README.md

Greedy Algorithms

always choosing the next piece that offers the most obvious and immediate benefit. 
So the problems where choosing locally optimal also leads to global solution are the best fit for Greedy.

0-1 Knapsack Problem

Given weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items respectively. Also given an integer W which represents knapsack capacity, find out the maximum value subset of val[] such that sum of the weights of this subset is smaller than or equal to W. You cannot break an item, either pick the complete item or don’t pick it (0-1 property).

method 1 :Recursion by Brute-Force algorithm OR Exhaustive Search.

Approach: A simple solution is to consider all subsets of items and calculate the total weight and value of all subsets. Consider the only subsets whose total weight is smaller than W. From all such subsets, pick the maximum value subset.
Optimal Sub-structure: To consider all subsets of items, there can be two cases for every item. 
Case 1: The item is included in the optimal subset.
Case 2: The item is not included in the optimal set.
Therefore, the maximum value that can be obtained from ‘n’ items is the max of the following two values. 
Maximum value obtained by n-1 items and W weight (excluding nth item).
Value of nth item plus maximum value obtained by n-1 items and W minus the weight of the nth item (including nth item).
If the weight of ‘nth’ item is greater than ‘W’, then the nth item cannot be included and Case 1 is the only possibility.
Below is the implementation of the above approach: 

C++ CODE

#include <bits/stdc++.h>
using namespace std;
 
// A utility function that returns
// maximum of two integers
int max(int a, int b) { return (a > b) ? a : b; }
 
// Returns the maximum value that
// can be put in a knapsack of capacity W
int knapSack(int W, int wt[], int val[], int n)
{
 
    // Base Case
    if (n == 0 || W == 0)
        return 0;
 
    // If weight of the nth item is more
    // than Knapsack capacity W, then
    // this item cannot be included
    // in the optimal solution
    if (wt[n - 1] > W)
        return knapSack(W, wt, val, n - 1);
 
    // Return the maximum of two cases:
    // (1) nth item included
    // (2) not included
    else
        return max(
            val[n - 1]
                + knapSack(W - wt[n - 1],
                           wt, val, n - 1),
            knapSack(W, wt, val, n - 1));
}
 
// Driver code
int main()
{
    int val[] = { 60, 100, 120 };
    int wt[] = { 10, 20, 30 };
    int W = 50;
    int n = sizeof(val) / sizeof(val[0]);
    cout << knapSack(W, wt, val, n);
    return 0;
}

Output:220

It should be noted that the above function computes the same sub-problems again and again. See the following recursion tree, K(1, 1) is being evaluated twice. The time complexity of this naive recursive solution is exponential (2^n)
Complexity Analysis Time Complexity: O(2^n). As there are redundant subproblems. Auxiliary Space :O(1) + O(N). As no extra data structure has been used for storing values but O(N) auxiliary stack space(ASS) has been used for recursion stack. Since subproblems are evaluated again, this problem has Overlapping Sub-problems property. So the 0-1 Knapsack problem has both properties (see this and this) of a dynamic programming problem.