LeetCode #1896 — HARD

Minimum Cost to Change the Final Value of Expression

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given a valid boolean expression as a string expression consisting of the characters '1','0','&' (bitwise AND operator),'|' (bitwise OR operator),'(', and ')'.

  • For example, "()1|1" and "(1)&()" are not valid while "1", "(((1))|(0))", and "1|(0&(1))" are valid expressions.

Return the minimum cost to change the final value of the expression.

  • For example, if expression = "1|1|(0&0)&1", its value is 1|1|(0&0)&1 = 1|1|0&1 = 1|0&1 = 1&1 = 1. We want to apply operations so that the new expression evaluates to 0.

The cost of changing the final value of an expression is the number of operations performed on the expression. The types of operations are described as follows:

  • Turn a '1' into a '0'.
  • Turn a '0' into a '1'.
  • Turn a '&' into a '|'.
  • Turn a '|' into a '&'.

Note: '&' does not take precedence over '|' in the order of calculation. Evaluate parentheses first, then in left-to-right order.

Example 1:

Input: expression = "1&(0|1)"
Output: 1
Explanation: We can turn "1&(0|1)" into "1&(0&1)" by changing the '|' to a '&' using 1 operation.
The new expression evaluates to 0.

Example 2:

Input: expression = "(0&0)&(0&0&0)"
Output: 3
Explanation: We can turn "(0&0)&(0&0&0)" into "(0|1)|(0&0&0)" using 3 operations.
The new expression evaluates to 1.

Example 3:

Input: expression = "(0|(1|0&1))"
Output: 1
Explanation: We can turn "(0|(1|0&1))" into "(0|(0|0&1))" using 1 operation.
The new expression evaluates to 0.

Constraints:

  • 1 <= expression.length <= 105
  • expression only contains '1','0','&','|','(', and ')'
  • All parentheses are properly matched.
  • There will be no empty parentheses (i.e: "()" is not a substring of expression).
Patterns Used
📊Dynamic Programming→📚Monotonic Stack→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a valid boolean expression as a string expression consisting of the characters '1','0','&' (bitwise AND operator),'|' (bitwise OR operator),'(', and ')'. For example, "()1|1" and "(1)&()" are not valid while "1", "(((1))|(0))", and "1|(0&(1))" are valid expressions. Return the minimum cost to change the final value of the expression. For example, if expression = "1|1|(0&0)&1", its value is 1|1|(0&0)&1 = 1|1|0&1 = 1|0&1 = 1&1 = 1. We want to apply operations so that the new expression evaluates to 0. The cost of changing the final value of an expression is the number of operations performed on the expression. The types of operations are described as follows: Turn a '1' into a '0'. Turn a '0' into a '1'. Turn a '&' into a '|'. Turn a '|' into a '&'. Note: '&' does not take precedence over '|' in the order of calculation. Evaluate parentheses first, then in left-to-right

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming · Stack

Example 1

"1&(0|1)"

Example 2

"(0&0)&(0&0&0)"

Example 3

"(0|(1|0&1))"
Step 02

Core Insight

What unlocks the optimal approach

  • How many possible states are there for a given expression?
  • Is there a data structure that we can use to solve the problem optimally?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1896: Minimum Cost to Change the Final Value of Expression
class Solution {
  public int minOperationsToFlip(String expression) {
    // [(the expression, the cost to toggle the expression)]
    Deque<Pair<Character, Integer>> stack = new ArrayDeque<>();
    Pair<Character, Integer> lastPair = null;

    for (final char e : expression.toCharArray()) {
      if (e == '(' || e == '&' || e == '|') {
        // These aren't expressions, so the cost is meaningless.
        stack.push(new Pair<>(e, 0));
        continue;
      }
      if (e == ')') {
        lastPair = stack.pop();
        stack.pop(); // Pop '('.
      } else {       // e == '0' || e == '1'
        // Store the '0' or '1'. The cost to change their values is just 1,
        // whether it's changing '0' to '1' or '1' to '0'.
        lastPair = new Pair<>(e, 1);
      }
      if (!stack.isEmpty() && (stack.peek().getKey() == '&' || stack.peek().getKey() == '|')) {
        final char op = stack.pop().getKey();
        final char a = stack.peek().getKey();
        final int costA = stack.pop().getValue();
        final char b = lastPair.getKey();
        final int costB = lastPair.getValue();
        // Determine the cost to toggle op(a, b).
        if (op == '&') {
          if (a == '0' && b == '0')
            // Change '&' to '|' and a|b to '1'.
            lastPair = new Pair<>('0', 1 + Math.min(costA, costB));
          else if (a == '0' && b == '1')
            // Change '&' to '|'.
            lastPair = new Pair<>('0', 1);
          else if (a == '1' && b == '0')
            // Change '&' to '|'.
            lastPair = new Pair<>('0', 1);
          else // a == '1' and b == '1'
            // Change a|b to '0'.
            lastPair = new Pair<>('1', Math.min(costA, costB));
        } else { // op == '|'
          if (a == '0' && b == '0')
            // Change a|b to '1'.
            lastPair = new Pair<>('0', Math.min(costA, costB));
          else if (a == '0' && b == '1')
            // Change '|' to '&'.
            lastPair = new Pair<>('1', 1);
          else if (a == '1' && b == '0')
            // Change '|' to '&'.
            lastPair = new Pair<>('1', 1);
          else // a == '1' and b == '1'
            // Change '|' to '&' and a|b to '0'.
            lastPair = new Pair<>('1', 1 + Math.min(costA, costB));
        }
      }
      stack.push(lastPair);
    }

    return stack.peek().getValue();
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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