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.length
and0 <= j < nums2.length
).
Implement the FindSumPairs
class:
FindSumPairs(int[] nums1, int[] nums2)
Initializes theFindSumPairs
object with two integer arraysnums1
andnums2
.void add(int index, int val)
Addsval
tonums2[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 <= 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 toadd
andcount
each.