Hướng dẫn triplet sum in python - tổng gấp ba trong python

Triplet Sum

Cho một số nguyên n mảng, tìm và in tất cả các bộ ba duy nhất (a, b, c) trong mảng cho tổng K. (a+b+c = k) .input

Dòng đầu tiên của đầu vào sẽ là các số nguyên được phân tách không gian, biểu thị các phần tử của mảng. Dòng thứ hai của đầu vào sẽ là một số nguyên biểu thị tổng số koutput cần thiết

Đầu ra phải là nhiều dòng, mỗi dòng chứa một bộ ba duy nhất. Các yếu tố của bộ ba phải được sắp xếp theo thứ tự tăng lên và tất cả các bộ ba được in phải được sắp xếp theo thứ tự tăng lên. In "Không tìm thấy bộ ba phù hợp" nếu không có bộ ba với tổng số đã cho.

Khi mảng đã cho là [0, 1, 2, 3, 5, 7, 13, 17, 19, 19] và tổng yêu cầu là 22, bộ ba (0, 3, 19), (0, 5, 17) , (1, 2, 19), (2, 3, 17) và (2, 7, 13) có tổng số 22 đã cho.

Đầu vào mẫu 1

0 12 17 8 9 21

29

Đầu ra mẫu 1

(0, 8, 21)

(0, 12, 17)

(8, 9, 12)

Đầu vào mẫu 2

0 1 2 3 5 7 13 17 19 19

22

Đầu ra mẫu 2

(0, 3, 19)

(0, 5, 17)

(1, 2, 19)

(2, 3, 17)

(2, 7, 13)

Triplet Sum

Cho một số nguyên n mảng, tìm và in tất cả các bộ ba duy nhất (a, b, c) trong mảng cho tổng K. (a+b+c = k) .input

Dòng đầu tiên của đầu vào sẽ là các số nguyên được phân tách không gian, biểu thị các phần tử của mảng. Dòng thứ hai của đầu vào sẽ là một số nguyên biểu thị tổng số koutput cần thiết

Đầu ra phải là nhiều dòng, mỗi dòng chứa một bộ ba duy nhất. Các yếu tố của bộ ba phải được sắp xếp theo thứ tự tăng lên và tất cả các bộ ba được in phải được sắp xếp theo thứ tự tăng lên. In "Không tìm thấy bộ ba phù hợp" nếu không có bộ ba với tổng số đã cho.

Đầu vào mẫu 1

0 12 17 8 9 21

29

Đầu ra mẫu 1

(0, 8, 21)

(0, 12, 17)

(8, 9, 12)

Đầu vào mẫu 2

0 1 2 3 5 7 13 17 19 19

22

Đầu ra mẫu 2

(0, 3, 19)

(0, 5, 17)

(1, 2, 19)

(2, 3, 17)

(2, 7, 13)

Đầu ra nên được in chính xác như thể hiện trong các trường hợp thử nghiệm trên.

# Program to find a triplet returns true if there is triplet
# with sum of a, b, c equal to 'K' present in arr[].
def TripletSum(arr, K):
    arr_size = len(arr)
    found=False
    arr.sort()
 
    for a in range(0, arr_size - 2):    
        b = a + 1
        c = arr_size - 1
        while (b < c):        
            if( arr[a] + arr[b] + arr[c] == K):
                print(( arr[a], arr[b], arr[c]))
                b+=1
                c-=1
                found = True          
            elif (arr[a] + arr[b] + arr[c] < K):
                b += 1
            else:
                c -= 1
    if (found == False):
        print("No Matching Triplets Found")
 
 
# TEST
arr = [0, 1, 2, 3, 5, 7, 13, 17, 19, 19]
K = 22
TripletSum(arr, K)

# OUTPUT
(0, 3, 19)
(0, 5, 17)
(1, 2, 19)
(2, 3, 17)
(2, 7, 13)

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    Examples:   
     

    Bàn luận array = {12, 3, 4, 1, 6, 9}, sum = 24; 
    Output: 12, 3, 9 
    Explanation: There is a triplet (12, 3 and 9) present 
    in the array whose sum is 24. 
    Input: array = {1, 2, 3, 4, 5}, sum = 9 
    Output: 5, 3, 1 
    Explanation: There is a triplet (5, 3 and 1) present 
    in the array whose sum is 9.

