1137. N-th Tribonacci Number
Problem Description
The Tribonacci sequence is a variation of the Fibonacci sequence where each number is the sum of the three preceding ones. The sequence begins with T_0 = 0
, T_1 = 1
, and T_2 = 1
. For all integers n >= 0
, the n+3
rd number of the sequence is calculated as T_n+3 = T_n + T_n+1 + T_n+2
. The problem asks for the T_n
value of the sequence given an integer n
.
Intuition
The intuitive way to calculate the n
th Tribonacci number would be to use recursion or iteration, starting from the base cases T_0
, T_1
, and T_2
. However, this can be computationally expensive for larger n
due to repeated calculations.
To optimize, one can use matrix exponentiation, realizing that each Tribonacci number can be obtained by multiplying a transformation matrix with the vector of the three previous Tribonacci numbers. The transformation matrix is:
[(1, 1, 0), (1, 0, 1), (1, 0, 0)]
By successively squaring this matrix and reducing the power, we apply an exponentiation by squaring technique which significantly reduces the number of multiplications needed. We only multiply the res
vector by the transformation matrix when the current bit in the binary representation of n
is set (using the bitwise AND operation n & 1
). The operation n >>= 1
is used to right shift the bits of n
by one position to check the next bit in the next iteration. The intuition behind this approach comes from a realization that n
th power of the transformation matrix can be broken down into powers of 2, which allows the sequence to be computed in a time complexity that is logarithmic to n
instead of linear, making it much more efficient for large values of n
.
Learn more about Memoization, Math and Dynamic Programming patterns.
Solution Approach
The solution is implemented using a class named Solution
with a method tribonacci
that takes an integer n
as its parameter and returns the n
th Tribonacci number.
To calculate the Tribonacci number efficiently, the solution applies matrix exponentiation. Here is a step-by-step explanation of the implementation:
-
Base Cases: For
n == 0
, return0
. Forn < 3
(which coversn == 1
orn == 2
), return1
. -
Transformation Matrix: The
factor
variable is defined as a 3x3 transformation matrix usingnumpy
:factor = np.mat([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
Each multiplication of this matrix by a vector [T[n], T[n+1], T[n+2]] will result in [T[n+1], T[n+2], T[n+3]].
-
Result Vector: The
res
variable is defined as a 1x3 matrix to keep track of the current values of [T[n], T[n+1], T[n+2]]:res = np.mat([(1, 1, 0)], np.dtype("O"))
Initially, it's set to represent [T[0], T[1], T[2]] (0, 1, 1) before the loop.
-
Exponentiation by Squaring: Before entering the loop,
n
is decremented by 3 as the initialres
already accounts for the first three numbers. The loop then iterates, applying the following steps untiln
becomes 0:-
Conditional Multiplication: If the least significant bit of
n
is 1 (n & 1
),res
is multiplied by thefactor
matrix. -
Matrix Squaring: The
factor
matrix is squared which is equivalent to doubling the exponent. -
Right Shift: The bits of
n
are shifted to the right (n >>= 1
) to process the next bit in the next iteration of the loop.
The loop effectively multiplies
res
byfactor
raised to the power ofn
, but it does so in logarithmic time relative to the value ofn
. -
-
Final Sum: After the loop ends, the sum of the
res
array will contain the value of T[n], and this is returned as the final result:return res.sum()
In terms of data structures, the solution utilizes numpy
matrices for efficient manipulation and multiplication of large arrays, which is pivotal in matrix exponentiation. The use of bit manipulation to check individual bits of the integer n
is an efficient pattern often used in algorithms involving repetitive multiplication, as it greatly reduces the number of calculations required.
Ready to land your dream job?
Unlock your dream job with a 2-minute evaluator for a personalized learning plan!
Start EvaluatorExample Walkthrough
Let's use n = 4
as a small example to illustrate the solution approach.
-
Initial Setup: We are given the Tribonacci sequence with
T_0 = 0
,T_1 = 1
, andT_2 = 1
. -
Base Cases Handling: For
n = 4
, it's not less than 3, so we don't return the base cases directly. -
Transformation Matrix: We have a transformation matrix
factor
, which when multiplied by[T_n, T_{n+1}, T_{n+2}]
, gives[T_{n+1}, T_{n+2}, T_{n+3}]
. -
Result Vector Initialization: We define the initial
res
vector as[T_0, T_1, T_2]
, which is[0, 1, 1]
. -
Exponentiation by Squaring Process: Since
n = 4
andres
covers the first three numbers we deduct 3 fromn
. Nown
is1
. The iteration would proceed as follows:- Start the loop with
n = 1
. - Since
n & 1
is1
(4 & 1
equals1
), we multiply ourres
vector[0, 1, 1]
by ourfactor
matrix to get[1, 1, 0]
. - We square our
factor
matrix, which doesn't affect theres
vector for this small value ofn
. - Right-shift
n
by 1 (n >>= 1
). Nown
is0
, which means our loop will terminate.
