LeetCode 1896: Minimum Cost to Change the Final Value of Expression Solution

Master LeetCode problem 1896 (Minimum Cost to Change the Final Value of Expression), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1896. Minimum Cost to Change the Final Value of Expression

Hard

Math String Dynamic Programming Stack

Problem Explanation

Explanation

To solve this problem, we can use a recursive approach to evaluate the expression and determine the cost of changing the final value to 0. We will consider different cases for each operator and recursively evaluate the sub-expressions. We will track the cost of changing the value for each sub-expression and return the minimum cost.

For each operator, we will consider changing the value to 0 or 1 and recursively evaluate the sub-expressions.
We will keep track of the cost of changing the value for each sub-expression.
We will return the minimum cost among the different possibilities.

Time complexity: O(n) where n is the length of the expression Space complexity: O(n) for the recursive call stack

Solution Code

class Solution {
    public int minOperationsToFlip(String expression) {
        return evaluateExpression(expression, 0, expression.length() - 1, 0)[1];
    }
    
    private int[] evaluateExpression(String expression, int start, int end, int target) {
        if (start == end) {
            return new int[] {expression.charAt(start) == target ? 0 : 1, 1};
        }
        
        int[] res = new int[] {Integer.MAX_VALUE, 0};
        for (int i = start + 1; i < end; i++) {
            if (expression.charAt(i) == '&') {
                int[] left = evaluateExpression(expression, start, i - 1, 1);
                int[] right = evaluateExpression(expression, i + 1, end, 1);
                if (target == 1) {
                    res[0] = Math.min(res[0], Math.min(left[0] + right[0], left[1] + right[1]) + 1);
                    res[1] += left[1] + right[1];
                } else {
                    res[0] = Math.min(res[0], Math.min(left[0] + right[0), Math.min(left[1] + right[1] + 1);
                    res[1] += left[1] + right[1];
                }
            } else if (expression.charAt(i) == '|') {
                int[] left = evaluateExpression(expression, start, i - 1, 0);
                int[] right = evaluateExpression(expression, i + 1, end, 0);
                if (target == 1) {
                    res[0] = Math.min(res[0], Math.min(left[0] + right[0], left[1] + right[1] + 1));
                    res[1] += left[1] + right[1];
                } else {
                    res[0] = Math.min(res[0], Math.min(left[0] + right[0), Math.min(left[1] + right[1));
                    res[1] += left[1] + right[1];
                }
            }
        }
        
        return res;
    }
}

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