1995 Finding Pairs With A Certain Sum

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:

**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**.

class FindSumPairs:

    def __init__(self, nums1: List[int], nums2: List[int]):
        """
        sort nums1
        for each nums1 < total , find in nums2 tot-i
        as # of elements in nums1 < 1000, then count runs in 1000
        """
        self.arr1 = nums1
        self.arr2 = {}
        self.arr3 = {}
        for i,  k in enumerate(nums2):
            self.arr3[i] = k
        for i,k in enumerate(nums2):
            if k in self.arr2:
                self.arr2[k].append(i)
            else:
                self.arr2[k] = [i]

    def add(self, index: int, val: int) -> None:
        k = self.arr3[index]
        if k + val in self.arr2:
            self.arr2[k + val].append(index)
        else:
            self.arr2[k + val] = [index]
        self.arr2[k].remove(index)
        self.arr3[index] = k + val
        

        

    def count(self, tot: int) -> int:
        res = 0
        for k in self.arr1:
            if tot - k in self.arr2:
                res += len(self.arr2[tot-k])
        return res
        


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