Introduction

In this blog post, we will explore how to solve the LeetCode problem "923" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer array arr, and an integer target, return the number of tuples i, j, k such that i < j < k and arr[i] + arr[j] + arr[k] == target. As the answer can be very large, return it modulo 109 + 7. Example 1: Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8 Output: 20 Explanation: Enumerating by the values (arr[i], arr[j], arr[k]): (1, 2, 5) occurs 8 times; (1, 3, 4) occurs 8 times; (2, 2, 4) occurs 2 times; (2, 3, 3) occurs 2 times. Example 2: Input: arr = [1,1,2,2,2,2], target = 5 Output: 12 Explanation: arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times: We choose one 1 from [1,1] in 2 ways, and two 2s from [2,2,2,2] in 6 ways. Example 3: Input: arr = [2,1,3], target = 6 Output: 1 Explanation: (1, 2, 3) occured one time in the array so we return 1. Constraints: 3 <= arr.length <= 3000 0 <= arr[i] <= 100 0 <= target <= 300

Explanation

Here's a solution to the problem, focusing on efficiency and clarity:

• Counting: The core idea is to count the occurrences of each number in the input array. This allows us to efficiently determine how many ways we can pick specific numbers to form a tuple.
• Iterating and Checking: We iterate through all possible combinations of i, j, and k values (where i <= j <= k based on the problem requirements regarding duplicates), formed by the unique numbers in the input. Then, we use the counts to calculate the number of tuples that sum up to the target.

• Combinations: The most important part is to handle different scenarios based on if i == j == k, i == j != k, i != j == k, or i != j != k, with appropriate combinations formulas.

• Time Complexity: O(N + M^3), where N is the length of the input array and M is the number of unique elements in the array (M <= 101). Space Complexity: O(M), for storing the counts of each number.

Code

    def threeSumMulti(arr, target):
    """
    Finds the number of tuples (i, j, k) such that i < j < k and arr[i] + arr[j] + arr[k] == target.

    Args:
        arr: An integer array.
        target: An integer target value.

    Returns:
        The number of tuples modulo 10^9 + 7.
    """

    MOD = 10**9 + 7
    count = {}
    for num in arr:
        count[num] = count.get(num, 0) + 1

    unique_nums = sorted(count.keys())
    n = len(unique_nums)
    ans = 0

    for i in range(n):
        for j in range(i, n):
            for k in range(j, n):
                if unique_nums[i] + unique_nums[j] + unique_nums[k] == target:
                    if i == j == k:
                        ans += count[unique_nums[i]] * (count[unique_nums[i]] - 1) * (count[unique_nums[i]] - 2) // 6
                    elif i == j != k:
                        ans += count[unique_nums[i]] * (count[unique_nums[i]] - 1) // 2 * count[unique_nums[k]]
                    elif i != j == k:
                        ans += count[unique_nums[i]] * count[unique_nums[j]] * (count[unique_nums[j]] - 1) // 2
                    else:
                        ans += count[unique_nums[i]] * count[unique_nums[j]] * count[unique_nums[k]]
                    ans %= MOD

    return ans