Program to find nth sequence after following the given string sequence rules in Python

Suppose we have two strings s, t and another positive number n is also given, we have to find the nth term of the sequence A where:

• A[0] = s
• A[1] = t
• A[n] = A[n - 1] + A[n - 2] when n is even, otherwise A[n] = A[n - 2] + A[n - 1]

As an example, if s = "a" and t = "b", then the sequence A would be: ["a", "b", "ba" ("b" + "a"), "bba" ("b" + "ba"), "bbaba" ("bba" + "ba")]

So, if the input is like s = "pk", t = "r", n = 4, then the output will be "rrpkrpk"

Algorithm

To solve this, we will follow these steps:

• If n is 0, return s
• If n is 1, return t
• Initialize a = s, b = t
• For i from 2 to n:
  • If i is even, concatenate: c = b + a
  • If i is odd, concatenate: c = a + b
  • Update: a = b, b = c
• Return c

Example

class Solution:
    def solve(self, s, t, n):
        if n == 0:
            return s
        elif n == 1:
            return t
        
        a = s
        b = t
        
        for i in range(2, n + 1):
            if i % 2 == 0:
                c = b + a
            else:
                c = a + b
            a = b
            b = c
        
        return c

# Test the solution
ob = Solution()
print(ob.solve("pk", "r", 4))

rrpkrpk

Step-by-Step Execution

Let's trace through the example with s = "pk", t = "r", n = 4:

def trace_sequence(s, t, n):
    print(f"A[0] = {s}")
    print(f"A[1] = {t}")
    
    if n <= 1:
        return
    
    a, b = s, t
    
    for i in range(2, n + 1):
        if i % 2 == 0:
            c = b + a
            print(f"A[{i}] = A[{i-1}] + A[{i-2}] = {b} + {a} = {c}")
        else:
            c = a + b
            print(f"A[{i}] = A[{i-2}] + A[{i-1}] = {a} + {b} = {c}")
        a, b = b, c

trace_sequence("pk", "r", 4)

A[0] = pk
A[1] = r
A[2] = A[1] + A[0] = r + pk = rpk
A[3] = A[1] + A[2] = r + rpk = rrpk
A[4] = A[3] + A[2] = rrpk + rpk = rrpkrpk

Alternative Implementation

Here's a more compact version without using a class:

def find_nth_sequence(s, t, n):
    if n == 0:
        return s
    if n == 1:
        return t
    
    prev2, prev1 = s, t
    
    for i in range(2, n + 1):
        if i % 2 == 0:
            current = prev1 + prev2
        else:
            current = prev2 + prev1
        prev2, prev1 = prev1, current
    
    return current

# Test cases
test_cases = [
    ("pk", "r", 4),
    ("a", "b", 3),
    ("x", "y", 0),
    ("hello", "world", 2)
]

for s, t, n in test_cases:
    result = find_nth_sequence(s, t, n)
    print(f"s='{s}', t='{t}', n={n} ? '{result}'")

s='pk', t='r', n=4 ? 'rrpkrpk'
s='a', t='b', n=3 ? 'ab'
s='x', t='y', n=0 ? 'x'
s='hello', t='world', n=2 ? 'worldhello'

Conclusion

This problem follows a modified Fibonacci-like pattern where string concatenation order depends on whether the index is even or odd. The solution uses iterative approach with O(n) time complexity to build the sequence step by step.