方法一：哈希表

我们注意到，数组 $\textit{nums1}$ 的长度不超过 ${10}^3$，数组 $\textit{nums2}$ 的长度达到 ${10}^5$，因此，如果直接暴力枚举所有下标对 $(i, j)$，计算 $\textit{nums1}[i] + \textit{nums2}[j]$ 是否等于指定值 $\textit{tot}$，那么会超出时间限制。

能否只枚举长度较短的数组 $\textit{nums1}$ 呢？答案是可以的。我们用一个哈希表 $\textit{cnt}$ 统计数组 $\textit{nums2}$ 中每个元素出现的次数，然后枚举数组 $\textit{nums1}$ 中的每个元素 $x$，计算 $\textit{cnt}[\textit{tot} - x]$ 的值之和即可。

在调用 $\text{add}$ 方法时，我们需要先将 $\textit{nums2}[index]$ 对应的值从 $\textit{cnt}$ 中减去 $1$，然后将 $\textit{nums2}[index]$ 的值加上 $\textit{val}$，最后将 $\textit{nums2}[index]$ 对应的值加上 $1$。

在调用 $\text{count}$ 方法时，我们只需要遍历数组 $\textit{nums1}$，对于每个元素 $x$，计算 $\textit{cnt}[\textit{tot} - x]$ 的值之和即可。

###python

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);
 */

###cpp

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);
 */

###ts

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)
 */

###rust

use std::collections::HashMap;

struct FindSumPairs {
    nums1: Vec<i32>,
    nums2: Vec<i32>,
    cnt: HashMap<i32, i32>,
}

impl FindSumPairs {
    fn new(nums1: Vec<i32>, nums2: Vec<i32>) -> Self {
        let mut cnt = HashMap::new();
        for &x in &nums2 {
            *cnt.entry(x).or_insert(0) += 1;
        }
        Self { nums1, nums2, cnt }
    }

    fn add(&mut self, index: i32, val: i32) {
        let i = index as usize;
        let old_val = self.nums2[i];
        *self.cnt.entry(old_val).or_insert(0) -= 1;
        if self.cnt[&old_val] == 0 {
            self.cnt.remove(&old_val);
        }

        self.nums2[i] += val;
        let new_val = self.nums2[i];
        *self.cnt.entry(new_val).or_insert(0) += 1;
    }

    fn count(&self, tot: i32) -> i32 {
        let mut ans = 0;
        for &x in &self.nums1 {
            let target = tot - x;
            if let Some(&c) = self.cnt.get(&target) {
                ans += c;
            }
        }
        ans
    }
}

/**
 * Your FindSumPairs object will be instantiated and called as such:
 * let mut obj = FindSumPairs::new(nums1, nums2);
 * obj.add(index, val);
 * let ret_2: i32 = obj.count(tot);
 */

###js

/**
 * @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)
 */

###cs

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);
 */

时间复杂度 $O(n \times q)$，空间复杂度 $O(m)$。其中 $n$ 和 $m$ 分别是数组 $\textit{nums1}$ 和 $\textit{nums2}$ 的长度，而 $q$ 是调用 $\text{count}$ 方法的次数。

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