Toggle Bitwise Expression - Google Top Interview Questions
Problem Statement :
You are given a string s representing a bitwise expression with the following characters: "0", "1", "&", "|", "(", and ")".
In one operation, you can:
Toggle "0" to "1"
Toggle "1" to "0"
Change "&" to "|"
Change "|" to "&"
Return the minimum number of operations required to toggle the value of the expression.
Note: the precedence of the operators don't matter, since they'll be wrapped in brackets when
necessary. For example, "1&0" and "1&(0&1)" are possible inputs but "1&0&1" will not occur.
Constraints
1 ≤ n ≤ 100,000 where n is the length of s
Example 1
Input
s = "0&(0&(0&0))"
Output
2
Explanation
The input expression evaluates to 0. To toggle it to 1, we can change the expression to "1|(0&(0&0))".
Example 2
Input
s = "(1|1)|(1|1)"
Output
3
Explanation
The input expression evaluates to 1. To toggle it to 0, we can change the expression to "(0&1)&(1|1)".
Example 3
Input
s = "0&0"
Output
2
Solution :
Solution in C++ :
int pairing[1000005];
unordered_map<long long, int> dp;
int force(string& s, int start, int end, int want) {
if (start == end) {
if (s[start] - '0' == want) return 0;
return 1;
}
long long key = (1000000LL * start) + (10 * end) + want;
if (dp.count(key)) return dp[key];
int ret = 1e9;
int lhs[2];
int rhs[2];
if (s[start] == '(' && s[end] == ')' && pairing[start] == end) {
return dp[key] = force(s, start + 1, end - 1, want);
}
if (s[start] == '(') {
int end = pairing[start];
lhs[0] = force(s, start + 1, end - 1, 0);
lhs[1] = force(s, start + 1, end - 1, 1);
start = end + 1;
} else {
int have = s[start++] - '0';
if (have == 0) {
lhs[0] = 0;
lhs[1] = 1;
} else {
lhs[0] = 1;
lhs[1] = 0;
}
}
char op = s[start++];
rhs[0] = force(s, start, end, 0);
rhs[1] = force(s, start, end, 1);
for (int a = 0; a < 2; a++) {
for (int b = 0; b < 2; b++) {
if ((a | b) == want) {
int cost = lhs[a] + rhs[b];
if (op == '&') cost++;
ret = min(ret, cost);
}
if ((a & b) == want) {
int cost = lhs[a] + rhs[b];
if (op == '|') cost++;
ret = min(ret, cost);
}
}
}
dp[key] = ret;
return ret;
}
int eval(string& s, int start, int end) {
assert(start <= end);
if (start == end) {
assert(s[start] == '0' || s[start] == '1');
return s[start] - '0';
}
if (s[start] == '(' && s[end] == ')' && pairing[start] == end) {
return eval(s, start + 1, end - 1);
}
int lhs;
if (s[start] == '(') {
int end = pairing[start];
lhs = eval(s, start + 1, end - 1);
start = end + 1;
} else {
lhs = s[start++] - '0';
}
char op = s[start++];
int rhs = eval(s, start, end);
if (op == '&') {
if (lhs == 0 || rhs == 0) return 0;
return 1;
}
if (op == '|') {
if (lhs == 1 || rhs == 1) return 1;
return 0;
}
assert(false);
}
int solve(string s) {
stack<int> paren;
for (int i = 0; i < s.size(); i++) pairing[i] = -1;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '(') {
paren.push(i);
} else if (s[i] == ')') {
int lhs = paren.top();
paren.pop();
pairing[lhs] = i;
pairing[i] = lhs;
}
}
int want = eval(s, 0, s.size() - 1);
dp.clear();
return force(s, 0, s.size() - 1, want ^ 1);
}
Solution in Java :
import java.util.*;
class Solution {
private static final class Tree {
private char op = ' ';
private int val = -1;
private Tree left, right;
public Tree(char op) {
this.op = op;
}
public Tree(int val) {
this.val = val;
}
}
private int idx = -1;
public int solve(String s) {
idx = 0;
Stack<Tree> stack = new Stack<>();
build(s, stack);
final Tree ROOT = stack.pop();
