Change Making Problem using Brute Force

 The Change Making Problem is a classic optimization problem where the goal is to make change for a specific amount using the fewest number of coins from a given set of denominations. When approaching this problem with a brute force method, it involves examining all possible combinations of coins that add up to the given amount and selecting the combination with the minimum number of coins.

 Brute Force Solution for Change Making Problem

 Problem Setup: 

Given a set of coin denominations (e.g., `[1, 5, 10]`) and a total amount (e.g., `12`), find the minimum number of coins that make up that amount.

 Brute Force Approach:

    1. Generate All Combinations: 

Produce all combinations of coins that sum up to the target amount. This usually requires recursion, where you explore each possible way to reach the amount using the given denominations.

    2. Select Minimum Coin Combination: 

For each combination that sums to the target amount, count the number of coins used. Keep track of the combination that uses the fewest coins.

    3. Return Result: 

The combination with the minimum number of coins is the solution.

 Pseudocode Example:

    function minCoins(coins, numberOfCoins, amount):

        if amount == 0:

            return 0  // No coins needed for zero amount

        minCoinsRequired = INFINITY

        // Try every coin that has a value smaller than amount

        for i from 0 to numberOfCoins:

            if coins[i] <= amount:

                subResult = minCoins(coins, numberOfCoins, amount  coins[i])

                // Check if the subResult needs fewer coins

                if subResult != INFINITY and subResult + 1 < minCoinsRequired:

                    minCoinsRequired = subResult + 1

        return minCoinsRequired

 

    // Example call

    minCoins([1, 5, 10], 3, 12)

  Time Complexity: 

The brute force approach is highly inefficient for this problem. Its time complexity is approximately O(m^n), where m is the amount and n is the number of different denominations. The reason for this high complexity is the repetitive calculation of the same subproblems.

 Important Considerations:

 Inefficiency and Redundancy: 

The brute force solution involves a lot of redundant computations, as it doesn't store results of subproblems for future reference.

 Dynamic Programming as an Alternative: 

A more efficient approach is to use dynamic programming, which solves each subproblem only once and stores its result to avoid redundant calculations.

In summary, while the brute force method can theoretically solve the Change Making problem by exploring every possible combination of coins, it is not practical for larger amounts or a greater number of denominations due to its exponential time complexity. For more efficient solutions, methods like dynamic programming are highly recommended.

Detailed Time Complexity analysis

 The `minCoins` function provided follows a recursive approach to solve the minimum coin change problem. To formulate its recurrence relation, we need to express how the solution to the problem can be constructed from the solutions to its subproblems.

 Recurrence Relation

Let ( C(i, a) ) represent the minimum number of coins needed to make change for amount ( a ) using the first ( i ) coin denominations.



 Explanation

1. Base Cases:

   - If ( a = 0 ), no coins are needed, so ( C(i, 0) = 0 ) for any ( i ).

   - If ( i = 0 ) and ( a > 0 ), it's impossible to make change (as no denominations are available), so ( C(0, a) = infinity ) for ( a > 0 ).

2. Recursive Cases:

   - If the value of the ( i )-th coin (denoted as `coins[i]`) is greater than the amount ( a ), we cannot use this coin, so the problem remains the same as ( C(i-1, a) ).

   - Otherwise, we have two choices:

       - Exclude the ( i )-th coin and solve for the rest of the coins, which is ( C(i-1, a) ).

       - Include the ( i )-th coin, reduce the amount by the value of this coin, and solve the subproblem with the new amount ( a - text{coins}[i] ), which adds 1 coin to the solution of ( C(i, a - text{coins}[i]) ).

   - The minimum of these two choices gives us the minimum number of coins.

 Time Complexity Implication

The recurrence relation highlights the overlapping subproblems and optimal substructure properties of the problem, making it a good candidate for optimization through Dynamic Programming. However, in its basic recursive form (as presented in the `minCoins` function), the algorithm has an exponential time complexity due to the computation of the same subproblems multiple times.


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