LeetCode #3699 — HARD

Number of ZigZag Arrays I

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given three integers n, l, and r.

A ZigZag array of length n is defined as follows:

Each element lies in the range [l, r].
No two adjacent elements are equal.
No three consecutive elements form a strictly increasing or strictly decreasing sequence.

Return the total number of valid ZigZag arrays.

Since the answer may be large, return it modulo 10⁹ + 7.

A sequence is said to be strictly increasing if each element is strictly greater than its previous one (if exists).

A sequence is said to be strictly decreasing if each element is strictly smaller than its previous one (if exists).

Example 1:

Input: n = 3, l = 4, r = 5

Output: 2

Explanation:

There are only 2 valid ZigZag arrays of length n = 3 using values in the range [4, 5]:

[4, 5, 4]
[5, 4, 5]

Example 2:

Input: n = 3, l = 1, r = 3

Output: 10

Explanation:

There are 10 valid ZigZag arrays of length n = 3 using values in the range [1, 3]:

[1, 2, 1], [1, 3, 1], [1, 3, 2]
[2, 1, 2], [2, 1, 3], [2, 3, 1], [2, 3, 2]
[3, 1, 2], [3, 1, 3], [3, 2, 3]

All arrays meet the ZigZag conditions.

Constraints:

3 <= n <= 2000
1 <= l < r <= 2000

Step 01

Brute Force Baseline

Problem summary: You are given three integers n, l, and r. A ZigZag array of length n is defined as follows: Each element lies in the range [l, r]. No two adjacent elements are equal. No three consecutive elements form a strictly increasing or strictly decreasing sequence. Return the total number of valid ZigZag arrays. Since the answer may be large, return it modulo 109 + 7. A sequence is said to be strictly increasing if each element is strictly greater than its previous one (if exists). A sequence is said to be strictly decreasing if each element is strictly smaller than its previous one (if exists).

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

3
4
5

Example 2

3
1
3

Step 02

Core Insight

What unlocks the optimal approach

Use dynamic programming: let <code>dp[i][dir][x]</code> be the count of length-<code>i</code> sequences ending at value <code>x</code> where <code>dir</code> is the required next comparison (0 = down, 1 = up).
If the required move is <code>up</code> (dir=1) do <code>dp[i+1][0][y] += sum(dp[i][1][x]) for x < y</code>; if the required move is <code>down</code> (dir=0) do <code>dp[i+1][1][y] += sum(dp[i][0][x]) for x > y</code>.
Speed up with prefix/suffix sums so each layer updates in O(<code>m</code>) instead of O(<code>m</code><sup>2</sup>); take values mod <code>10<sup>9</sup>+7</code>.

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

Largest constraint values

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 #3699: Number of ZigZag Arrays I
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3699: Number of ZigZag Arrays I
// package main
// 
// import "slices"
// 
// // https://space.bilibili.com/206214
// func zigZagArrays1(n, l, r int) (ans int) {
// 	const mod = 1_000_000_007
// 	k := r - l + 1
// 	f0 := make([]int, k) // 后两个数递增
// 	f1 := make([]int, k) // 后两个数递减
// 	for i := range f0 {
// 		f0[i] = 1
// 		f1[i] = 1
// 	}
// 
// 	s0 := make([]int, k+1)
// 	s1 := make([]int, k+1)
// 	for range n - 1 {
// 		for j, v := range f0 {
// 			s0[j+1] = s0[j] + v
// 			s1[j+1] = s1[j] + f1[j]
// 		}
// 		for j := range f0 {
// 			f0[j] = s1[j] % mod
// 			f1[j] = (s0[k] - s0[j+1]) % mod
// 		}
// 	}
// 
// 	for j, v := range f0 {
// 		ans += v + f1[j]
// 	}
// 	return ans % mod
// }
// 
// func zigZagArrays(n, l, r int) (ans int) {
// 	const mod = 1_000_000_007
// 	k := r - l + 1
// 	f := make([]int, k)
// 	for i := range f {
// 		f[i] = 1
// 	}
// 
// 	for i := 1; i < n; i++ {
// 		pre := 0
// 		for j, v := range f {
// 			f[j] = pre % mod
// 			pre += v
// 		}
// 		slices.Reverse(f)
// 	}
// 
// 	for _, v := range f {
// 		ans += v
// 	}
// 	return ans * 2 % mod
// }

// Accepted solution for LeetCode #3699: Number of ZigZag Arrays I
package main

import "slices"

