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
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