Given three integers N, M and X, the task is to find the number of ways to form an array, such that all consecutive numbers of the array are distinct, and the value at any index of the array from 2 to N – 1(Considering 1 based indexing) lies between 1 and M, while the value at index 1 is X and the value at index N is 1.
Note: Value of X lies between 1 and M.
Input: N = 4, M = 3, X = 2
The following arrays are possible:
1) 2, 1, 2, 1
2) 2, 1, 3, 1
3) 2, 3, 2, 1
Input: N = 2, M = 3, X = 2
The only possible array is: 2, 1
Approach: The problem can be solved using Dynamic Programming. Let, f(i) represent the number of ways to form the array till the ith index, such that every consecutive element of the array is distinct. Let f(i, One) represent the number of ways to form the array till the i-th index such that every consecutive element of the array is distinct and arri = 1.
Similarly, let f(i, Non-One) represent the number of ways to form the array till the ith index, such that every consecutive element of the array is distinct and arri is not equal to 1.
The following recurrence is formed:
f(i, Non-One) = f(i - 1, One) * (M - 1) + f(i - 1, Non-One) * (M - 2)
which means that the number of ways to form the array till the ith index with arrayi not equal to 1 is formed using two cases:
- If the number at arrayi – 1 is 1, then opt one number out of (M – 1) options to place at the ith index, since arrayi is not equal to 1.
- If the number at arrayi – 1 is not 1, then we need to opt one number out of (M – 2) options, since arrayi is not equal to 1 and arrayi ≠ arrayi – 1.
Similarly, f(i, One) = f(i – 1, Non-One), since the number of ways to form the array till the ith index with arrayi = 1, is same as number of ways to form the array till the (i – 1)th index with arrayi – 1 ≠ 1, thus at the ith index there is only one option. At the end the required answer if f(N, One) since arrayN needs to be equal to 1.
Below is the implementation of the above approach:
Time Complexity: O(N), where N is the size of the array
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