Program to determine the minimum cost to build a given string in python

Suppose we have to build a string str of length n. To build the string, we can perform two operations:

• Add a character to the end of str for cost a
• Add a substring that already exists in the current string for cost r

We need to calculate the minimum cost of building the string str using dynamic programming.

Example

If the input is a = 5, r = 4, str = 'tpoint', then the output will be 29.

To build the string 'tpoint', the cost breakdown is:

str = 't'; new character added, cost = 5
str = 'tp'; new character added, cost = 5  
str = 'tpo'; new character added, cost = 5
str = 'tpoi'; new character added, cost = 5
str = 'tpoin'; new character added, cost = 5
str = 'tpoint'; substring 't' copied, cost = 4

Total cost: 5 + 5 + 5 + 5 + 5 + 4 = 29

Algorithm

The approach follows these steps:

• Find the longest substring ending at each position that exists earlier in the string
• Use dynamic programming to calculate minimum cost at each position
• Choose between adding a new character (cost a) or copying a substring (cost r)

Implementation

def solve(a, r, string):
    size = len(string)
    largest = []
    low = 0
    
    # Find longest matching substring for each position
    for upp in range(1, size + 1):
        while string[low:upp] not in string[:low]:
            low += 1
        largest.append(upp - low)
    
    # Dynamic programming to find minimum cost
    cost = [a]  # Cost to build first character
    
    for i in range(1, size):
        if largest[i] == 0:
            # No matching substring, must add new character
            cost.append(cost[-1] + a)
        else:
            # Choose minimum between adding character or copying substring
            add_char = cost[-1] + a
            copy_substring = cost[i - largest[i]] + r
            cost.append(min(add_char, copy_substring))
    
    return cost[-1]

# Test the function
result = solve(5, 4, 'tpoint')
print(f"Minimum cost to build 'tpoint': {result}")

The output of the above code is:

Minimum cost to build 'tpoint': 29

How It Works

The algorithm works in two phases:

• Phase 1: For each position, find the longest substring ending at that position which already exists earlier in the string
• Phase 2: Use dynamic programming where cost[i] represents the minimum cost to build the string up to position i

At each position, we choose the minimum between:

• Adding a new character: cost[i-1] + a
• Copying a substring: cost[i-length] + r

Conclusion

This dynamic programming approach efficiently finds the minimum cost to build a string by choosing between adding new characters or reusing existing substrings. The time complexity is O(n³) due to substring operations.