Logically Consistent Book - Google Top Interview Questions

Problem Statement :

```You are given a two-dimensional list of integers lists.

Each element contains [start, end, type], meaning that from pages [start, end] inclusive, there's either an even number of words (type = 0) or an odd number of words (type = 1).

Return whether lists is logically consistent.

Constraints

n ≤ 100,000 where n is the length of lists

Example 1

Input

lists = [

[1, 5, 1],

[6, 10, 0],

[1, 10, 0]

]

Output

False

Explanation

If there are an odd number of words from pages [1, 5] and an even number of words from pages [6, 10],
that must mean there's an odd number from pages [1, 10]. So this is a contradiction.

Example 2

Input

lists = [

[0, 2, 0],

[2, 4, 0],

[0, 4, 0]

]

Output

True

Explanation

If there are an even number of words from pages [0, 2] and an even number of words from pages [2, 4],
that means there can be an even number from pages [0, 4]. So there's no contradiction.

Example 3
Input
lists = [
[0, 2, 1],
[3, 4, 1],
[0, 4, 0]
]
Output
True
Explanation
If there are an odd number of words from pages [0, 2] and an odd number of words from pages [3, 4], that must mean there's an even number from pages [0, 4]. So there's no contradiction.```

Solution :

```                        ```Solution in C++ :

unordered_map<int, vector<pair<int, int>>> edges;
bool solve(vector<vector<int>>& lists) {
unordered_map<int, int> parity;
edges.clear();
for (auto& edge : lists) {
edges[edge[0] - 1].emplace_back(edge[1], edge[2]);
edges[edge[1]].emplace_back(edge[0] - 1, edge[2]);
}
for (auto edge : edges) {
int src = edge.first;
if (parity.count(src)) continue;
parity[src] = 0;
queue<int> q;
q.push(src);
while (q.size()) {
int curr = q.front();
q.pop();
for (auto edge : edges[curr]) {
int nparity = parity[curr] ^ edge.second;
if (parity.count(edge.first) && parity[edge.first] != nparity) return false;
if (!parity.count(edge.first)) {
parity[edge.first] = nparity;
q.push(edge.first);
}
}
}
}
return true;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(int[][] nums) {
Comparator c = new Comparator<int[]>() {
@Override
public int compare(int[] a, int[] b) {
if (a[0] != b[0])
return a[0] - b[0];
else
return a[1] - b[1];
}
};
PriorityQueue<int[]> pq = new PriorityQueue<int[]>(c);
for (int[] range : nums) pq.add(range);

while (pq.size() >= 2) {
int[] r1 = pq.poll();
int[] r2 = pq.peek();
if (r1[0] == r2[0]) {
if (r1[1] == r2[1]) {
if (r1[2] != r2[2])
return false;
} else {
pq.add(new int[] {r1[1] + 1, r2[1], Math.abs(r1[2] - r2[2])});
}
}
}
return true;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, lists):
graph = defaultdict(list)
for i, j, p in lists:
graph[i].append([j + 1, p])
graph[j + 1].append([i, p])

state = {}

def dfs(node, color):
state[node] = color
for nei, w in graph[node]:
if nei not in state:
if not dfs(nei, color ^ w):
return False
elif state[nei] != color ^ w:
return False

return True

for node in graph:
if node not in state and dfs(node, 0) is False:
return False
return True```
```

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis