Today I have done leetcode's July LeetCoding Challenge with cpp.

July LeetCoding Challenge 6

Description

Finding Pairs With a Certain Sum

You are given two integer arrays nums1 and nums2. You are tasked to implement a data structure that supports queries of two types:

1. Add a positive integer to an element of a given index in the array nums2.
2. 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.

Solution

class FindSumPairs {
  vector<int> n1, n2;
  map<int, int> count1, count2;
public:
  FindSumPairs(vector<int>& nums1, vector<int>& nums2): n1(nums1), n2(nums2) {
    for(auto n : n1) count1[n] += 1;
    for(auto n : n2) count2[n] += 1;
  }
  
  void add(int index, int val) {
    count2[n2[index]] -= 1;
    if(!count2[n2[index]]) count2.erase(n2[index]);
    n2[index] += val;
    count2[n2[index]] += 1;
  }
  
  int count(int tot) {
    int result = 0;
    for(auto [n, c] : count1) {
      if(count2.count(tot - n)) result += c * count2[tot - n];
    }
    return result;
  }
};

// Accepted
// 26/26 cases passed (241 ms)
// Your runtime beats 43.21 % of cpp submissions
// Your memory usage beats 41.54 % of cpp submissions (83.4 MB)