Leetcode Problem 1896. Minimum Cost to Change the Final Value of Expression

1896. Minimum Cost to Change the Final Value of Expression

Leetcode Solutions

Dynamic Programming with Stack

  1. Initialize a stack to keep track of sub-expressions and their costs.
  2. Iterate over the characters in the expression.
    • If the character is an operand ('0' or '1'), push its cost to change to 0 and 1 onto the stack.
    • If the character is an operator ('&' or '|'), push it onto the stack as a placeholder for future operations.
    • If the character is an opening parenthesis '(', push a special marker onto the stack to indicate the start of a sub-expression.
    • If the character is a closing parenthesis ')', pop elements from the stack until the corresponding opening parenthesis is found, evaluate the sub-expression, and push the result back onto the stack.
  3. After processing the entire expression, the stack will contain the result of the expression and the costs to change it to 0 and 1.
  4. The minimum cost to change the final value of the expression is the non-zero cost from the result on the stack.
UML Thumbnail

Recursive Depth-First Search

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