1865

1865. Finding Pairs With a Certain Sum (Medium)

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.

Companies:
Quora

Related Topics:
Hash Table, Design, Ordered Map

Solution 1.

// OJ: https://leetcode.com/problems/finding-pairs-with-a-certain-sum/
// Author: github.com/lzl124631x
// Time:
//      FindSumPairs: O(B)
//      add: O(1)
//      count: O(A)
// Space: O(A + B)
class FindSumPairs {
    vector<int> A, B;
    unordered_map<int, int> m;
public:
    FindSumPairs(vector<int>& A, vector<int>& B): A(A), B(B) {
        for (int n : B) m[n]++;
    }
    void add(int index, int val) {
        int n = B[index];
        if (--m[n] == 0) m.erase(n);
        B[index] += val;
        m[B[index]]++;
    }
    int count(int tot) {
        int ans = 0;
        for (int a : A) {
            if (m.count(tot - a)) ans += m[tot - a];
        }
        return ans;
    }
};