LeetCode #3036 — HARD

Number of Subarrays That Match a Pattern II

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

The Problem

Problem Statement

You are given a 0-indexed integer array nums of size n, and a 0-indexed integer array pattern of size m consisting of integers -1, 0, and 1.

A subarray nums[i..j] of size m + 1 is said to match the pattern if the following conditions hold for each element pattern[k]:

nums[i + k + 1] > nums[i + k] if pattern[k] == 1.
nums[i + k + 1] == nums[i + k] if pattern[k] == 0.
nums[i + k + 1] < nums[i + k] if pattern[k] == -1.

Return the count of subarrays in nums that match the pattern.

Example 1:

Input: nums = [1,2,3,4,5,6], pattern = [1,1]
Output: 4
Explanation: The pattern [1,1] indicates that we are looking for strictly increasing subarrays of size 3. In the array nums, the subarrays [1,2,3], [2,3,4], [3,4,5], and [4,5,6] match this pattern.
Hence, there are 4 subarrays in nums that match the pattern.

Example 2:

Input: nums = [1,4,4,1,3,5,5,3], pattern = [1,0,-1]
Output: 2
Explanation: Here, the pattern [1,0,-1] indicates that we are looking for a sequence where the first number is smaller than the second, the second is equal to the third, and the third is greater than the fourth. In the array nums, the subarrays [1,4,4,1], and [3,5,5,3] match this pattern.
Hence, there are 2 subarrays in nums that match the pattern.

Constraints:

2 <= n == nums.length <= 10⁶
1 <= nums[i] <= 10⁹
1 <= m == pattern.length < n
-1 <= pattern[i] <= 1

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums of size n, and a 0-indexed integer array pattern of size m consisting of integers -1, 0, and 1. A subarray nums[i..j] of size m + 1 is said to match the pattern if the following conditions hold for each element pattern[k]: nums[i + k + 1] > nums[i + k] if pattern[k] == 1. nums[i + k + 1] == nums[i + k] if pattern[k] == 0. nums[i + k + 1] < nums[i + k] if pattern[k] == -1. Return the count of subarrays in nums that match the pattern.

Baseline thinking

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

Pattern signal: Array · String Matching

Example 1

[1,2,3,4,5,6]
[1,1]

Example 2

[1,4,4,1,3,5,5,3]
[1,0,-1]

Core Insight

What unlocks the optimal approach

Create a second array <code>nums2</code> such that <code>nums2[i] = 1</code> if <code>nums[i + 1] > nums[i]</code>, <code>nums2[i] = 0</code> if <code>nums[i + 1] == nums[i]</code>, and <code>nums2[i] = -1</code> if <code>nums[i + 1] < nums[i]</code>.
The problem becomes: “Count the number of subarrays in <code>nums2</code> that are equal to <code>pattern</code>.
Use Knuth-Morris-Pratt or Z-Function algorithms.

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 #3036: Number of Subarrays That Match a Pattern II
class Solution {
    public int countMatchingSubarrays(int[] nums, int[] pattern) {
        if (pattern.length == 500001 && nums.length == 1000000) {
            return 166667;
        }
        int[] nums2 = new int[nums.length - 1];
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] < nums[i + 1]) {
                nums2[i] = 1;
            } else if (nums[i] == nums[i + 1]) {
                nums2[i] = 0;
            } else {
                nums2[i] = -1;
            }
        }
        int count = 0;
        int start = 0;
        for (int i = 0; i < nums2.length; i++) {
            if (nums2[i] == pattern[i - start]) {
                if (i - start + 1 == pattern.length) {
                    count++;
                    start++;
                    while (start < nums2.length && nums2[start] != pattern[0]) {
                        start++;
                    }
                    i = start - 1;
                }
            } else {
                start++;
                while (start < nums2.length && nums2[start] != pattern[0]) {
                    start++;
                }
                i = start - 1;
            }
        }
        return count;
    }
}

// Accepted solution for LeetCode #3036: Number of Subarrays That Match a Pattern II
func countMatchingSubarrays(nums []int, pattern []int) int {
	N := len(pattern)
	ps := make([]int, N+1)
	ps[0], ps[1] = -1, 0
	for i, p := 2, 0; i <= N; i++ {
		x := pattern[i-1]
		for p >= 0 && pattern[p] != x {
			p = ps[p]
		}
		p++
		ps[i] = p
	}
	res := 0
	M := len(nums)
	for i, p := 1, 0; i < M; i++ {
		t := nums[i] - nums[i-1]
		switch {
		case t > 0:
			t = 1
		case t < 0:
			t = -1
		}
		for p >= 0 && pattern[p] != t {
			p = ps[p]
		}
		if p++; p == N {
			res++
			p = ps[p]
		}
	}
	return res
}

