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: Hash Table

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 */
const FindSumPairs = function (nums1, nums2) {
  this.nums1 = nums1;
  this.nums2 = nums2;
  this.num2Map = nums2.reduce((map, num) => {
    const count = map.get(num) ?? 0;

    return map.set(num, count + 1);
  }, new Map());
};

/**
 * @param {number} index
 * @param {number} val
 * @return {void}
 */
FindSumPairs.prototype.add = function (index, val) {
  const num2 = this.nums2[index];
  const sum = num2 + val;
  const count = this.num2Map.get(sum) ?? 0;

  this.nums2[index] = sum;
  this.num2Map.set(num2, this.num2Map.get(num2) - 1);
  this.num2Map.set(sum, count + 1);
};

/**
 * @param {number} tot
 * @return {number}
 */
FindSumPairs.prototype.count = function (tot) {
  return this.nums1.reduce((result, num1) => {
    const num2 = tot - num1;
    if (num2 < 0) return result;
    const num2Count = this.num2Map.get(num2) ?? 0;

    return result + num2Count;
  }, 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)
 */