Minimum Cost to Change the Final Value of Expression LeetCode solution in Java

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class Solution {
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  Deque<int[]> stack = new ArrayDeque<>();
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  Deque<Character> opStack = new ArrayDeque<>();
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  public int minOperationsToFlip(String expression) {
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    for (char ch : expression.toCharArray()) {
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      if (ch == '(') {
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        opStack.push(ch);
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      } else if (ch == ')') {
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        while (opStack.peek() != '(') {
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          go();
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        }
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        opStack.pop(); // remove '('
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      } else if (ch == '&' || ch == '|') {
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        while (!opStack.isEmpty() && opStack.peek() != '(') {
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          go();
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        }
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        opStack.push(ch);
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      } else { // num
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        stack.push(new int[] {ch, 1});
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      }
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    }
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    while (!opStack.isEmpty()) {
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      go();
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    }
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    return stack.peek()[1];
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  }
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  private void go() {
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    stack.push(check(opStack.pop(), stack.pop(), stack.pop()));
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  }
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  private int[] check(char op, int[] r, int[] l) {
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    int[] ans;
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    int rval = r[0], rcost = r[1];
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    int lval = l[0], lcost = l[1];
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    if (op == '|' && rval == '0' && lval == '0') {
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      ans = new int[] {'0', Math.min(rcost, lcost)};
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    } else if (op == '|' && rval == '1' && lval == '0') {
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      ans = new int[] {'1', 1};
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    } else if (op == '|' && rval == '0' && lval == '1') {
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      ans = new int[] {'1', 1};
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    } else if (op == '|' && rval == '1' && lval == '1') {
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      ans = new int[] {'1', 1 + Math.min(rcost, lcost)};
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    } else if (op == '&' && rval == '0' && lval == '0') {
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      ans = new int[] {'0', 1 + Math.min(rcost, lcost)};
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    } else if (op == '&' && rval == '1' && lval == '0') {
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      ans = new int[] {'0', 1};
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    } else if (op == '&' && rval == '0' && lval == '1') {
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      ans = new int[] {'0', 1};
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    } else { // &, 1, 1
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      ans = new int[] {'1', Math.min(rcost, lcost)};
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    }
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    return ans;
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  }
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}