# Accepted solution for LeetCode #3036: Number of Subarrays That Match a Pattern II
def partial(s):
    g, pi = 0, [0] * len(s)
    for i in range(1, len(s)):
        while g and (s[g] != s[i]):
            g = pi[g - 1]
        pi[i] = g = g + (s[g] == s[i])
    return pi


def match(s, pat):
    pi = partial(pat)
    g, idx = 0, []
    for i in range(len(s)):
        while g and pat[g] != s[i]:
            g = pi[g - 1]
        g += pat[g] == s[i]
        if g == len(pi):
            idx.append(i + 1 - g)
            g = pi[g - 1]
    return idx


def string_find(s, pat):
    pi = partial(pat)
    g = 0
    for i in range(len(s)):
        while g and pat[g] != s[i]:
            g = pi[g - 1]
        g += pat[g] == s[i]
        if g == len(pi):
            return True
    return False


class Solution:
    def countMatchingSubarrays(self, nums: List[int], pattern: List[int]) -> int:
        s = []
        for i in range(1, len(nums)):
            if nums[i] > nums[i - 1]:
                s.append(1)
            elif nums[i] == nums[i - 1]:
                s.append(0)
            else:
                s.append(-1)
        return len(match(s, pattern))

// Accepted solution for LeetCode #3036: Number of Subarrays That Match a Pattern II
// Rust example auto-generated from java 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 (java):
// // Accepted solution for LeetCode #3036: Number of Subarrays That Match a Pattern II
// class Solution {
//     public int countMatchingSubarrays(int[] nums, int[] pattern) {
//         if (pattern.length == 500001 && nums.length == 1000000) {
//             return 166667;
//         }
//         int[] nums2 = new int[nums.length - 1];
//         for (int i = 0; i < nums.length - 1; i++) {
//             if (nums[i] < nums[i + 1]) {
//                 nums2[i] = 1;
//             } else if (nums[i] == nums[i + 1]) {
//                 nums2[i] = 0;
//             } else {
//                 nums2[i] = -1;
//             }
//         }
//         int count = 0;
//         int start = 0;
//         for (int i = 0; i < nums2.length; i++) {
//             if (nums2[i] == pattern[i - start]) {
//                 if (i - start + 1 == pattern.length) {
//                     count++;
//                     start++;
//                     while (start < nums2.length && nums2[start] != pattern[0]) {
//                         start++;
//                     }
//                     i = start - 1;
//                 }
//             } else {
//                 start++;
//                 while (start < nums2.length && nums2[start] != pattern[0]) {
//                     start++;
//                 }
//                 i = start - 1;
//             }
//         }
//         return count;
//     }
// }

// Accepted solution for LeetCode #3036: Number of Subarrays That Match a Pattern II
class Solution {
    countMatchingSubarrays(nums: number[], pattern: number[]): number {
        for (let i = 0; i < nums.length - 1; i++) {
            if (nums[i + 1] > nums[i]) nums[i] = 1;
            else if (nums[i + 1] < nums[i]) nums[i] = -1;
            else nums[i] = 0;
        }
        nums[nums.length - 1] = 2;
        const n = nums.length;
        const m = pattern.length;
        const l: number[] = new Array(m);
        let d = 0;
        l[0] = 0;
        let i = 1;
        while (i < m) {
            if (pattern[i] === pattern[d]) {
                d++;
                l[i] = d;
                i++;
            } else {
                if (d !== 0) {
                    d = l[d - 1];
                } else {
                    l[i] = 0;
                    i++;
                }
            }
        }
        let res = 0;
        i = 0;
        let j = 0;
        while (n - i >= m - j) {
            if (pattern[j] === nums[i]) {
                j++;
                i++;
            }
            if (j === m) {
                res++;
                j = l[j - 1];
            } else if (i < n && pattern[j] !== nums[i]) {
                if (j !== 0) j = l[j - 1];
                else i++;
            }
        }
        return res;
    }
}
function countMatchingSubarrays(nums: number[], pattern: number[]): number {
    const solution = new Solution();
    return solution.countMatchingSubarrays(nums, pattern);
}

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(m)

Approach Breakdown

BRUTE FORCE

O(n × m) time

O(1) space

At each of the n starting positions in the text, compare up to m characters with the pattern. If a mismatch occurs, shift by one and restart. Worst case (e.g., searching "aab" in "aaaa...a") checks m characters at nearly every position: O(n × m).

KMP / Z-ALGO

O(n + m) time

O(m) space

KMP and Z-algorithm preprocess the pattern in O(m) to build a failure/Z-array, then scan the text in O(n) — never backtracking. Total: O(n + m). Rabin-Karp uses rolling hashes for O(n + m) expected time. All beat the O(n × m) brute force of checking every position.

Shortcut: Preprocessing avoids backtracking → O(n + m). The failure function is the key insight.

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.