    Hướng dẫn triplet sum in python - tổng gấp ba trong python

    Cho một mảng và một giá trị, hãy tìm nếu có một bộ ba trong mảng có tổng bằng với giá trị đã cho. Nếu có một bộ ba có mặt trong mảng, thì hãy in bộ ba và trả về đúng. Khác trả về sai. This is the naive approach towards solving the above problem.  

    • Đầu vào: mảng = {12, 3, 4, 1, 6, 9}, sum = 24; & nbsp; đầu ra: 12, 3, 9 & nbsp; giải thích: có một bộ ba (12, 3 và 9) hiện tại & nbsp; trong mảng có tổng là 24. & nbsp; đầu vào: mảng = {1, 2, 3, 4, 5}, sum = 9 & nbsp; đầu ra: 5, 3, 1 & nbsp; giải thích: có một bộ ba (5, 3 và 1) hiện tại & nbsp; Trong mảng có tổng là 9. A simple method is to generate all possible triplets and compare the sum of every triplet with the given value. The following code implements this simple method using three nested loops.
    • Algorithm:  
      1. Phương pháp 1: Đây là cách tiếp cận ngây thơ đối với việc giải quyết vấn đề trên. & Nbsp; & nbsp;
      2. Cách tiếp cận: Một phương pháp đơn giản là tạo ra tất cả các bộ ba có thể và so sánh tổng của mỗi bộ ba với giá trị đã cho. Mã sau đây thực hiện phương pháp đơn giản này bằng cách sử dụng ba vòng lồng nhau.
      3. Đưa ra một mảng có độ dài n và một tổng
      4. Tạo ba vòng lặp đầu tiên lồng nhau chạy từ đầu đến cuối (bộ đếm vòng I), vòng lặp thứ hai chạy từ I+1 đến cuối (vòng lặp vòng J) và vòng lặp thứ ba chạy từ J+1 đến cuối (Bộ đếm vòng K)
      5. Bộ đếm của các vòng này đại diện cho chỉ số của 3 yếu tố của bộ ba.
    • Implementation:

    C++

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    Tìm tổng của phần tử Ith, JTH và KTH. Nếu tổng bằng với tổng đã cho. In bộ ba và phá vỡ.

    Nếu không có bộ ba, thì in rằng không có bộ ba tồn tại.

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    using93

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    Đầu ra

    Triplet is 4, 10, 8
    • Phân tích độ phức tạp: & NBSP; 
      • Độ phức tạp về thời gian: O (N3). & NBSP; Có ba vòng lồng nhau đi qua mảng, do đó độ phức tạp thời gian là o (n^3) O(n3). 
        There are three nested loops traversing the array, so the time complexity is O(n^3)
      • Độ phức tạp không gian: O (1). & NBSP; vì không cần thêm không gian.O(1). 
        As no extra space is required.

    Phương pháp 2: Phương pháp này sử dụng sắp xếp để tăng hiệu quả của mã. & NBSP; This method uses sorting to increase the efficiency of the code. 

    • Cách tiếp cận: Bằng cách sắp xếp mảng, hiệu quả của thuật toán có thể được cải thiện. Cách tiếp cận hiệu quả này sử dụng kỹ thuật hai con trỏ. Đi qua mảng và sửa phần tử đầu tiên của bộ ba. Bây giờ sử dụng thuật toán hai con trỏ để tìm nếu có một cặp có tổng bằng x - mảng [i]. Hai thuật toán con trỏ mất thời gian tuyến tính để nó tốt hơn một vòng lặp lồng nhau. By Sorting the array the efficiency of the algorithm can be improved. This efficient approach uses the two-pointer technique. Traverse the array and fix the first element of the triplet. Now use the Two Pointers algorithm to find if there is a pair whose sum is equal to x – array[i]. Two pointers algorithm take linear time so it is better than a nested loop.
    • Thuật toán: & nbsp; 
      1. Sắp xếp mảng đã cho.
      2. Vòng lặp qua mảng và sửa phần tử đầu tiên của bộ ba có thể, mảng [i].
      3. Sau đó sửa hai con trỏ, một ở I + 1 và một ở n - 1. và nhìn vào tổng, & nbsp;
        1. Nếu tổng nhỏ hơn tổng yêu cầu, hãy tăng con trỏ đầu tiên.
        2. Khác, nếu tổng lớn hơn, giảm con trỏ cuối để giảm tổng.
        3. Khác, nếu tổng các phần tử ở hai con trỏ bằng với tổng trước thì in bộ ba và ngắt.
    • Implementation:

