Recall the activity$-$selection problem. Suppose that instead of always selecting the first activity to finish,

you instead select the last activity to start that is compatible with all previous selected activities. Describe

how this approach is a greedy algorithm, and prove that it yields an optimal solution.Does the Greedy approach solve the $0/1$ knapsack problem? If your answer is no then propose a suitable algorithmic approach to determine the optimum solution to the $0/1$ knapsack problem.