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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:

  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.

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.

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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)

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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);
 */

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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);
 */

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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);
 */

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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)
 */

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/**
 * @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)
 */

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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);
 */

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