Minimum Runes to Add to Cast Spell

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ArraysStringsRecursion

You are a wizard, and you want to cast a powerful spell. To do this, you need to gather enough runes. Each rune has a value, and you want to find the minimum number of additional runes you need to add to reach a specific total value.

You are given a string word representing a mathematical expression containing digits, '+' (addition) and '=' (equal sign). You need to replace all the letters with digits from 0 to 9 such that the equation is valid. You are not allowed to have leading zeros.

Return the minimum number of runes to add to the string so that the equation is valid. If it is not possible to make the equation valid after adding runes, return -1.

A rune is a character that can be added to the string.

Note:

You must replace all occurrences of the same letter with the same digit.
You must replace different letters with different digits.
Each number must not have leading zeros.

Example 1:

Input: word = "SEND+MORE=MONEY"
Output: 1
Explanation: One possible solution is SEND = 9567, MORE = 1085, MONEY = 10652.
The length of MONEY is 6, and the length of SEND, MORE is 4. We need to add a rune to SEND or MORE.

Example 2:

Input: word = "SIX+SIX=SEVEN"
Output: 5
Explanation: One possible solution is SIX = 650, SEVEN = 10810. 
The length of SEVEN is 5, and the length of SIX is 3. We need to add 2 runes to SIX and 3 runes to SIX.

Example 3:

Input: word = "THIS+IS=TOO"
Output: 0
Explanation: One possible solution is THIS = 7806, IS = 06, TOO = 786. 
The equation is valid without adding any runes.

Example 4:

Input: word = "LEET+CODE=POINT"
Output: -1
Explanation: It is not possible to make the equation valid after adding runes.

Constraints:

3 <= word.length <= 35
word consists of digits, '+', '=', and uppercase English letters.
It is guaranteed that the equation contains exactly one '+' and one '='.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible characters in the input string, and are spaces allowed?
Can the integers in the expression be negative, and what is the maximum possible value for any integer?
Is an empty string a valid input, and if so, what should I return?
Does the input string always start with a number, and can it end with an operator?
If the expression is already valid, should I return 0?

Brute Force Solution

Approach

The brute force strategy for this problem is all about trying every possible combination. We'll systematically check all the ways we can modify the original string until we find one that satisfies the spell requirements.

Here's how the algorithm would work step-by-step:

Start by making the smallest possible change to the original string. This could be adding a single rune at the beginning.
Next, try adding a single rune at the second position, then the third, and so on, until you've tried adding a single rune at every possible spot.
Now, try adding two runes. Start by adding them at the beginning, then shift them along the string, considering all positions for both runes.
Continue this process, adding three runes, then four, and so on. For each number of runes to add, try all the possible positions where those runes could be placed.
For each modified string you create in the steps above, check if it now follows the spell rules.
Keep track of the strings that follow the rules, along with the number of runes you had to add to make them work.
Once you've tried all reasonable possibilities, pick the solution that required adding the fewest runes. That's your answer!

Code Implementation

def minimum_runes_to_add_brute_force(original_spell, spell_rules):    minimum_runes_added = float('inf')
    rune_options = ['A', 'B', 'C']
    max_runes_to_add = 3
    for num_runes_added in range(max_runes_to_add + 1):
        possible_insert_locations = range(len(original_spell) + 1)
        if num_runes_added == 0:
            modified_spell = original_spell
            if spell_rules(modified_spell):
                return 0
        elif num_runes_added == 1:
            for insert_location in possible_insert_locations:
                for rune in rune_options:
                    modified_spell = original_spell[:insert_location] + rune + original_spell[insert_location:]
                    if spell_rules(modified_spell):
                        minimum_runes_added = min(minimum_runes_added, 1)
        elif num_runes_added == 2:
            for first_insert_location in possible_insert_locations:
                for second_insert_location in possible_insert_locations:
                    for first_rune in rune_options:
                        for second_rune in rune_options:
                            # Need to handle inserting at different indices
                            if first_insert_location <= second_insert_location:
                                temp_spell = original_spell[:first_insert_location] + first_rune + original_spell[first_insert_location:]
                                modified_spell = temp_spell[:second_insert_location + 1] + second_rune + temp_spell[second_insert_location + 1:]
                            else:
                                temp_spell = original_spell[:second_insert_location] + second_rune + original_spell[second_insert_location:]
                                modified_spell = temp_spell[:first_insert_location + 1] + first_rune + temp_spell[first_insert_location + 1:]
                            
                            if spell_rules(modified_spell):
                                minimum_runes_added = min(minimum_runes_added, 2)
        elif num_runes_added == 3:
            for first_insert_location in possible_insert_locations:
                for second_insert_location in possible_insert_locations:
                    for third_insert_location in possible_insert_locations:
                        for first_rune in rune_options:
                            for second_rune in rune_options:
                                for third_rune in rune_options:
                                    # Need to handle inserting at different indices
                                    temp_spell = original_spell[:first_insert_location] + first_rune + original_spell[first_insert_location:]
                                    if first_insert_location <= second_insert_location:
                                        second_spell = temp_spell[:second_insert_location + 1] + second_rune + temp_spell[second_insert_location + 1:]
                                    else:
                                        second_spell = temp_spell[:second_insert_location] + second_rune + temp_spell[second_insert_location:]
                                    if second_insert_location <= third_insert_location:
                                        modified_spell = second_spell[:third_insert_location + 2] + third_rune + second_spell[third_insert_location + 2:]
                                    else:
                                        modified_spell = second_spell[:third_insert_location] + third_rune + second_spell[third_insert_location:]

                                    # Check if the spell is now valid after changes
                                    if spell_rules(modified_spell):
                                        minimum_runes_added = min(minimum_runes_added, 3)
    # If no solution, return the largest possible number
    if minimum_runes_added == float('inf'):
        return -1
    return minimum_runes_added

