How to Solve Fractional Knapsack Problem?

Are you struggling to find the best solution for the Fractional Knapsack Problem? Look no further! With the power of greedy algorithms, solving this problem has never been easier.

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You're faced with a knapsack that can only hold a certain weight and you need to determine the maximum value of the fractions of items that can fit inside. How do you solve this puzzle?

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Consider a set of N items with each having a value V, a weight W, and the total capacity knapsack. You must find the maximal value of the fractions of items that fit in the knapsack.

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Example: Input: A[] = {{60, 20} , {100, 50}, {120, 30}}, Total_capacity = 50 Output = 180.00 Explanation = Take the 1st item & the 3rd item. Total value = 60 + 120 = 180 with a total capacity of 20 + 30 = 50.

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The simplest solution to this problem is to try every possible combination of items and find the maximum value among them. But, is it the most efficient solution?

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With the Greedy Algorithm, you can sort the items by their value-to-weight ratio, making it a much more efficient solution to the Fractional Knapsack Problem.

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This solution has a time complexity of O(N *log N) and a space complexity of O(1), making it a quick and space-saving solution.

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Ready to dive into Fractional Knapsack Problem ?

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How to Solve Fractional Knapsack Problem?

A Quick Overview

Are you struggling to find the best solution for the Fractional Knapsack Problem? Look no further! With the power of greedy algorithms, solving this problem has never been easier.

You're faced with a knapsack that can only hold a certain weight and you need to determine the maximum value of the fractions of items that can fit inside. How do you solve this puzzle?

The Fractional Knapsack Dilemma

Consider a set of N items with each having a value V, a weight W, and the total capacity knapsack. You must find the maximal value of the fractions of items that fit in the knapsack.

Problem Statement

Example: Input: A[] = {{60, 20} , {100, 50}, {120, 30}}, Total_capacity = 50 Output = 180.00 Explanation = Take the 1st item & the 3rd item. Total value = 60 + 120 = 180 with a total capacity of 20 + 30 = 50.

The simplest solution to this problem is to try every possible combination of items and find the maximum value among them. But, is it the most efficient solution?

Brute Force Approach - The Traditional Approach

With the Greedy Algorithm, you can sort the items by their value-to-weight ratio, making it a much more efficient solution to the Fractional Knapsack Problem.

A Better Solution: The Greedy Algorithm

This solution has a time complexity of O(N *log N) and a space complexity of O(1), making it a quick and space-saving solution.

Optimized Performance

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