content: package codingquestions.leetcode;
  
  import java.util.HashMap;
  import java.util.Map;
  
  /**
   * 1865. Finding Pairs With a Certain Sum
   * Solved
   * Medium
   * Topics
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   * Hint
   * You are given two integer arrays nums1 and nums2. You are tasked to implement a data structure that supports queries of two types:
   * <p>
   * 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:
   * <p>
   * 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.
   * <p>
   * <p>
   * Example 1:
   * <p>
   * 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]
   * <p>
   * 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
   * <p>
   * <p>
   * Constraints:
   * <p>
   * 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.
   */
  public class FindingPairsWithACertainSum {
      int[] nums1;
      int[] nums2;
      Map<Integer, Integer> m = new HashMap<>();
  
      public FindingPairsWithACertainSum(int[] nums1, int[] nums2) {
          this.nums1 = nums1;
          this.nums2 = nums2;
          for (int j = 0; j < nums2.length; j++) {
              m.put(nums2[j], m.getOrDefault(nums2[j], 0) + 1);
          }
      }
  
      public static void main(String[] args) {
  
      }
  
      public void add(int index, int val) {
          m.put(nums2[index], m.get(nums2[index]) - 1);
          nums2[index] = nums2[index] + val;
          m.put(nums2[index], m.getOrDefault(nums2[index], 0) + 1);
      }
  
      public int count(int tot) {
          int count = 0;
  
          for (int i = 0; i < nums1.length; i++) {
              count += m.getOrDefault(tot - nums1[i], 0);
          }
          return count;
      }
  
  }
  