LeetCode #1630 — MEDIUM

Arithmetic Subarrays

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

A sequence of numbers is called arithmetic if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence s is arithmetic if and only if s[i+1] - s[i] == s[1] - s[0] for all valid i.

For example, these are arithmetic sequences:

1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic:

1, 1, 2, 5, 7

You are given an array of n integers, nums, and two arrays of m integers each, l and r, representing the m range queries, where the i^th query is the range [l[i], r[i]]. All the arrays are 0-indexed.

Return a list of boolean elements answer, where answer[i] is true if the subarray nums[l[i]], nums[l[i]+1], ... , nums[r[i]] can be rearranged to form an arithmetic sequence, and false otherwise.

Example 1:

Input: nums = [4,6,5,9,3,7], l = [0,0,2], r = [2,3,5]
Output: [true,false,true]
Explanation:
In the 0^th query, the subarray is [4,6,5]. This can be rearranged as [6,5,4], which is an arithmetic sequence.
In the 1^st query, the subarray is [4,6,5,9]. This cannot be rearranged as an arithmetic sequence.
In the 2^nd query, the subarray is [5,9,3,7]. This can be rearranged as [3,5,7,9], which is an arithmetic sequence.

Example 2:

Input: nums = [-12,-9,-3,-12,-6,15,20,-25,-20,-15,-10], l = [0,1,6,4,8,7], r = [4,4,9,7,9,10]
Output: [false,true,false,false,true,true]

Constraints:

n == nums.length
m == l.length
m == r.length
2 <= n <= 500
1 <= m <= 500
0 <= l[i] < r[i] < n
-10⁵ <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: A sequence of numbers is called arithmetic if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence s is arithmetic if and only if s[i+1] - s[i] == s[1] - s[0] for all valid i. For example, these are arithmetic sequences: 1, 3, 5, 7, 9 7, 7, 7, 7 3, -1, -5, -9 The following sequence is not arithmetic: 1, 1, 2, 5, 7 You are given an array of n integers, nums, and two arrays of m integers each, l and r, representing the m range queries, where the ith query is the range [l[i], r[i]]. All the arrays are 0-indexed. Return a list of boolean elements answer, where answer[i] is true if the subarray nums[l[i]], nums[l[i]+1], ... , nums[r[i]] can be rearranged to form an arithmetic sequence, and false otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[4,6,5,9,3,7]
[0,0,2]
[2,3,5]

Example 2

[-12,-9,-3,-12,-6,15,20,-25,-20,-15,-10]
[0,1,6,4,8,7]
[4,4,9,7,9,10]

Core Insight

What unlocks the optimal approach

To check if a given sequence is arithmetic, just check that the difference between every two consecutive elements is the same.
If and only if a set of numbers can make an arithmetic sequence, then its sorted version makes an arithmetic sequence. So to check a set of numbers, sort it, and check if that sequence is arithmetic.
For each query, get the corresponding set of numbers which will be the sub-array represented by the query, sort it, and check if the result sequence is arithmetic.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1630: Arithmetic Subarrays
class Solution {
    public List<Boolean> checkArithmeticSubarrays(int[] nums, int[] l, int[] r) {
        List<Boolean> ans = new ArrayList<>();
        for (int i = 0; i < l.length; ++i) {
            ans.add(check(nums, l[i], r[i]));
        }
        return ans;
    }

    private boolean check(int[] nums, int l, int r) {
        Set<Integer> s = new HashSet<>();
        int n = r - l + 1;
        int a1 = 1 << 30, an = -a1;
        for (int i = l; i <= r; ++i) {
            s.add(nums[i]);
            a1 = Math.min(a1, nums[i]);
            an = Math.max(an, nums[i]);
        }
        if ((an - a1) % (n - 1) != 0) {
            return false;
        }
        int d = (an - a1) / (n - 1);
        for (int i = 1; i < n; ++i) {
            if (!s.contains(a1 + (i - 1) * d)) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #1630: Arithmetic Subarrays
func checkArithmeticSubarrays(nums []int, l []int, r []int) (ans []bool) {
	check := func(nums []int, l, r int) bool {
		s := map[int]struct{}{}
		n := r - l + 1
		a1, an := 1<<30, -(1 << 30)
		for _, x := range nums[l : r+1] {
			s[x] = struct{}{}
			if a1 > x {
				a1 = x
			}
			if an < x {
				an = x
			}
		}
		if (an-a1)%(n-1) != 0 {
			return false
		}
		d := (an - a1) / (n - 1)
		for i := 1; i < n; i++ {
			if _, ok := s[a1+(i-1)*d]; !ok {
				return false
			}
		}
		return true
	}
	for i := range l {
		ans = append(ans, check(nums, l[i], r[i]))
	}
	return
}

# Accepted solution for LeetCode #1630: Arithmetic Subarrays
class Solution:
    def checkArithmeticSubarrays(
        self, nums: List[int], l: List[int], r: List[int]
    ) -> List[bool]:
        def check(nums, l, r):
            n = r - l + 1
            s = set(nums[l : l + n])
            a1, an = min(nums[l : l + n]), max(nums[l : l + n])
            d, mod = divmod(an - a1, n - 1)
            return mod == 0 and all((a1 + (i - 1) * d) in s for i in range(1, n))

        return [check(nums, left, right) for left, right in zip(l, r)]

// Accepted solution for LeetCode #1630: Arithmetic Subarrays
impl Solution {
    pub fn check_arithmetic_subarrays(nums: Vec<i32>, l: Vec<i32>, r: Vec<i32>) -> Vec<bool> {
        let m = l.len();
        let mut res = vec![true; m];
        for i in 0..m {
            let mut arr = nums[l[i] as usize..=r[i] as usize].to_vec();
            arr.sort();
            for j in 2..arr.len() {
                if arr[j - 2] - arr[j - 1] != arr[j - 1] - arr[j] {
                    res[i] = false;
                    break;
                }
            }
        }
        res
    }
}

// Accepted solution for LeetCode #1630: Arithmetic Subarrays
function checkArithmeticSubarrays(nums: number[], l: number[], r: number[]): boolean[] {
    const check = (nums: number[], l: number, r: number): boolean => {
        const s = new Set<number>();
        const n = r - l + 1;
        let a1 = 1 << 30;
        let an = -a1;
        for (let i = l; i <= r; ++i) {
            s.add(nums[i]);
            a1 = Math.min(a1, nums[i]);
            an = Math.max(an, nums[i]);
        }
        if ((an - a1) % (n - 1) !== 0) {
            return false;
        }
        const d = Math.floor((an - a1) / (n - 1));
        for (let i = 1; i < n; ++i) {
            if (!s.has(a1 + (i - 1) * d)) {
                return false;
            }
        }
        return true;
    };
    const ans: boolean[] = [];
    for (let i = 0; i < l.length; ++i) {
        ans.push(check(nums, l[i], r[i]));
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.