Number of Arithmetic Triplets

Number of Arithmetic Triplets - Problem

You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:

i < j < k,
nums[j] - nums[i] == diff, and
nums[k] - nums[j] == diff.

Return the number of unique arithmetic triplets.

Input & Output

Example 1 — Basic Case

$ Input: nums = [0,1,4,6,7,10], diff = 3

› Output: 2

💡 Note: Two arithmetic triplets: (1,4,7) with differences 4-1=3 and 7-4=3, and (4,7,10) with differences 7-4=3 and 10-7=3

Example 2 — Single Triplet

$ Input: nums = [4,5,6,7,8,9], diff = 2

› Output: 2

💡 Note: Two arithmetic triplets: (4,6,8) and (5,7,9), both with difference 2

Example 3 — No Triplets

$ Input: nums = [1,2,3], diff = 5

› Output: 0

💡 Note: No arithmetic triplets possible with difference 5 in this small array

Constraints

3 ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ 200
The given array is strictly increasing
1 ≤ diff ≤ 50

Visualization

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Asked in

M Meta 15 G Google 12

The key insight is to use each element as a potential middle element of an arithmetic triplet. For optimal solution, use a hash set to check if required partners exist in O(1) time. Best approach: Hash Set with Time: O(n), Space: O(n)

Common Approaches

✓ Hash Set Lookup

⏱️ Time: O(n) Space: O(n)

Store all numbers in a hash set, then for each element as middle, check if both required partners exist using constant time lookups.

Triple Nested Loop

⏱️ Time: O(n³) Space: O(1)

Iterate through all possible combinations of three indices i < j < k and check if they form an arithmetic triplet with the given difference.

Hash Set Lookup — Algorithm Steps

Step 1: Store all numbers in a hash set
Step 2: For each number as middle element, check if first and third elements exist
Step 3: Count valid triplets

Visualization

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Step-by-Step Walkthrough

Build Set

Store all numbers in hash set

Check Middle

For each number, check if partners exist

Count Valid

Count elements that can be middle of triplet

Code -

solution.c — C

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

#define HASH_SIZE 10007

typedef struct HashNode {
    int val;
    struct HashNode* next;
} HashNode;

int hash(int val) {
    return abs(val) % HASH_SIZE;
}

void insert(HashNode** table, int val) {
    int idx = hash(val);
    HashNode* node = (HashNode*)malloc(sizeof(HashNode));
    node->val = val;
    node->next = table[idx];
    table[idx] = node;
}

int contains(HashNode** table, int val) {
    int idx = hash(val);
    HashNode* node = table[idx];
    while (node) {
        if (node->val == val) return 1;
        node = node->next;
    }
    return 0;
}

int solution(int* nums, int numsSize, int diff) {
    HashNode* table[HASH_SIZE] = {NULL};
    
    for (int i = 0; i < numsSize; i++) {
        insert(table, nums[i]);
    }
    
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (contains(table, nums[i] - diff) && contains(table, nums[i] + diff)) {
            count++;
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[200];
    int numsSize;
    parseArray(line, nums, &numsSize);
    
    int diff;
    scanf("%d", &diff);
    
    printf("%d\n", solution(nums, numsSize, diff));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to build set plus single pass to check each element

✓ Linear Growth

Space Complexity

O(n)

Hash set stores all n elements from the array

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Hash Set Lookup — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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