int[] res = dfs(ROOT);
return Math.abs(res[0] - res[1]);
}
private int[] dfs(Tree node) {
if (node.val != -1) {
return new int[] {node.val, 1 ^ node.val};
}
int[] left = dfs(node.left);
int[] right = dfs(node.right);
if (node.op == '&') {
int zero = Math.min(left[0], right[0]);
int one = Math.min(left[1] + right[1], 1 + Math.min(left[1], right[1]));
return new int[] {zero, one};
} else {
int zero = Math.min(left[0] + right[0], 1 + Math.min(left[0], right[0]));
int one = Math.min(left[1], right[1]);
return new int[] {zero, one};
}
}
private void build(String s, Stack<Tree> stack) {
if (idx == s.length())
return;
final char ch = s.charAt(idx);
if (ch == '0' || ch == '1') {
idx++;
final Tree node = new Tree(ch - '0');
if (stack.isEmpty())
stack.push(node);
else
stack.peek().right = node;
} else if (ch == '(') {
idx++;
Stack<Tree> tmp = new Stack<>();
build(s, tmp);
if (stack.isEmpty())
stack.push(tmp.pop());
else
stack.peek().right = tmp.pop();
idx++;
} else if (ch == ')') {
return;
} else {
final Tree node = new Tree(ch);
node.left = stack.pop();
stack.push(node);
idx++;
}
build(s, stack);
}
}
Solution in Python :
class Solution:
def solve(self, st):
if len(st) == 1:
return 1
st = "(" + st + ")"
s = []
for x in st:
if x == ")":
# az = min number of edits get to zero on the "a" side
# ao = min number of edits to get to one on the "a" side
az, ao = s.pop()
op = s.pop()
# bz = min number of edits get to zero on the "a" side
# bo = min number of edits to get to one on the "a" side
bz, bo = s.pop()
nz, no = None, None
if op == "|":
# These are just the "standard" way without edits, the new way to get to zero is if both the left and the right are zeroes
nz = az + bz
# We take the cheapest way to get to one by enumerating all the possibilities of a|b ->
# the cost of (a = 0, b = 1), (a = 1, b = 1), (a = 1, b = 0)
no = min(az + bo, ao + bo, ao + bz)
# These are the cost of changing either the left or the right
# another way to get to zero is starting from (a = 0, b = 1), and changing b = 0, with a cost of bo, or (a = 1, b = 0), and changing a = 0, with a cost of ao
nz = min(nz, az + bo + bz, ao + bz + az)
# if a = 0, b = 0, we either change b to 1 (with the cost of bo) or we change a to 1 (with cost of ao)
no = min(no, az + bz + bo, az + bz + ao)
# we can also change the sign - we can get to zero if we change to "and", and it will cost "1" to change the operator
nz = min(nz, min(az + bz, az + bo, ao + bz) + 1)
# we can get to one if we already have ones and change the signs to "and". (Note, this is never going to be true as 1|1 == 1&1, but here for symmetry)
no = min(no, ao + bo + 1)
else:
# Same logic, but for "and"
nz = min(az + bz, az + bo, ao + bz)
no = ao + bo
nz = min(nz, ao + bo + az, ao + bo + bz)
no = min(no, az + bo + ao, ao + bz + bo)
nz = min(nz, az + bz + 1)
no = min(no, min(az + bo, ao + bo, ao + bz) + 1)
s.append((nz, no))
elif x in "01":
if int(x) == 0:
s.append((0, 1))
else:
s.append((1, 0))
elif x not in "(":
s.append(x)
return max(s[0])
View More Similar Problems
Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →Find Merge Point of Two Lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share
View Solution →Inserting a Node Into a Sorted Doubly Linked List
Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function
View Solution →Reverse a doubly linked list
This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.
View Solution →