Class FindSumPairs
Medium
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 thatnums1[i] + nums2[j]equals a given value (0 <= i < nums1.lengthand0 <= j < nums2.length). 
Implement the FindSumPairs class:
FindSumPairs(int[] nums1, int[] nums2)Initializes theFindSumPairsobject with two integer arraysnums1andnums2.void add(int index, int val)Addsvaltonums2[index], i.e., applynums2[index] += val.int count(int tot)Returns the number of pairs(i, j)such thatnums1[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 <= 10001 <= nums2.length <= 1051 <= nums1[i] <= 1091 <= nums2[i] <= 1050 <= index < nums2.length1 <= val <= 1051 <= tot <= 109- At most 
1000calls are made toaddandcounteach. 
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Constructor Summary
Constructors - 
Method Summary
 
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Constructor Details
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FindSumPairs
public FindSumPairs(int[] nums1, int[] nums2)  
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Method Details
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add
public void add(int index, int val)  - 
count
public int count(int tot)  
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