Math Expression Evaluator (MEE)

I think the use of a full-fledge Math Expression Evaluator (MEE) might be too overkill for solving this particular problem, but it is always good to have an MEE in your portfolio of tools in case it is needed.

The hard part of this problem is actually the parsing of the expression to create the final one - not actually hard, just a little annoying with the placement of the indexes. Once you have it, use the MEE and save the smallest val. Code is down below, cheers, ACC.

Minimize Result by Adding Parentheses to Expression - LeetCode

2232. Minimize Result by Adding Parentheses to Expression
Medium

You are given a 0-indexed string expression of the form "<num1>+<num2>" where <num1> and <num2> represent positive integers.

Add a pair of parentheses to expression such that after the addition of parentheses, expression is a valid mathematical expression and evaluates to the smallest possible value. The left parenthesis must be added to the left of '+' and the right parenthesis must be added to the right of '+'.

Return expression after adding a pair of parentheses such that expression evaluates to the smallest possible value. If there are multiple answers that yield the same result, return any of them.

The input has been generated such that the original value of expression, and the value of expression after adding any pair of parentheses that meets the requirements fits within a signed 32-bit integer.

Example 1:

Input: expression = "247+38"
Output: "2(47+38)"
Explanation: The expression evaluates to 2 * (47 + 38) = 2 * 85 = 170.
Note that "2(4)7+38" is invalid because the right parenthesis must be to the right of the '+'.
It can be shown that 170 is the smallest possible value.

Example 2:

Input: expression = "12+34"
Output: "1(2+3)4"
Explanation: The expression evaluates to 1 * (2 + 3) * 4 = 1 * 5 * 4 = 20.

Example 3:

Input: expression = "999+999"
Output: "(999+999)"
Explanation: The expression evaluates to 999 + 999 = 1998.

Constraints:

  • 3 <= expression.length <= 10
  • expression consists of digits from '1' to '9' and '+'.
  • expression starts and ends with digits.
  • expression contains exactly one '+'.
  • The original value of expression, and the value of expression after adding any pair of parentheses that meets the requirements fits within a signed 32-bit integer.

public class Solution {
 public string MinimizeResult(string expression)
 {
 double smallest = Double.MaxValue;
 string retVal = "";
 int midIndex = expression.IndexOf('+');
 string firstHalf = expression.Substring(0, midIndex);
 string secondHalf = expression.Substring(midIndex + 1);
 for (int i = 0; i < firstHalf.Length; i++)
 {
 string exp1 = "";
 if (i == 0)
 {
 exp1 = "(" + firstHalf;
 }
 else
 {
 exp1 = firstHalf.Substring(0, i) + "*(" + firstHalf.Substring(i);
 }
 for (int j = secondHalf.Length - 1; j >= 0; j--)
 {
 string exp2 = "";
 if (j == secondHalf.Length - 1)
 {
 exp2 = secondHalf + ")";
 }
 else
 {
 exp2 = secondHalf.Substring(0, j + 1) + ")*" + secondHalf.Substring(j + 1);
 }
 string finalExpression = exp1 + "+" + exp2;
 double current = 0;
 MathExpressionEvaluator(finalExpression, out current);
 if (current < smallest)
 {
 smallest = current;
 retVal = finalExpression;
 }
 }
 }
 return retVal.Replace("*", "");
 }
 