- Start the loop with
-
Final Result: The loop has ended, and
res
vector is[1, 1, 0]
. The sum ofres
isT_4 = 1 + 1 + 0 = 2
.
Therefore, the 4th Tribonacci number T_4
is 2
.
Solution Implementation
1import numpy as np
2
3class Solution:
4 def tribonacci(self, n: int) -> int:
5 # Base case: return 0 for the zero-th element in the sequence.
6 if n == 0:
7 return 0
8
9 # Base cases: return 1 for the first or second element in the sequence.
10 if n < 3:
11 return 1
12
13 # Define the transition matrix for the Tribonacci sequence.
14 # This matrix represents the relationship between successive elements.
15 transition_matrix = np.matrix([[1, 1, 0], [1, 0, 1], [1, 0, 0]], np.dtype("O"))
16
17 # Initialize the result matrix as the identity matrix aligned with the sequence order.
18 result_matrix = np.matrix([[1, 1, 0]], np.dtype("O"))
19 n -= 3
20
21 # Use the Fast Exponentiation algorithm to raise the matrix to the power of (n-3).
22 while n:
23 # When the current power is odd, multiply result_matrix with transition_matrix.
24 if n & 1:
25 result_matrix *= transition_matrix
26
27 # Square the transition_matrix to get the next higher power of 2.
28 transition_matrix *= transition_matrix
29
30 # Shift right to divide n by 2, flooring it.
31 n >>= 1
32
33 # Return the sum of the elements in the resulting matrix as the answer.
34 return result_matrix.sum()
35
1class Solution {
2 // Calculates the n-th Tribonacci number
3 public int tribonacci(int n) {
4 // Base cases for n = 0, 1, 2
5 if (n == 0) {
6 return 0;
7 }
8 if (n < 3) {
9 return 1;
10 }
11 // Transformation matrix for Tribonacci sequence
12 int[][] transformationMatrix = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
13 // Calculate the power of the matrix to (n-3), since we know the first three values
14 int[][] resultingMatrix = matrixPower(transformationMatrix, n - 3);
15 // Initialize answer
16 int answer = 0;
17 // Add the top row of the matrix to get the answer
18 for (int element : resultingMatrix[0]) {
19 answer += element;
20 }
21 return answer;
22 }
23
24 // Multiplies two matrices and returns the result
25 private int[][] matrixMultiply(int[][] a, int[][] b) {
26 int rows = a.length, cols = b[0].length;
27 int[][] resultMatrix = new int[rows][cols];
28 // Perform matrix multiplication
29 for (int i = 0; i < rows; ++i) {
30 for (int j = 0; j < cols; ++j) {
31 for (int k = 0; k < b.length; ++k) {
32 resultMatrix[i][j] += a[i][k] * b[k][j];
33 }
34 }
35 }
36 return resultMatrix;
37 }
38
39 // Calculates the matrix exponentiation of matrix 'a' raised to the power of 'n'
40 private int[][] matrixPower(int[][] a, int n) {
41 // Create an identity matrix for initial result
42 int[][] result = {{1, 1, 0}};
43 // Loop to do binary exponentiation
44 while (n > 0) {
45 // Multiply with 'a' when the least significant bit is 1
46 if ((n & 1) == 1) {
47 result = matrixMultiply(result, a);
48 }
49 // Square the matrix 'a'
50 a = matrixMultiply(a, a);
51 // Right shift 'n' to process the next bit
52 n >>= 1;
53 }
54 return result;
55 }
56}
57
1#include <vector>
2#include <numeric>
3
4class Solution {
5public:
6 // Computes the n-th Tribonacci number using matrix exponentiation
7 int tribonacci(int n) {
8 // Base case when n is 0
9 if (n == 0) {
10 return 0;
11 }
12 // Base cases for n being 1 or 2
13 if (n < 3) {
14 return 1;
15 }
16 // Initial transformation matrix
17 std::vector<std::vector<long long>> transformationMatrix = {
18 {1, 1, 0},
19 {1, 0, 1},
20 {1, 0, 0}
21 };
22 // Raise the transformation matrix to the power (n - 3)
23 std::vector<std::vector<long long>> resultMatrix = matrixPower(transformationMatrix, n - 3);
24 // The sum of the first row of the result matrix gives us the nth Tribonacci number
25 return std::accumulate(resultMatrix[0].begin(), resultMatrix[0].end(), 0);
26 }
27
28private:
29 using ll = long long; // Alias for long long type for easier reading
30
31 // Multiplies two matrices and returns the result
32 std::vector<std::vector<ll>> matrixMultiply(std::vector<std::vector<ll>>& a, std::vector<std::vector<ll>>& b) {
33 int rows = a.size(), columns = b[0].size();
34 std::vector<std::vector<ll>> product(rows, std::vector<ll>(columns));
35
36 for (int i = 0; i < rows; ++i) {
37 for (int j = 0; j < columns; ++j) {
38 for (int k = 0; k < b.size(); ++k) {
39 product[i][j] += a[i][k] * b[k][j];
40 }
41 }
42 }
43 return product;
44 }
45
46 // Calculates the power of a matrix to the given exponent n
47 std::vector<std::vector<ll>> matrixPower(std::vector<std::vector<ll>>& a, int n) {
48 // Starting with the identity matrix of size 3x3
49 std::vector<std::vector<ll>> result = {
50 {1, 0, 0},
51 {0, 1, 0},
52 {0, 0, 1}
53 };
54
55 while (n) {
56 if (n & 1) { // If n is odd, multiply the result by the current matrix a
57 result = matrixMultiply(result, a);
58 }
59 a = matrixMultiply(a, a); // Square the matrix a
60 n >>= 1; // Divide n by 2
61 }
62 return result;
63 }
64};
65
1// Calculates the n-th Tribonacci number using matrix exponentiation.
