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1995. Finding Pairs With A Certain Sum

Difficulty: Medium

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1995. 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.

Solution

import java.util.HashMap;

class FindSumPairs {
    private HashMap<Integer, Integer> map;
    private int arr[];
    private int brr[];
    public FindSumPairs(int[] nums1, int[] nums2) {
        map = new HashMap<>();
        arr = new int[nums1.length];
        brr = new int[nums2.length];
        for (int ele : nums2)
            map.put(ele, map.getOrDefault(ele, 0) + 1);
        for (int i = 0; i < nums1.length; i++)
            arr[i] = nums1[i];
        for (int i = 0; i < nums2.length; i++)
            brr[i] = nums2[i];
    }

    public void add(int index, int val) {
        int lastVal = brr[index];
        map.put(lastVal, map.getOrDefault(lastVal, 0) -1);
        if (map.getOrDefault(lastVal, 0) == 0)
            map.remove(lastVal);
        brr[index] += val;
        map.put(brr[index], map.getOrDefault(brr[index], 0) + 1);
    }

    public int count(int tot) {
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            int current = arr[i];
            int req = tot - current;
            count += map.getOrDefault(req, 0);
        }
        return count;
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here