Yet Another Query Problem

Summary

Given an array of integers, your task is to find all pairs that sum up to a given target value. This problem involves using prefix sums and arrays, making it a medium-level difficulty query problem.

Detailed Explanation

The basic idea is to use the property of prefix sums: if we have two indices i and j, such that arr[i] + arr[j] = target, then their prefix sums should be equal. We iterate over the array once, calculate the prefix sum for each element, and store it in a hash map.

Here's a step-by-step breakdown of the solution:

Initialize an empty hash map prefixSumMap.
Iterate over the array from left to right:
- Calculate the prefix sum ps for each element.
- Check if there exists another element j such that arr[j] + ps = target. If found, return the pair (arr[i], arr[j]).
Return an empty list if no pairs are found.

Time complexity: O(n), where n is the length of the array. We only iterate over the array once. Space complexity: O(n), for storing the prefix sums in the hash map.

Here's a simple ASCII art diagram to illustrate the process:

  arr = [1, 2, 3, 4]
  ps_map = {}
  i = 0
  while i < len(arr):
    ps = sum(arr[:i+1])  # prefix sum up to i
    if target - ps in ps_map:
      return [(arr[i], x) for x in ps_map[target-ps]]
    ps_map[ps] = arr[i]
    i += 1

Optimized Solutions

Java

public List<int[]> findPairs(int[] arr, int target) { Map<Integer, Integer> prefixSumMap = new HashMap<>(); for (int i = 0; i < arr.length; i++) { int ps = 0; for (int j = 0; j <= i; j++) { ps += arr[j]; } if (prefixSumMap.containsKey(target - ps)) { return Arrays.stream(prefixSumMap.get(target - ps)).boxed().collect(Collectors.toList()); } prefixSumMap.put(ps, arr[i]); } return new ArrayList<>(); }