1865. Finding Pairs With a Certain Sum 

Description

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.

Solutions

Solution 1: Hash Table

We note that the length of the array $\textit{nums1}$ does not exceed ${10}^3$, while the length of the array $\textit{nums2}$ reaches ${10}^5$. Therefore, if we directly enumerate all index pairs $(i, j)$ and check whether $\textit{nums1}[i] + \textit{nums2}[j]$ equals the specified value $\textit{tot}$, it will exceed the time limit.

Can we only enumerate the shorter array $\textit{nums1}$? The answer is yes. We use a hash table $\textit{cnt}$ to count the occurrences of each element in the array $\textit{nums2}$, then enumerate each element $x$ in the array $\textit{nums1}$ and calculate the sum of $\textit{cnt}[\textit{tot} - x]$.

When calling the $\text{add}$ method, we need to first decrement the value corresponding to $\textit{nums2}[index]$ in $\textit{cnt}$ by $1$, then add $\textit{val}$ to the value of $\textit{nums2}[index]$, and finally increment the value corresponding to $\textit{nums2}[index]$ in $\textit{cnt}$ by $1$.

When calling the $\text{count}$ method, we only need to traverse the array $\textit{nums1}$ and calculate the sum of $\textit{cnt}[\textit{tot} - x]$ for each element $x$.

The time complexity is $O(n \times q)$, and the space complexity is $O(m)$. Here, $n$ and $m$ are the lengths of the arrays $\textit{nums1}$ and $\textit{nums2}$, respectively, and $q$ is the number of times the $\text{count}$ method is called.

Python3

class FindSumPairs:

    def __init__(self, nums1: List[int], nums2: List[int]):
        self.cnt = Counter(nums2)
        self.nums1 = nums1
        self.nums2 = nums2

    def add(self, index: int, val: int) -> None:
        self.cnt[self.nums2[index]] -= 1
        self.nums2[index] += val
        self.cnt[self.nums2[index]] += 1

    def count(self, tot: int) -> int:
        return sum(self.cnt[tot - x] for x in self.nums1)


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

Java

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

    public FindSumPairs(int[] nums1, int[] nums2) {
        this.nums1 = nums1;
        this.nums2 = nums2;
        for (int x : nums2) {
            cnt.merge(x, 1, Integer::sum);
        }
    }

    public void add(int index, int val) {
        cnt.merge(nums2[index], -1, Integer::sum);
        nums2[index] += val;
        cnt.merge(nums2[index], 1, Integer::sum);
    }

    public int count(int tot) {
        int ans = 0;
        for (int x : nums1) {
            ans += cnt.getOrDefault(tot - x, 0);
        }
        return ans;
    }
}

/**
 * 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);
 */

C++

class FindSumPairs {
public:
    FindSumPairs(vector<int>& nums1, vector<int>& nums2) {
        this->nums1 = nums1;
        this->nums2 = nums2;
        for (int x : nums2) {
            ++cnt[x];
        }
    }

    void add(int index, int val) {
        --cnt[nums2[index]];
        nums2[index] += val;
        ++cnt[nums2[index]];
    }

    int count(int tot) {
        int ans = 0;
        for (int x : nums1) {
            ans += cnt[tot - x];
        }
        return ans;
    }

private:
    vector<int> nums1;
    vector<int> nums2;
    unordered_map<int, int> cnt;
};

/**
 * 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);
 */

Go

type FindSumPairs struct {
	nums1 []int
	nums2 []int
	cnt   map[int]int
}

func Constructor(nums1 []int, nums2 []int) FindSumPairs {
	cnt := map[int]int{}
	for _, x := range nums2 {
		cnt[x]++
	}
	return FindSumPairs{nums1, nums2, cnt}
}

func (this *FindSumPairs) Add(index int, val int) {
	this.cnt[this.nums2[index]]--
	this.nums2[index] += val
	this.cnt[this.nums2[index]]++
}

func (this *FindSumPairs) Count(tot int) (ans int) {
	for _, x := range this.nums1 {
		ans += this.cnt[tot-x]
	}
	return
}

/**
 * Your FindSumPairs object will be instantiated and called as such:
 * obj := Constructor(nums1, nums2);
 * obj.Add(index,val);
 * param_2 := obj.Count(tot);
 */

TypeScript

class FindSumPairs {
    private nums1: number[];
    private nums2: number[];
    private cnt: Map<number, number>;

    constructor(nums1: number[], nums2: number[]) {
        this.nums1 = nums1;
        this.nums2 = nums2;
        this.cnt = new Map();
        for (const x of nums2) {
            this.cnt.set(x, (this.cnt.get(x) || 0) + 1);
        }
    }

    add(index: number, val: number): void {
        const old = this.nums2[index];
        this.cnt.set(old, this.cnt.get(old)! - 1);
        this.nums2[index] += val;
        const now = this.nums2[index];
        this.cnt.set(now, (this.cnt.get(now) || 0) + 1);
    }

    count(tot: number): number {
        return this.nums1.reduce((acc, x) => acc + (this.cnt.get(tot - x) || 0), 0);
    }
}

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

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 */
var FindSumPairs = function (nums1, nums2) {
    this.nums1 = nums1;
    this.nums2 = nums2;
    this.cnt = new Map();
    for (const x of nums2) {
        this.cnt.set(x, (this.cnt.get(x) || 0) + 1);
    }
};

/**
 * @param {number} index
 * @param {number} val
 * @return {void}
 */
FindSumPairs.prototype.add = function (index, val) {
    const old = this.nums2[index];
    this.cnt.set(old, this.cnt.get(old) - 1);
    this.nums2[index] += val;
    const now = this.nums2[index];
    this.cnt.set(now, (this.cnt.get(now) || 0) + 1);
};

/**
 * @param {number} tot
 * @return {number}
 */
FindSumPairs.prototype.count = function (tot) {
    return this.nums1.reduce((acc, x) => acc + (this.cnt.get(tot - x) || 0), 0);
};

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

C#

public class FindSumPairs {
    private int[] nums1;
    private int[] nums2;
    private Dictionary<int, int> cnt = new Dictionary<int, int>();

    public FindSumPairs(int[] nums1, int[] nums2) {
        this.nums1 = nums1;
        this.nums2 = nums2;
        foreach (int x in nums2) {
            if (cnt.ContainsKey(x)) {
                cnt[x]++;
            } else {
                cnt[x] = 1;
            }
        }
    }

    public void Add(int index, int val) {
        int oldVal = nums2[index];
        if (cnt.TryGetValue(oldVal, out int oldCount)) {
            if (oldCount == 1) {
                cnt.Remove(oldVal);
            } else {
                cnt[oldVal] = oldCount - 1;
            }
        }
        nums2[index] += val;
        int newVal = nums2[index];
        if (cnt.TryGetValue(newVal, out int newCount)) {
            cnt[newVal] = newCount + 1;
        } else {
            cnt[newVal] = 1;
        }
    }

    public int Count(int tot) {
        int ans = 0;
        foreach (int x in nums1) {
            int target = tot - x;
            if (cnt.TryGetValue(target, out int count)) {
                ans += count;
            }
        }
        return ans;
    }
}

/**
 * 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);
 */