 public static bool MathExpressionEvaluator(string expression,
 out double result)
 {
 result = 0;
 if (expression == null)
 {
 return false;
 }
 if (expression.Trim().Length == 0)
 {
 return true;
 }
 if (expression.Length > 2 &&
 expression.StartsWith("(") &&
 expression.EndsWith(")") &&
 IsWellFormedBrackets(expression.Substring(1, expression.Length - 2)))
 {
 return MathExpressionEvaluator(expression.Substring(1, expression.Length - 2), out result);
 }
 if (Double.TryParse(expression, out result))
 {
 return true;
 }
 char[] operators = new char[] { '+', '-', '*', '/' };
 foreach (char op in operators)
 {
 int bracketsCount = 0;
 for (int i = expression.Length - 1; i >= 0; i--)
 {
 if (expression[i] == '(')
 {
 bracketsCount++;
 }
 else if (expression[i] == ')')
 {
 bracketsCount--;
 }
 if (expression[i] == op &&
 bracketsCount == 0)
 {
 double value1 = 0;
 double value2 = 0;
 string part1 = "";
 string part2 = "";
 if (i > 0)
 {
 part1 = expression.Substring(0, i);
 }
 if (i + 1 < expression.Length)
 {
 part2 = expression.Substring(i + 1);
 }
 if (!MathExpressionEvaluator(part1, out value1))
 {
 return false;
 }
 if (!MathExpressionEvaluator(part2, out value2))
 {
 return false;
 }
 switch (op)
 {
 case '+':
 result = value1 + value2;
 break;
 case '-':
 result = value1 - value2;
 break;
 case '*':
 result = value1 * value2;
 break;
 case '/':
 if (value2 == 0)
 {
 return false;
 }
 result = value1 / value2;
 break;
 }
 return true;
 }
 }
 }
 return false;
 }
 public static bool IsWellFormedBrackets(string expression)
 {
 if (expression == null ||
 expression.Trim().Length == 0)
 {
 return true;
 }
 int bracketsCount = 0;
 for (int i = 0; i < expression.Length; i++)
 {
 if (expression[i] == '(')
 {
 bracketsCount++;
 }
 else if (expression[i] == ')')
 {
 bracketsCount--;
 }
 if (bracketsCount < 0)
 {
 return false;
 }
 }
 return bracketsCount == 0;
 }
}

Comments

Post a Comment

[フレーム]

Popular posts from this blog

Quasi FSM (Finite State Machine) problem + Vibe

Not really an FSM problem since the state isn't changing, it is just defined by the current input. Simply following the instructions should do it. Using VSCode IDE you can also engage the help of Cline or Copilot for a combo of coding and vibe coding, see below screenshot. Cheers, ACC. Process String with Special Operations I - LeetCode You are given a string  s  consisting of lowercase English letters and the special characters:  * ,  # , and  % . Build a new string  result  by processing  s  according to the following rules from left to right: If the letter is a  lowercase  English letter append it to  result . A  '*'   removes  the last character from  result , if it exists. A  '#'   duplicates  the current  result  and  appends  it to itself. A  '%'   reverses  the current  result . Return the final string  result  after processing all char...

Shortest Bridge – A BFS Story (with a Twist)

Here's another one from the Google 30 Days challenge on LeetCode — 934. Shortest Bridge . The goal? Given a 2D binary grid where two islands (groups of 1s) are separated by water (0s), flip the fewest number of 0s to 1s to connect them. Easy to describe. Sneaky to implement well. 🧭 My Approach My solution follows a two-phase Breadth-First Search (BFS) strategy: Find and mark one island : I start by scanning the grid until I find the first 1 , then use BFS to mark all connected land cells as 2 . I store their positions for later use. Bridge-building BFS : For each cell in the marked island, I run a BFS looking for the second island. Each BFS stops as soon as it hits a cell with value 1 . The minimum distance across all these searches gives the shortest bridge. πŸ” Code Snippet Here's the core logic simplified: public int ShortestBridge(int[][] grid) { // 1. Mark one island as '2' and gather its coordinates List<int> island = FindAndMark...

Classic Dynamic Programming IX

A bit of vibe code together with OpenAI O3. I asked O3 to just generate the sieve due to laziness. Sieve is used to calculate the first M primes (when I was using Miller-Rabin, was giving me TLE). The DP follows from that in a straightforward way: calculate the numbers from i..n-1, then n follows by calculating the min over all M primes. Notice that I made use of Goldbach's Conjecture as a way to optimize the code too. Goldbach's Conjecture estates that any even number greater than 2 is the sum of 2 primes. The conjecture is applied in the highlighted line. Cheers, ACC. PS: the prompt for the sieve was the following, again using Open AI O3 Advanced Reasoning: " give me a sieve to find the first M prime numbers in C#. The code should produce a List<int> with the first M primes " Minimum Number of Primes to Sum to Target - LeetCode You are given two integers  n  and  m . You have to select a multiset of  prime numbers  from the  first   m  pri...