    C++

    #include

    using namespace std;

    bool find3Numbers(

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    Đầu ra

    Phân tích độ phức tạp: & NBSP;

    Độ phức tạp về thời gian: O (N3). & NBSP; Có ba vòng lồng nhau đi qua mảng, do đó độ phức tạp thời gian là o (n^3)

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    Độ phức tạp không gian: O (1). & NBSP; vì không cần thêm không gian.

    using7

    Phương pháp 2: Phương pháp này sử dụng sắp xếp để tăng hiệu quả của mã. & NBSP;

    find3Numbers(1

    std;41

    std;42

    Cách tiếp cận: Bằng cách sắp xếp mảng, hiệu quả của thuật toán có thể được cải thiện. Cách tiếp cận hiệu quả này sử dụng kỹ thuật hai con trỏ. Đi qua mảng và sửa phần tử đầu tiên của bộ ba. Bây giờ sử dụng thuật toán hai con trỏ để tìm nếu có một cặp có tổng bằng x - mảng [i]. Hai thuật toán con trỏ mất thời gian tuyến tính để nó tốt hơn một vòng lặp lồng nhau.

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    Thuật toán: & nbsp;

    using7

    bool find3Numbers(

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    Đầu ra

    Phân tích độ phức tạp: & NBSP;

    Độ phức tạp về thời gian: O (N3). & NBSP; Có ba vòng lồng nhau đi qua mảng, do đó độ phức tạp thời gian là o (n^3)

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    Độ phức tạp không gian: O (1). & NBSP; vì không cần thêm không gian.

    using7

    Phương pháp 2: Phương pháp này sử dụng sắp xếp để tăng hiệu quả của mã. & NBSP;

    Cách tiếp cận: Bằng cách sắp xếp mảng, hiệu quả của thuật toán có thể được cải thiện. Cách tiếp cận hiệu quả này sử dụng kỹ thuật hai con trỏ. Đi qua mảng và sửa phần tử đầu tiên của bộ ba. Bây giờ sử dụng thuật toán hai con trỏ để tìm nếu có một cặp có tổng bằng x - mảng [i]. Hai thuật toán con trỏ mất thời gian tuyến tính để nó tốt hơn một vòng lặp lồng nhau.

    Thuật toán: & nbsp;

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    Vòng lặp qua mảng và sửa phần tử đầu tiên của bộ ba có thể, mảng [i].

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    Đầu ra

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    • Phân tích độ phức tạp: & NBSP; 
      • Độ phức tạp về thời gian: O (n^2). & Nbsp; chỉ có hai vòng lồng nhau đi qua mảng, do đó độ phức tạp về thời gian là o (n^2). Hai thuật toán con trỏ mất thời gian O (N) và phần tử đầu tiên có thể được sửa bằng cách sử dụng một đường truyền khác. O(N^2). 
        There are only two nested loops traversing the array, so time complexity is O(n^2). Two pointers algorithm takes O(n) time and the first element can be fixed using another nested traversal.
      • Độ phức tạp không gian: O (1). & NBSP; vì không cần thêm không gian. O(1). 
        As no extra space is required.

    Phương pháp 3: Đây là một giải pháp dựa trên băm. & NBSP; This is a Hashing-based solution. 

    • Cách tiếp cận: Cách tiếp cận này sử dụng thêm không gian nhưng đơn giản hơn cách tiếp cận hai con trỏ. Chạy hai vòng lặp bên ngoài từ đầu đến cuối và vòng bên trong từ I+1 đến cuối. Tạo Hashmap hoặc đặt để lưu trữ các phần tử ở giữa I+1 đến J-1. Vì vậy, nếu tổng đã cho là x, hãy kiểm tra xem có một số trong tập hợp bằng x - mảng [i] - mảng [j]. Nếu có in bộ ba. & Nbsp; & nbsp;This approach uses extra space but is simpler than the two-pointers approach. Run two loops outer loop from start to end and inner loop from i+1 to end. Create a hashmap or set to store the elements in between i+1 to j-1. So if the given sum is x, check if there is a number in the set which is equal to x – arr[i] – arr[j]. If yes print the triplet. 
       
    • Algorithm:  
      1. Đi qua mảng từ đầu đến cuối. (Loop Counter I)
      2. Tạo Hashmap hoặc đặt để lưu trữ các cặp độc đáo.
      3. Chạy một vòng lặp khác từ I+1 đến cuối mảng. (Loop Counter J)
      4. Nếu có một phần tử trong tập hợp bằng x- arr [i]-arr [j], thì in bộ ba (mảng [i], mảng [j], x-arr [i] -arr [j] ) và phá vỡ
      5. Chèn phần tử thứ j trong tập hợp.
    • Implementation:

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    967

    #include 3using85

    #include 3using3 using4using5

    Triplet is 4, 8, 10
    8using7

    Triplet is 4, 8, 10
    8
    Triplet is 4, 10, 8
    705

    Triplet is 4, 8, 10
    1using7

    Triplet is 4, 8, 10
    4using7

    Triplet is 4, 8, 10
    4using3 namespace6using5

    Triplet is 4, 10, 8
    7using7

    Triplet is 4, 10, 8
    7
    Triplet is 4, 10, 8
    989

    Triplet is 4, 10, 8
    7namespace49

    Triplet is 4, 10, 8
    7namespace51

    Triplet is 4, 10, 8
    7
    Triplet is 4, 8, 10
    39
    Triplet is 4, 8, 10
    40
    Triplet is 4, 8, 10
    41
    Triplet is 4, 8, 10
    42

    namespace54

    Output:

    Triplet is 4, 8, 10

    Triplet is 4, 8, 10
    4
    Triplet is 4, 10, 8
    0
    Triplet is 4, 8, 10
    51
    Triplet is 4, 10, 8
    98
    Triplet is 4, 8, 10
    53
    Triplet is 4, 8, 10
    54
    Triplet is 4, 8, 10
    53
    Triplet is 4, 8, 10
    56
    Triplet is 4, 8, 10
    53__
    O(N^2) 
    Auxiliary Space: O(N), since n extra space has been taken.

    Triplet is 4, 8, 10
    4
    Triplet is 4, 10, 8
    0
    Triplet is 4, 8, 10
    66
    Triplet is 4, 8, 10
    67using5

    Triplet is 4, 8, 10
    4
    Triplet is 4, 10, 8
    0
    Triplet is 4, 8, 10
    71
    How to print all triplets with given sum? 
    Refer Find all triplets with zero sum. 
    Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


    Làm thế nào để bạn đếm các bộ ba trong Python?

    res: = 0 ..

    Đối với i trong phạm vi 0 đến kích thước của num - 1, làm. Đối với J trong phạm vi I+1 đến kích thước của num - 1, làm. Đối với k trong phạm vi J+1 đến kích thước của num - 1, làm. if | nums [i] - nums [j] |

    trả lại res ..

    Triplet Python là gì?

    Khoa học dữ liệu thực tế sử dụng Python trong danh sách các số mà chúng tôi muốn tìm ra ba yếu tố nào có thể tham gia để đưa ra một khoản tiền nhất định. Chúng tôi gọi nó là một bộ ba. Và trong danh sách có thể có nhiều bộ ba như vậy. Ví dụ: tổng 10 có thể được tạo ra số mẫu 1,6,3 cũng như 1,5,4.three elements can join to give a certain sum. We call it a triplet. And in the list there can be many such triplets. For example, the sum 10 can be generated form numbers 1,6,3 as well as 1,5,4.

    Làm thế nào để bạn tìm thấy bộ ba trong một mảng python?

    Bộ ba có thể được tìm thấy bằng cách sử dụng kỹ thuật băm ...

    Đi qua mảng từ i = 0 đến n - 2 ..

    Tạo một bảng băm trống ..

    Traverse từ J = I+ 1 đến N -1 ..

    sum = mảng [i] + mảng [j].

    if (-sum) có mặt trong bảng băm,.

    sau đó in mảng [i], mảng [j] và -sum dưới dạng ba lần ..

    Khác, chèn ARR [J] trong bảng băm và tiến hành ..

    Triplet trong lập trình là gì?

    Ngoài ra, được gọi là ba byte, một bộ ba là một byte được tạo thành từ ba bit thay vì tám bit tiêu chuẩn.a byte that is made up of three bits instead of the standard eight bits.