2function tribonacci(n: number): number {
3 // Base cases for n = 0, 1, and 2.
4 if (n === 0) {
5 return 0;
6 }
7 if (n < 3) {
8 return 1;
9 }
10
11 // Matrix representation of the Tribonacci relation.
12 const tribonacciMatrix = [
13 [1, 1, 0],
14 [1, 0, 1],
15 [1, 0, 0],
16 ];
17
18 // Raise the matrix to the (n-3)-th power and sum the top row for the result.
19 return matrixPower(tribonacciMatrix, n - 3)[0].reduce((sum, element) => sum + element);
20}
21
22// Multiplies two matrices and returns the result.
23function multiplyMatrices(matrix1: number[][], matrix2: number[][]): number[][] {
24 const rows = matrix1.length;
25 const cols = matrix2[0].length;
26 // Create an empty matrix for the result.
27 const result = Array.from({ length: rows }, () => Array.from({ length: cols }, () => 0));
28
29 // Calculate the product of the two matrices.
30 for (let i = 0; i < rows; ++i) {
31 for (let j = 0; j < cols; ++j) {
32 for (let k = 0; k < matrix2.length; ++k) {
33 result[i][j] += matrix1[i][k] * matrix2[k][j];
34 }
35 }
36 }
37
38 return result;
39}
40
41// Raises a matrix to the power of n and returns the result.
42function matrixPower(matrix: number[][], n: number): number[][] {
43 // Initialize the result as an identity matrix in the shape of a 1 x 3 matrix.
44 let result = [
45 [1, 1, 0],
46 ];
47
48 while (n > 0) {
49 // If the exponent is odd, multiply the result by the matrix.
50 if (n & 1) {
51 result = multiplyMatrices(result, matrix);
52 }
53 // Square the matrix.
54 matrix = multiplyMatrices(matrix, matrix);
55 // Halve the exponent by right-shifting it.
56 n >>= 1;
57 }
58
59 return result;
60}
61
Time and Space Complexity
The given Python code implements a fast matrix exponentiation approach to find the n
-th number in the Tribonacci sequence. The Tribonacci sequence is a generalization of the Fibonacci sequence, where each number is the sum of the preceding three numbers in the sequence.
Time Complexity
The time complexity of this solution mainly depends on the matrix exponentiation part of the algorithm. The algorithm uses a loop that runs in O(log n) time, which accounts for the repeated squaring of the matrix factor
. In each iteration of the loop, the matrix is either multiplied by itself (squaring) which takes O(1) time since the matrix size is fixed at 3x3, or the result matrix res
is multiplied by factor
, which also takes O(1) time for the same reason. Thus, the overall time complexity of the algorithm is O(log n)
.
Space Complexity
The space complexity of the algorithm is determined by the size of the matrices used for the calculations. The matrices factor
and res
have a constant size of 3x3 and 1x3, respectively, so the space used does not depend on n
. Consequently, the space complexity is O(1)
, which means it uses constant space.
Learn more about how to find time and space complexity quickly using problem constraints.
Which of the tree traversal order can be used to obtain elements in a binary search tree in sorted order?
Recommended Readings
Memoization Prereq Backtracking problems backtracking Memoization is a fancy word for a simple concept so is the case for a lot of things we learn in school It means saving the previous function call result in a dictionary and reading from it when we do the exact same call again
Math for Technical Interviews How much math do I need to know for technical interviews The short answer is about high school level math Computer science is often associated with math and some universities even place their computer science department under the math faculty However the reality is that you
What is Dynamic Programming Prerequisite DFS problems dfs_intro Backtracking problems backtracking Memoization problems memoization_intro Pruning problems backtracking_pruning Dynamic programming is an algorithmic optimization technique that breaks down a complicated problem into smaller overlapping sub problems in a recursive manner and uses solutions to the sub problems to construct a solution
Want a Structured Path to Master System Design Too? Don’t Miss This!