Hướng dẫn minimum coin change problem python - vấn đề thay đổi tiền xu tối thiểu python

Cho một giá trị v, nếu chúng ta muốn thực hiện thay đổi cho v xu và chúng ta có nguồn cung cấp vô hạn của mỗi c = {c1, c2, .., cm} các đồng tiền có giá trị, số tiền tối thiểu là bao nhiêu biến đổi? Nếu nó không thể thực hiện thay đổi, in -1.V, if we want to make a change for V cents, and we have an infinite supply of each of C = { C1, C2, .., Cm} valued coins, what is the minimum number of coins to make the change? If it’s not possible to make a change, print -1.

Examples:    

Đầu vào: Tiền xu [] = {25, 10, 5}, v = 30Output: Tối thiểu 2 đồng xu cần thiết Chúng ta có thể sử dụng một đồng tiền 25 xu và một trong 5 xu & nbsp; coins[] = {25, 10, 5}, V = 30
Output: Minimum 2 coins required We can use one coin of 25 cents and one of 5 cents 

Đầu vào: Tiền xu [] = {9, 6, 5, 1}, v = 11Output: Tối thiểu 2 Tiền yêu cầu Chúng ta có thể sử dụng một đồng tiền 6 xu và 1 đồng xu 5 xu 5 xu coins[] = {9, 6, 5, 1}, V = 11
Output: Minimum 2 coins required We can use one coin of 6 cents and 1 coin of 5 cents

Vấn đề này là một biến thể của vấn đề được thảo luận về vấn đề thay đổi tiền xu. Ở đây thay vì tìm tổng số giải pháp có thể, chúng ta cần tìm giải pháp với số lượng tiền tối thiểu.Coin Change Problem. Here instead of finding the total number of possible solutions, we need to find the solution with the minimum number of coins.

Số lượng tiền tối thiểu cho giá trị V có thể được tính toán bằng cách sử dụng công thức đệ quy dưới đây. & NBSP;

If V == 0, then 0 coins required.
If V > 0
   minCoins(coins[0..m-1], V) = min {1 + minCoins(V-coin[i])} 
                               where i varies from 0 to m-1 
                               and coin[i] <= V

Dưới đây là một giải pháp đệ quy dựa trên công thức đệ quy ở trên. & NBSP;

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Độ phức tạp về thời gian của giải pháp trên là theo cấp số nhân và độ phức tạp không gian lớn hơn O (n). Nếu chúng ta vẽ cây đệ quy hoàn chỉnh, chúng ta có thể quan sát rằng nhiều vấn đề phụ được giải quyết nhiều lần. Ví dụ: khi chúng ta bắt đầu từ V = 11, chúng ta có thể đạt 6 bằng cách trừ một 5 lần và trừ 5 lần một lần. Vì vậy, biểu tượng con cho 6 được gọi là hai lần. & Nbsp;

Vì các vấn đề nhỏ tương tự được gọi là lại, vấn đề này có thuộc tính Subroprosplips chồng chéo. Vì vậy, vấn đề Min Coins có cả hai thuộc tính (xem điều này và điều này) của một vấn đề lập trình động. Giống như các vấn đề lập trình động (DP) điển hình khác, các phép tính lại của cùng một vấn đề phụ có thể được tránh bằng cách xây dựng bảng mảng tạm thời [] [] theo cách từ dưới lên. Dưới đây là giải pháp dựa trên lập trình động. & NBSP;

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Độ phức tạp về thời gian: O (M*V) .AUXILIARY Không gian: O (v) vì sử dụng thêm không gian cho bảng mảng & nbsp; O(m*V).
Auxiliary space: O(V) because using extra space for array table
 

Cảm ơn Goku đã đề xuất giải pháp trên trong một bình luận ở đây và cảm ơn Vignesh Mohan vì đã đề xuất vấn đề này và giải pháp ban đầu. Xin vui lòng viết bình luận nếu bạn tìm thấy bất cứ điều gì không chính xác, hoặc bạn muốn chia sẻ thêm thông tin về chủ đề được thảo luận ở trên & NBSP;
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 


Số lượng đồng tiền tối thiểu phải được lật lại là bao nhiêu?

Giải thích: Tối thiểu 4 xu cần thiết.4 coins required.

Vấn đề thay đổi xu cho ví dụ là gì?

Ví dụ 1: Giả sử bạn được cho các đồng tiền 1 cent, 5 xu và 10 xu với n = 8 xu, tổng số kết hợp của các đồng tiền bạn có thể sắp xếp để có được 8 xu.Đầu vào: n = 8 đồng tiền: 1, 5, 10 đầu ra: 2 Giải thích: 1 cách: 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8 xu.2 cách: 1 + 1 + 1 + 5 = 8 xu.Suppose you are given the coins 1 cent, 5 cents, and 10 cents with N = 8 cents, what are the total number of combinations of the coins you can arrange to obtain 8 cents. Input: N=8 Coins : 1, 5, 10 Output: 2 Explanation: 1 way: 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8 cents. 2 way: 1 + 1 + 1 + 5 = 8 cents.

Sự phức tạp về thời gian của vấn đề thay đổi tiền xu là gì?

Độ phức tạp về thời gian của vấn đề thay đổi tiền xu là (trong mọi trường hợp) (n*c) và độ phức tạp không gian là (n*c) (n).(n*c), and the space complexity is (n*c) (n).

Bạn có thể thay đổi bao nhiêu cách với tiền xu và tổng số tiền trong Python?

Giải thích: Có bốn giải pháp: {1, 1, 1, 1}, {1, 1, 2}, {2, 2}, {1, 3}.four solutions: {1, 1, 1, 1}, {1, 1, 2}, {2, 2}, {1, 3}.