Longest Arithmetic Sequence Tree Path - Google Top Interview Questions
Problem Statement :
You are given a binary tree root. Return the length of the longest path starting from a node that goes top to bottom where its values form an arithmetic sequence. An arithmetic sequence is a sequence of numbers where the difference between each consecutive two numbers is the same. For example, [2, 4, 6, 8] is an arithmetic sequence as is [1, 4, 7, 10]. A sequence of length two or fewer is considered an arithmetic sequence. Constraints 0 ≤ n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [1, [4, [7, [0, null, null], null], null], null] Output 3
Solution :
Solution in C++ :
int maxS = 0;
void dfs(Tree* root, int diff, int l) {
maxS = max(maxS, l);
if (!root) {
return;
}
if (root->left) {
if (root->left->val - root->val == diff) {
dfs(root->left, diff, l + 1);
} else {
dfs(root->left, root->left->val - root->val, 2);
}
}
if (root->right) {
if (root->right->val - root->val == diff) {
dfs(root->right, diff, l + 1);
} else {
dfs(root->right, root->right->val - root->val, 2);
}
}
}
int solve(Tree* root) {
if (root) {
maxS = 1;
} else {
return 0;
}
dfs(root, 0, 1);
return maxS;
}
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
int ans;
public int solve(Tree root) {
if (root == null) {
return 0;
}
ans = 1;
sol(root);
return ans;
}
HashMap<Integer, Integer> sol(Tree root) {
if (root == null) {
return new HashMap<>();
}
HashMap<Integer, Integer> l = sol(root.left);
HashMap<Integer, Integer> r = sol(root.right);
HashMap<Integer, Integer> mp = new HashMap<>();
if (root.left != null) {
int ldiff = root.val - root.left.val;
int x = l.getOrDefault(ldiff, 1);
int y = mp.getOrDefault(ldiff, 1);
mp.put(ldiff, Math.max(x + 1, y));
ans = Math.max(ans, mp.get(ldiff));
}
if (root.right != null) {
int rdiff = root.val - root.right.val;
int x = r.getOrDefault(rdiff, 1);
int y = mp.getOrDefault(rdiff, 1);
mp.put(rdiff, Math.max(x + 1, y));
ans = Math.max(ans, mp.get(rdiff));
}
return mp;
}
}
Solution in Python :
class Solution:
# global ans to use same solve function for recursion 😅
ans = 0
# diff is diffrence of cur value with prev to check airth seq
# ltill is longest airth seq ending before cur node
def solve(self, cur, ltill=0, diff=None):
# Intialize a global ans ans is max of prev longest airthmetic seq and cur longest
self.ans = max(self.ans, ltill)
# Check if ther is left child
if cur.left:
# compute diff with left child
ldiff = cur.left.val - cur.val
self.solve(
cur.left,
# if diff with previous child is diffrent
# from diff of cur from par
# or it's root new airthmetic sequence
# start from here so ltill is 1
# else extend prev max length ltill by 1
1 if diff == None or ldiff != diff else ltill + 1,
ldiff,
)
# same for right
if cur.right:
rdiff = cur.right.val - cur.val
self.solve(cur.right, 1 if diff == None or rdiff != diff else ltill + 1, rdiff)
# we start count from 0 so +1 solve single legth seq also
return self.ans + 1
View More Similar Problems
Direct Connections
Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do
View Solution →Subsequence Weighting
A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =
View Solution →Kindergarten Adventures
Meera teaches a class of n students, and every day in her classroom is an adventure. Today is drawing day! The students are sitting around a round table, and they are numbered from 1 to n in the clockwise direction. This means that the students are numbered 1, 2, 3, . . . , n-1, n, and students 1 and n are sitting next to each other. After letting the students draw for a certain period of ti
View Solution →Mr. X and His Shots
A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M
View Solution →Jim and the Skyscrapers
Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space
View Solution →Palindromic Subsets
Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t
View Solution →