Finding Pairs With a Certain Sum

Problem

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 <= 10^5
1 <= nums1[i] <= 10^9
1 <= nums2[i] <= 10^5
0 <= index < nums2.length
1 <= val <= 10^5
1 <= tot <= 10^9
At most 1000 calls are made to add and count each.

Solution (Java)

class FindSumPairs {
    private Map<Integer, Integer> numFreq = new HashMap<>();
    private int[] nums1;
    private int[] nums2;

    public FindSumPairs(int[] nums1, int[] nums2) {
        this.nums1 = nums1;
        this.nums2 = nums2;
        for (int num : nums2) {
            numFreq.put(num, numFreq.getOrDefault(num, 0) + 1);
        }
    }

    public void add(int index, int val) {
        numFreq.put(nums2[index], numFreq.getOrDefault(nums2[index], 0) - 1);
        nums2[index] += val;
        numFreq.put(nums2[index], numFreq.getOrDefault(nums2[index], 0) + 1);
    }

    public int count(int tot) {
        int res = 0;
        for (int num : nums1) {
            res += numFreq.getOrDefault(tot - num, 0);
        }
        return res;
    }
}

/**
 * Your FindSumPairs object will be instantiated and called as such:
 * FindSumPairs obj = new FindSumPairs(nums1, nums2);
 * obj.add(index,val);
 * int param_2 = obj.count(tot);
 */

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).