Prison Escape States: Counting Possible Scenarios with O(n) Algorithm

1. Problem Statement:

There are N rooms in a prison, one for each prisoner, and M religions. Each prisoner follows one of these religions. An escape can occur if prisoners in adjacent rooms share the same religion. Design an O(n) algorithm to find the number of escape states possible. For example, with 3 rooms and 2 religions, there are 6 different escape states.

2. Algorithm:

• Initialization:

  • Set a counter variable 'states' to 0.
  • Create an array 'religions' of size N to store the religion of each prisoner.

• Iteration:

  • Iterate through the 'religions' array from index 1 to N-1.
  • For each index 'i':
    • Check if the religion of the current prisoner (religions[i]) is the same as the religion of the previous prisoner (religions[i-1]).
    • If they match, increment 'states' by 1.

• Return:

  • Return the value of 'states'.

Pseudo-code:

function countEscapeStates(N, M):
    states = 0
    religions = new Array of size N
    
    for i from 1 to N-1:
        if religions[i] == religions[i-1]:
            states = states + 1
    
    return states

3. Correctness Proof:

The algorithm iterates through the 'religions' array, comparing adjacent prisoners' religions. Each time adjacent prisoners share the same religion, an escape state is identified, and the 'states' counter is incremented. This ensures that the algorithm accurately counts all possible escape states based on the given condition.

4. Complexity Analysis:

• The algorithm iterates through the 'religions' array once, which has a size of N.
• The operations performed inside the loop are constant time.
• Therefore, the overall time complexity of the algorithm is O(N), which is equivalent to O(n) since N represents the number of rooms (prisoners).