Introduction

In this blog post, we will explore how to solve the LeetCode problem "1865" 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

You are given two integer arrays nums1 and nums2. You are tasked to implement a data structure that supports queries of two types: Add a positive integer to an element of a given index in the array nums2. Count the number of pairs (i, j) such that nums1[i] + nums2[j] equals a given value (0 <= i < nums1.length and 0 <= j < nums2.length). Implement the FindSumPairs class: FindSumPairs(int[] nums1, int[] nums2) Initializes the FindSumPairs object with two integer arrays nums1 and nums2. void add(int index, int val) Adds val to nums2[index], i.e., apply nums2[index] += val. int count(int tot) Returns the number of pairs (i, j) such that nums1[i] + nums2[j] == tot. Example 1: Input ["FindSumPairs", "count", "add", "count", "count", "add", "add", "count"] [[[1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]], [7], [3, 2], [8], [4], [0, 1], [1, 1], [7]] Output [null, 8, null, 2, 1, null, null, 11] Explanation FindSumPairs findSumPairs = new FindSumPairs([1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]); findSumPairs.count(7); // return 8; pairs (2,2), (3,2), (4,2), (2,4), (3,4), (4,4) make 2 + 5 and pairs (5,1), (5,5) make 3 + 4 findSumPairs.add(3, 2); // now nums2 = [1,4,5,4,5,4] findSumPairs.count(8); // return 2; pairs (5,2), (5,4) make 3 + 5 findSumPairs.count(4); // return 1; pair (5,0) makes 3 + 1 findSumPairs.add(0, 1); // now nums2 = [2,4,5,4,5,4] findSumPairs.add(1, 1); // now nums2 = [2,5,5,4,5,4] findSumPairs.count(7); // return 11; pairs (2,1), (2,2), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,4) make 2 + 5 and pairs (5,3), (5,5) make 3 + 4 Constraints: 1 <= nums1.length <= 1000 1 <= nums2.length <= 105 1 <= nums1[i] <= 109 1 <= nums2[i] <= 105 0 <= index < nums2.length 1 <= val <= 105 1 <= tot <= 109 At most 1000 calls are made to add and count each.

Explanation

Here's a breakdown of the approach and the Python code:

• High-Level Approach:

  • Use a hash map (dictionary in Python) to store the frequencies of elements in nums2. This allows for efficient counting of occurrences.
  • When add(index, val) is called, update nums2 and adjust the frequencies in the hash map accordingly.
  • When count(tot) is called, iterate through nums1 and check how many elements in nums2 can sum up to tot using the frequency information from the hash map.

• Complexity:

  • Runtime: O(n) for initialization, O(1) for add, O(m) for count (where n is the length of nums2 and m is the length of nums1).
  • Storage: O(n) to store the frequency map of nums2.

Code

    class FindSumPairs:

    def __init__(self, nums1: list[int], nums2: list[int]):
        self.nums1 = nums1
        self.nums2 = nums2
        self.freq = {}
        for num in nums2:
            self.freq[num] = self.freq.get(num, 0) + 1

    def add(self, index: int, val: int) -> None:
        old_val = self.nums2[index]
        self.freq[old_val] -= 1
        if self.freq[old_val] == 0:
            del self.freq[old_val]

        self.nums2[index] += val
        new_val = self.nums2[index]
        self.freq[new_val] = self.freq.get(new_val, 0) + 1

    def count(self, tot: int) -> int:
        count = 0
        for num in self.nums1:
            complement = tot - num
            if complement in self.freq:
                count += self.freq[complement]
        return count