Number of Subarrays That Match a Pattern I - Problem

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.

Input & Output

Example 1 — Basic Pattern Match

$ Input: nums = [1,2,3,4,5,6], pattern = [1,1]

› Output: 4

💡 Note: Four subarrays match: [1,2,3] (1<2<3), [2,3,4] (2<3<4), [3,4,5] (3<4<5), [4,5,6] (4<5<6)

Example 2 — Mixed Pattern

$ Input: nums = [1,4,4,1,3,5,5,3], pattern = [1,0,-1]

› Output: 2

💡 Note: Two subarrays match: [1,4,4,1] (1<4=4>1) and [3,5,5,3] (3<5=5>3)

Example 3 — No Matches

$ Input: nums = [1,2,3,4], pattern = [-1,-1]

› Output: 0

💡 Note: No subarray of length 3 has consecutive decreasing elements

Constraints

1 ≤ nums.length ≤ 10⁶
1 ≤ pattern.length ≤ nums.length
1 ≤ nums[i] ≤ 10⁹
-1 ≤ pattern[i] ≤ 1

Visualization

Tap to expand

Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is to transform the comparison problem into pattern matching. Convert the numeric array into comparison symbols (1, 0, -1) and then use string matching algorithms like KMP for efficient pattern search. Best approach is KMP Pattern Matching with Time: O(n + m), Space: O(n + m).

Common Approaches

✓ Brute Force Pattern Matching

⏱️ Time: O(n × m) Space: O(1)

For each valid starting position, examine the subarray of length m+1 and compare consecutive elements according to the pattern requirements. Count how many subarrays satisfy all pattern conditions.

Pattern String Matching

⏱️ Time: O(n + m) Space: O(n + m)

Transform the numeric array into a pattern string by comparing consecutive elements, then use string matching techniques like KMP to find all occurrences of the target pattern efficiently.

Brute Force Pattern Matching — Algorithm Steps

Step 1: Iterate through each valid starting position
Step 2: For each position, check if subarray matches pattern
Step 3: Count all matching subarrays

Visualization

Tap to expand

Step-by-Step Walkthrough

Check Position 0

Compare nums[0..2] with pattern [1,-1]

Check Position 1

Compare nums[1..3] with pattern [1,-1]

Count Matches

Return total count of matching subarrays

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize, int* pattern, int patternSize) {
    int count = 0;
    
    // Check each possible starting position
    for (int i = 0; i <= numsSize - patternSize - 1; i++) {
        int matches = 1;
        
        // Check if subarray starting at i matches the pattern
        for (int k = 0; k < patternSize; k++) {
            if (pattern[k] == 1) {
                if (nums[i + k + 1] <= nums[i + k]) {
                    matches = 0;
                    break;
                }
            } else if (pattern[k] == 0) {
                if (nums[i + k + 1] != nums[i + k]) {
                    matches = 0;
                    break;
                }
            } else if (pattern[k] == -1) {
                if (nums[i + k + 1] >= nums[i + k]) {
                    matches = 0;
                    break;
                }
            }
        }
        
        if (matches) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse nums array
    int nums[100];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    // Parse pattern array
    fgets(line, sizeof(line), stdin);
    int pattern[100];
    int patternSize = 0;
    token = strtok(line, "[,]");
    while (token != NULL) {
        pattern[patternSize++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, numsSize, pattern, patternSize);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

Outer loop runs n-m times, inner loop runs m times for pattern checking

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

28.5K Views

Medium Frequency

~25 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Number of Subarrays That Match a Pattern I - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Pattern Matching — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler