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Leetcode #1865: Finding Pairs With a Certain Sum

In this guide, we solve Leetcode #1865 Finding Pairs With a Certain Sum in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given two integer arrays nums1 and nums2. You are tasked to implement a data structure that supports queries of two types: Add a positive integer to an element of a given index in the array nums2.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Design, Array, Hash Table

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

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

Python Solution

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)

Complexity

The time complexity is O(n×q)O(n \times q)O(n×q), and the space complexity is O(m)O(m)O(m). The space complexity is O(m)O(m)O(m).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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