Given two arrays arr1[] and arr2[] of pairs of the form {ID, value} of size N and M respectively and an integer target, the task is to find all the pairs of IDs from both the arrays such that the sum of the values of the pairs is maximum and has a value at most M.
Examples:
Input: arr1[] = [[1, 2000], [2, 3000], [3, 2000]], arr2[] = [[1, 2500], [2, 3000], [3, 3000]], target = 6000
Output:
2 2
2 3
Explanation:
Following are the pairs of elements from the two arrays arr1[] and arr2[]:
- (arr1[2], arr2[2]): The sum of elements 3000 + 3000 = 6000(= target) and closest to the value target. Print the IDs as (2, 2).
- (arr1[2], arr2[3]): The sum of elements 3000 + 3000 = 6000(= target) and closest to the value target. Print the IDs as (2, 3).
Input: arr1[] = [[1, 2000], [2, 3000], [3, 2000]], arr2[] = [[1, 3000], [2, 3000]], target = 5500
Output:
1 1
1 2
3 1
3 2
Approach: The given problem can be solved using the TreeMap Data Structure to store the array elements arr1[] and efficiently find the pair for every element in the other array arr2[]. Below are the steps:
- Create a TreeMap where the key is the value of the array element and the value is the list of IDs.
- Iterate the array arr1[] and insert its elements into the TreeMap.
- Initialize a variable, say closestTarget to keep track of the closest value to the target and not exceeding it.
- Initialize an ArrayList result to store all the possible pairs of IDs.
- Iterate the array arr2[] and for every element calculate the remaining value to be searched in the TreeMap.
- If the remaining value, say (target – arr2[i]) is less than 0, then the pair cannot be formed so continue the iteration.
- Use the floorKey() function of the TreeMap to find a value less than or equal to the remaining value. If the returned value of the above function is null, then no corresponding element is found.
- If the returned value, say currentTarget is greater than closestTarget, then update closestTarget and re-initialize the arrayList result[] to a new list.
- Iterate through the list of ids whose key is currentTarget and add all possible combinations of IDs pairs into the result list.
- After completing the above steps, print all the pairs of IDs stored in the ArrayList result[].
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// Function to find pairs of ids with// sum of values closest to the target// but not exceeding the targetvoid closestPair(vector<vector<int> > arr1, vector<vector<int> > arr2, int target){ // Initialize map having array // element value as key and list of // ids as value in the map map<int, vector<int> > valueToIds; // list to store all answer pairs vector<vector<int> > result; // Keeps the track of maximum sum // of pair values not exceeding target int closestTarget = 0; // Iterate through the array arr1 and // add all elements in the map for (vector<int> a : arr1) { int val = a[1], id = a[0]; if (valueToIds.find(val) == valueToIds.end()) valueToIds[val] = vector<int>(); valueToIds[val].push_back(id); } for (vector<int> b : arr2) { // Find the corresponding value // to be found int remaining = target - b[1]; if (remaining < 0) continue; // Find a value which is close to // desired value, not exceeding it map<int, vector<int> >::iterator it = valueToIds.lower_bound(remaining); if (it == valueToIds.end()) continue; int floor = it->first; int currentTarget = b[1] + floor; if (currentTarget >= closestTarget) { if (currentTarget > closestTarget) { // Update closestTarget and reset // result list closestTarget = currentTarget; result.clear(); } // Add all possible pairs in // result list for (int id : valueToIds[floor]) result.push_back({ id, b[0] }); } } // Print all the id pairs for (vector<int> ans : result) { cout << ans[0] << " " << ans[1] << endl; }}// Driver Codeint main(){ vector<vector<int> > arr1 = { { 1, 2000 }, { 2, 3000 }, { 3, 2000 } }; vector<vector<int> > arr2 = { { 1, 3000 }, { 2, 3000 } }; int target = 5500; closestPair(arr1, arr2, target); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to find pairs of ids with // sum of values closest to the target // but not exceeding the target public static void closestPair( int[][] arr1, int[][] arr2, int target) { // Initialize TreeMap having array // element value as key and list of // ids as value in the TreeMap TreeMap<Integer, List<Integer> > valueToIds = new TreeMap<>(); // list to store all answer pairs List<int[]> result = new ArrayList<>(); // Keeps the track of maximum sum // of pair values not exceeding target int closestTarget = 0; // Iterate through the array arr1 and // add all elements in the TreeMap for (int[] a : arr1) { int val = a[1], id = a[0]; valueToIds.putIfAbsent( val, new ArrayList<>()); valueToIds.get(val).add(id); } for (int[] b : arr2) { // Find the corresponding value // to be found int remaining = target - b[1]; if (remaining < 0) continue; // Find a value which is close to // desired value, not exceeding it Integer floor = valueToIds.floorKey( remaining); // No value found which is less // than or equal to floor if (floor == null) continue; int currentTarget = b[1] + floor; if (currentTarget >= closestTarget) { if (currentTarget > closestTarget) { // Update closeTarget and reset // result list closestTarget = currentTarget; result = new ArrayList<>(); } // Add all possible pairs in // result list for (int id : valueToIds.get(floor)) result.add( new int[] { id, b[0] }); } } // Print all the id pairs for (int[] ans : result) { System.out.println( ans[0] + " " + ans[1]); } } // Driver Code public static void main(String[] args) { int[][] arr1 = { { 1, 2000 }, { 2, 3000 }, { 3, 2000 } }; int[][] arr2 = { { 1, 3000 }, { 2, 3000 } }; int target = 5500; closestPair(arr1, arr2, target); }} |
