Arithmetic Sequence Queries ☃️ - Google Top Interview Questions
Problem Statement :
You are given a list of integers nums and a two-dimensional list of integers queries.
Each element in queries contains [i, j] and asks whether the sublist of nums from [i, j], inclusive, is an arithmetic sequence.
Return the number of queries that are true.
Constraints
0 ≤ n ≤ 100,000 where n is the length of nums
0 ≤ m ≤ 100,000 where m is the length of queries
Example 1
Input
nums = [1, 3, 5, 7, 6, 5, 4, 1]
queries = [
[0, 3],
[3, 4],
[2, 4]
]
Output
2
Explanation
[1, 3, 5, 7] is an arithmetic sequence, so query [0, 3] is true.
[7, 6] is an arithmetic sequence, so query [3, 4] is true.
[5, 7, 6] is not an arithmetic sequence, so query [2, 4] is false.
Example 2
Input
nums = [1, 2]
queries = [
[0, 0],
[0, 1],
[0, 1]
]
Output
3
Explanation
Any list with length less than or equal to 2 is an arithmetic sequence.
Solution :
Solution in C++ :
int dp[100005];
int solve(vector<int>& nums, vector<vector<int>>& queries) {
int n = nums.size();
for (int i = 0; i + 1 < n;) {
int j = i + 1;
while (j + 1 < n && nums[j + 1] - nums[j] == nums[i + 1] - nums[i]) j++;
for (int k = i; k < j; k++) dp[k] = j - k;
i = j;
}
int ret = 0;
for (auto& q : queries) {
if (dp[q[0]] >= q[1] - q[0]) ret++;
}
return ret;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int[][] queries) {
HashMap<Integer, Integer> hm = new HashMap();
// length of arithmetic subsequence dp
int n = nums.length;
hm.put(n - 1, 1);
hm.put(n - 2, 2);
for (int i = n - 3; i >= 0; i--) {
int this_diff = nums[i] - nums[i + 1];
int next_diff = nums[i + 1] - nums[i + 2];
if (this_diff == next_diff) {
hm.put(i, 1 + hm.get(i + 1));
} else {
hm.put(i, 2);
}
}
int res = 0;
for (int[] q : queries) {
int left = q[0];
int right = q[1];
if (hm.get(left) >= (right - left + 1)) {
res++;
}
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, nums, queries):
if not nums:
return 0
# preprocessing
n = len(nums)
# compute pairwise differences
diff = [nums[i + 1] - nums[i] for i in range(n - 1)]
# "run-length encoding": `rle[i]` is length of the longest constant sublist
# of `diff` ending at (and including) `i`)
rle = [0] * (n - 1)
for i in range(n - 1):
if i > 0 and diff[i] == diff[i - 1]:
rle[i] = rle[i - 1] + 1
else:
rle[i] = 1
# answer queries
ans = 0
for i, j in queries:
# `nums[i:j+1]` is an arithmetic sequence iff `diff[i:j]` is constant
if i == j:
ans += 1
else:
ans += rle[j - 1] >= (j - i)
return ans
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