// https://space.bilibili.com/206214
func zigZagArrays1(n, l, r int) (ans int) {
	const mod = 1_000_000_007
	k := r - l + 1
	f0 := make([]int, k) // 后两个数递增
	f1 := make([]int, k) // 后两个数递减
	for i := range f0 {
		f0[i] = 1
		f1[i] = 1
	}

	s0 := make([]int, k+1)
	s1 := make([]int, k+1)
	for range n - 1 {
		for j, v := range f0 {
			s0[j+1] = s0[j] + v
			s1[j+1] = s1[j] + f1[j]
		}
		for j := range f0 {
			f0[j] = s1[j] % mod
			f1[j] = (s0[k] - s0[j+1]) % mod
		}
	}

	for j, v := range f0 {
		ans += v + f1[j]
	}
	return ans % mod
}

func zigZagArrays(n, l, r int) (ans int) {
	const mod = 1_000_000_007
	k := r - l + 1
	f := make([]int, k)
	for i := range f {
		f[i] = 1
	}

	for i := 1; i < n; i++ {
		pre := 0
		for j, v := range f {
			f[j] = pre % mod
			pre += v
		}
		slices.Reverse(f)
	}

	for _, v := range f {
		ans += v
	}
	return ans * 2 % mod
}

# Accepted solution for LeetCode #3699: Number of ZigZag Arrays I
# Time:  O(n * (r - l))
# Space: O(r - l)

# prefix sum, dp
class Solution(object):
    def zigZagArrays(self, n, l, r):
        """
        :type n: int
        :type l: int
        :type r: int
        :rtype: int
        """
        MOD = 10**9+7
        r -= l
        dp = [1]*(r+1)
        for _ in xrange(n-1):
            prefix = 0
            for i in xrange(len(dp)):
                dp[i], prefix = prefix, (prefix+dp[i])%MOD
            dp.reverse()
        return (reduce(lambda accu, x: (accu+x)%MOD, dp, 0)*2)%MOD

// Accepted solution for LeetCode #3699: Number of ZigZag Arrays I
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3699: Number of ZigZag Arrays I
// package main
// 
// import "slices"
// 
// // https://space.bilibili.com/206214
// func zigZagArrays1(n, l, r int) (ans int) {
// 	const mod = 1_000_000_007
// 	k := r - l + 1
// 	f0 := make([]int, k) // 后两个数递增
// 	f1 := make([]int, k) // 后两个数递减
// 	for i := range f0 {
// 		f0[i] = 1
// 		f1[i] = 1
// 	}
// 
// 	s0 := make([]int, k+1)
// 	s1 := make([]int, k+1)
// 	for range n - 1 {
// 		for j, v := range f0 {
// 			s0[j+1] = s0[j] + v
// 			s1[j+1] = s1[j] + f1[j]
// 		}
// 		for j := range f0 {
// 			f0[j] = s1[j] % mod
// 			f1[j] = (s0[k] - s0[j+1]) % mod
// 		}
// 	}
// 
// 	for j, v := range f0 {
// 		ans += v + f1[j]
// 	}
// 	return ans % mod
// }
// 
// func zigZagArrays(n, l, r int) (ans int) {
// 	const mod = 1_000_000_007
// 	k := r - l + 1
// 	f := make([]int, k)
// 	for i := range f {
// 		f[i] = 1
// 	}
// 
// 	for i := 1; i < n; i++ {
// 		pre := 0
// 		for j, v := range f {
// 			f[j] = pre % mod
// 			pre += v
// 		}
// 		slices.Reverse(f)
// 	}
// 
// 	for _, v := range f {
// 		ans += v
// 	}
// 	return ans * 2 % mod
// }

// Accepted solution for LeetCode #3699: Number of ZigZag Arrays I
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3699: Number of ZigZag Arrays I
// package main
// 
// import "slices"
// 
// // https://space.bilibili.com/206214
// func zigZagArrays1(n, l, r int) (ans int) {
// 	const mod = 1_000_000_007
// 	k := r - l + 1
// 	f0 := make([]int, k) // 后两个数递增
// 	f1 := make([]int, k) // 后两个数递减
// 	for i := range f0 {
// 		f0[i] = 1
// 		f1[i] = 1
// 	}
// 
// 	s0 := make([]int, k+1)
// 	s1 := make([]int, k+1)
// 	for range n - 1 {
// 		for j, v := range f0 {
// 			s0[j+1] = s0[j] + v
// 			s1[j+1] = s1[j] + f1[j]
// 		}
// 		for j := range f0 {
// 			f0[j] = s1[j] % mod
// 			f1[j] = (s0[k] - s0[j+1]) % mod
// 		}
// 	}
// 
// 	for j, v := range f0 {
// 		ans += v + f1[j]
// 	}
// 	return ans % mod
// }
// 
// func zigZagArrays(n, l, r int) (ans int) {
// 	const mod = 1_000_000_007
// 	k := r - l + 1
// 	f := make([]int, k)
// 	for i := range f {
// 		f[i] = 1
// 	}
// 
// 	for i := 1; i < n; i++ {
// 		pre := 0
// 		for j, v := range f {
// 			f[j] = pre % mod
// 			pre += v
// 		}
// 		slices.Reverse(f)
// 	}
// 
// 	for _, v := range f {
// 		ans += v
// 	}
// 	return ans * 2 % mod
// }

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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.