Python3
from bisect import bisect_rightdef closestPair(arr1, arr2, target): # Create a dictionary to store the array elements value_to_ids = {} for a in arr1: val, id = a[1], a[0] if val not in value_to_ids: value_to_ids[val] = [] value_to_ids[val].append(id) # Create a list to store the closest pairs closest_pairs = [] closest_target = 0 # Iterate through arr2 for b in arr2: remaining = target - b[1] if remaining < 0: continue # Find the closest value to remaining in value_to_ids closest_val = bisect_right(sorted(value_to_ids.keys()), remaining) if closest_val == 0: continue closest_val = sorted(value_to_ids.keys())[closest_val - 1] current_target = b[1] + closest_val if current_target >= closest_target: if current_target > closest_target: closest_target = current_target closest_pairs = [] for id in value_to_ids[closest_val]: closest_pairs.append((id, b[0])) # Print all the id pairs for pair in closest_pairs: print(pair[0], pair[1]) # Driver Codeif __name__ == "__main__": arr1 = [[1, 2000], [2, 3000], [3, 2000]] arr2 = [[1, 3000], [2, 3000]] target = 5500 closestPair(arr1, arr2, target) |
C#
using System;using System.Collections.Generic;class GFG { // Function to find pairs of ids with // sum of values closest to the target // but not exceeding the target public static void closestPair(int[][] arr1, int[][] arr2, int target) { // Initialize Dictionary having array // element value as key and list of // ids as value in the Dictionary Dictionary<int, List<int>> valueToIds = new Dictionary<int, List<int>>(); // list to store all answer pairs List<int[]> result = new List<int[]>(); // Keeps the track of maximum sum // of pair values not exceeding target int closestTarget = 0; // Iterate through the array arr1 and // add all elements in the Dictionary foreach (int[] a in arr1) { int val = a[1], id = a[0]; if (!valueToIds.ContainsKey(val)) { valueToIds[val] = new List<int>(); } valueToIds[val].Add(id); } foreach (int[] b in arr2) { // Find the corresponding value // to be found int remaining = target - b[1]; if (remaining < 0) { continue; } // Find a value which is close to // desired value, not exceeding it int? floor = null; foreach (int key in valueToIds.Keys) { if (key <= remaining) { floor = floor == null ? key : Math.Max(floor.Value, key); } } // No value found which is less // than or equal to floor if (floor == null) { continue; } int currentTarget = b[1] + floor.Value; if (currentTarget >= closestTarget) { if (currentTarget > closestTarget) { // Update closeTarget and reset // result list closestTarget = currentTarget; result = new List<int[]>(); } // Add all possible pairs in // result list foreach (int id in valueToIds[floor.Value]) { result.Add(new int[] { id, b[0] }); } } } // Print all the id pairs foreach (int[] ans in result) { Console.WriteLine(ans[0] + " " + ans[1]); } } // Driver Code public static void Main(string[] args) { int[][] arr1 = new int[][] { new int[] { 1, 2000 }, new int[] { 2, 3000 }, new int[] { 3, 2000 } }; int[][] arr2 = new int[][] { new int[] { 1, 3000 }, new int[] { 2, 3000 } }; int target = 5500; closestPair(arr1, arr2, target); }} |
Javascript
function closestPair(arr1, arr2, target) { // Create a dictionary to store the array elements let value_to_ids = {}; for (let a of arr1) { let val = a[1], id = a[0]; if (!value_to_ids[val]) { value_to_ids[val] = []; } value_to_ids[val].push(id); } // Create a list to store the closest pairs let closest_pairs = []; let closest_target = 0; // Iterate through arr2 for (let b of arr2) { let remaining = target - b[1]; if (remaining < 0) { continue; } // Find the closest value to remaining in value_to_ids let keys = Object.keys(value_to_ids).sort((a, b) => a - b); let closest_val = keys.find(k => k >= remaining); if (!closest_val) { continue; } let current_target = b[1] + closest_val; if (current_target >= closest_target) { if (current_target > closest_target) { closest_target = current_target; closest_pairs = []; } for (let id of value_to_ids[closest_val]) { closest_pairs.push([id, b[0]]); } } } // Print all the id pairs for (let pair of closest_pairs) { console.log(pair[0], pair[1]); }}// Driver Codeif (typeof require === 'undefined' || require.main === module) { let arr1 = [[1, 2000], [2, 3000], [3, 2000]]; let arr2 = [[1, 3000], [2, 3000]]; let target = 5500; closestPair(arr1, arr2, target);}// This code is contributed by aadityamaharshi21. |
Output:
1 1 3 1 1 2 3 2
Time Complexity: O(N*log N + M)
Auxiliary Space: O(N*M)
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