows-1;
    int j=cols-1;
    while(MaxVal>0){

        if(i==0){
            if(MaxVal == val[0])
            result.push_back(wt[0]);
        }

        if(i-1>=0){
            if(T[i-1][j] == T[i][j]){
                //the item is not selected
                i = i-1;
                continue;
            }
            else{
                result.push_back(wt[i]);
                j = totalwt - wt[i];
                MaxVal -= val[i];
                i = i-1;
            }
        }
    }

    return result;
}


int knapsack(const vector<int> &val, const vector<int> &wt, const int &W){

    int rows = val.size();
    int cols = W + 1;

    vector<vector<int>> T (rows, vector<int>( cols , 0 ));

    for(int i=0; i<rows; ++i){
        for(int j=1; j< cols; ++j){
            // when considering item i of weight wt[i] and val[i] to fill the sack of capacity j
            // we have two option either pick up the item i or do not pick up
            // if you pick up the item, the sack is left with capacity : j-wt[i] and we already know how to fill it using earlier 0..(i-1) items.

            if(i==0){
                if(j>=wt[i])
                    T[0][j] = val[i];
                continue;
            }

            if(j < wt[i])
                T[i][j] = T[i-1][j];
            else
                T[i][j] = max(val[i] + T[i-1][j-wt[i]] , T[i-1][j]);
        }
    }

    for(int i=0; i<rows; ++i){
        for(int j=0; j<cols; ++j){
            cout << T[i][j] << " ";
        }
        cout << endl;
    }

    return T[rows-1][cols-1];
}


int main()
{
    vector<int> val = {1,4,5,7};
    vector<int> wt = {1,3,4,5};
    int W = 7;

    cout << knapsack(val,wt,W) << endl;
    return 0;

}