Big(O) Analysis

Time Complexity

O(k * a^k * n^k) – The brute force solution tries adding runes until a valid spell is found. Let 'n' be the length of the original string, 'a' be the number of possible runes, and 'k' be the minimum number of runes added to make the string a valid spell. We iterate through adding 1 rune, 2 runes, and so on, up to k runes. For each number of runes 'i', we have 'n+i choose i' possible positions to add them. For each possible string with 'i' added runes, we also need to check whether it satisfies the spell rules, which is assumed to take constant time in terms of string length given that the prompt provides no details on spell rule checking. Therefore, the total time complexity can be approximated as the sum of 'a^i * (n+i choose i)' for i=1 to k, where 'a^i' represents the possible combinations of runes and '(n+i choose i)' possible positions. The time complexity is dominated by the term where i = k and approximated by O(k * a^k * n^k).

Space Complexity

O(N!) – The brute force solution, as described, explores adding runes at every possible position and in every possible combination. The number of modified strings generated grows factorially with the number of runes added. In the worst case, if many runes need to be added to satisfy the spell rules, the algorithm has to store all the possible modified strings. Therefore, the auxiliary space required to store these modified strings in memory becomes proportional to N!, where N represents the maximum number of runes to add before a valid spell is found, leading to a space complexity of O(N!).

Optimal Solution

Approach

The goal is to find the fewest extra symbols needed to make mathematical equations true. The optimal approach focuses on analyzing the difference between the left and right sides of each equation and strategically adding symbols to balance them.

Here's how the algorithm would work step-by-step:

First, look at each equation and calculate the difference between the result on the right side and the evaluated value on the left side.
Next, for each equation, determine the minimum number of symbols required to bridge the gap between the sides.
When considering what symbol to add, focus on finding the factors or divisors of the difference between the sides. These factors can often be added as multiplication or division to reduce the difference more efficiently than simple addition or subtraction.
After adding symbols, recalculate the equation. If the equation is not balanced, repeat the symbol adding process until both sides are equal.
Keep a running count of the total symbols added across all equations to report the final result.

Code Implementation

def minimum_runes_to_add(equations):
    total_runes_added = 0
    for equation in equations:
        left_side, right_side = equation.split('=')
        try:
            left_value = eval(left_side)
            right_value = int(right_side)
        except (SyntaxError, NameError, TypeError, ValueError):
            return -1

        difference = right_value - left_value

        while difference != 0:
            # Need to balance the equation.
            total_runes_added += 1
            
            if difference > 0:
                # Add to the left side to increase its value
                if difference > 1:
                  left_side += '+ {}'.format(1)
                else:
                  left_side += '+{}'.format(1)
            else:
                 # Subtract from the left side to decrease its value
                if difference < -1:
                  left_side += '- {}'.format(1)
                else:
                  left_side += '-{}'.format(1)

            try:
                left_value = eval(left_side)
            except (SyntaxError, NameError, TypeError, ValueError):
                return -1
            
            difference = right_value - left_value
            
    return total_runes_added

Big(O) Analysis

Time Complexity

O(D * F) – The primary loop iterates through each equation (D equations) to evaluate and balance. Within each equation, we calculate the difference and find factors. The factor finding operation's complexity depends on the magnitude of this difference, but can be simplified to consider the number of potential factors F. The number of iterations needed to make each equation balance also depends on the magnitude of difference and the effectiveness of the factors found. Thus, the overall complexity is approximately O(D * F) where D is the number of equations and F represents the maximum number of factors considered across all equations, and iterations needed to balance.

Space Complexity

O(1) – The algorithm primarily calculates the difference between equation sides and adds symbols iteratively. The space used mainly consists of storing the difference, the number of symbols added for each equation, and a running count of total symbols. These variables consume constant space regardless of the number of equations (N) in the input. Thus, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input string	Return 0, as no runes need to be added to an empty expression.
String containing only a number (e.g., "123")	Return 0, as a single number is considered a valid, albeit simple, formula.
String containing only operators (e.g., "++**")	Return the length of the string since each character must be replaced by a number.
String starting with an operator (e.g., "+123")	Add one rune at the beginning, ensuring the formula starts with a number.
String ending with an operator (e.g., "123+")	Add one rune at the end, ensuring the formula ends with a number.
Consecutive numbers or operators (e.g., "1234" or "++")	Add runes in between to alternate numbers and operators, counting the necessary additions.
String with leading or trailing whitespace.	Trim the whitespace before processing the string to avoid incorrect character analysis.
Very long input string potentially causing integer overflow during length calculations.	Use appropriate data types (e.g., long) to store lengths and intermediate results to prevent overflow.

Null or empty input string

How to Handle:

Return 0, as no runes need to be added to an empty expression.

String containing only a number (e.g., "123")

How to Handle:

Return 0, as a single number is considered a valid, albeit simple, formula.

String containing only operators (e.g., "++**")

How to Handle:

Return the length of the string since each character must be replaced by a number.

String starting with an operator (e.g., "+123")

How to Handle:

Add one rune at the beginning, ensuring the formula starts with a number.

String ending with an operator (e.g., "123+")

How to Handle:

Add one rune at the end, ensuring the formula ends with a number.

Consecutive numbers or operators (e.g., "1234" or "++")

How to Handle:

Add runes in between to alternate numbers and operators, counting the necessary additions.

String with leading or trailing whitespace.

How to Handle:

Trim the whitespace before processing the string to avoid incorrect character analysis.

Very long input string potentially causing integer overflow during length calculations.

How to Handle:

Use appropriate data types (e.g., long) to store lengths and intermediate results to prevent overflow.