Given K arrays of different size. The task is to check if there exist any two arrays which have the same sum of elements after removing exactly one element from each of them. (Any element can be removed, but exactly one has to be removed). Print the indices of the array and the index of the removed elements if such pairs exist. If there are multiple pairs, print any one of them. If no such pairs exist, print -1. Examples:
Input: k = 3 a1 = {8, 1, 4, 7, 1} a2 = {10, 10} a3 = {1, 3, 4, 7, 3, 2, 2} Output: Array 1, index 4 Array 3, index 5 sum of Array 1{8, 1, 4, 7, 1} without index 4 is 20. sum of Array 3{1, 3, 4, 7, 3, 2, 2} without index 5 is 20. Input: k = 4 a1 = {2, 4, 6, 6} a2 = {1, 2, 4, 8, 16} a3 = {1, 3, 8} a4 = {1, 4, 16, 64} Output: -1
Brute Force: For every pair of arrays, for each element, find the sum excluding that element and compare it with the sum excluding each element one by one in the second array of the chosen pair. (Here maxl denotes the maximum length of an array in the set). Time Complexity : 
Space Complexity : 
Efficient Approach: Precompute all possible values of the sum obtained by removing one element from each of the arrays. Store the array index and element index which is removed with the computed sum. When these values are arranged in increasing order, it can easily be seen that if a solution exists, then both the sum values must be adjacent to the new arrangement. When two adjacent sum values are same, check if they belong to different arrays. If they do, print the array number and index of the element removed. If no such sum value is found, then no such pairs exist. Below is the implementation of the above approach:
CPP
// C++program to print the pair of arrays// whose sum are equal after removing// exactly one element from each#include <bits/stdc++.h>using namespace std;// Function to print the pair of array and index// of element to be removed to make sum equalvoid printPair(vector<int> a[], int k){ // stores the sum removing one element, // array number and index of removed element vector<pair<int, pair<int, int> > > ans; // traverse in every array for (int i = 0; i < k; i++) { // length of array int l = a[i].size(); int sum = 0; // compute total sum of array for (int j = 0; j < l; j++) { sum = sum + a[i][j]; } // remove each element once and insert sum in // ans vector along with index for (int j = 0; j < l; j++) { ans.push_back({ sum - a[i][j], { i + 1, j } }); } } // sort the ans vector so that // same sum values after removing // a single element comes together sort(ans.begin(), ans.end()); bool flag = false; // iterate and check if any adjacent sum are equal for (int p = 1; p < ans.size(); p++) { // check if the adjacent sum belong to different array // if the adjacent sum is equal if (ans[p - 1].first == ans[p].first && ans[p - 1].second.first != ans[p].second.first) { // first array number int ax = ans[p - 1].second.first; // element's index removed from first array int aidx = ans[p - 1].second.second; // second array number int bx = ans[p].second.first; // element's index removed from second array int bidx = ans[p].second.second; cout << "Array " << ax << ", index " << aidx << "\n"; cout << "Array " << bx << ", index " << bidx << "\n"; flag = true; break; } } // If no pairs are found if (!flag) cout << "No special pair exists\n";}// Driver Codeint main(){ // k sets of array vector<int> a[] = { { 8, 1, 4, 7, 1 }, { 10, 10 }, { 1, 3, 4, 7, 3, 2, 2 } }; int k = sizeof(a) / sizeof(a[0]); // Calling Function to print the pairs if any printPair(a, k); return 0;} |
Java
import java.util.ArrayList;import java.util.List;import java.util.Collections;class Pair<F, S> { private F first; private S second; public Pair(F first, S second) { this.first = first; this.second = second; } public F getFirst() { return first; } public S getSecond() { return second; }}public class Main { // Function to print the pair of array and index // of element to be removed to make the sum equal static void printPair(List<Integer>[] a, int k) { // Stores the sum removing one element, // array number, and index of the removed element List<Pair<Integer, Pair<Integer, Integer>>> ans = new ArrayList<>(); // Traverse in every array for (int i = 0; i < k; i++) { // Length of array int l = a[i].size(); int sum = 0; // Compute the total sum of the array for (int j = 0; j < l; j++) { sum += a[i].get(j); } // Remove each element once and insert the sum in // ans list along with index for (int j = 0; j < l; j++) { ans.add(new Pair<>(sum - a[i].get(j), new Pair<>(i + 1, j))); } } // Sort the ans list so that // the same sum values after removing // a single element come together Collections.sort(ans, (p1, p2) -> p1.getFirst().compareTo(p2.getFirst())); boolean flag = false; // Iterate and check if any adjacent sums are equal for (int p = 1; p < ans.size(); p++) { // Check if the adjacent sum belongs to different arrays // If the adjacent sums are equal if (ans.get(p - 1).getFirst().equals(ans.get(p).getFirst()) && !ans.get(p - 1).getSecond().getFirst().equals(ans.get(p).getSecond().getFirst())) { // First array number int ax = ans.get(p - 1).getSecond().getFirst(); // Element's index removed from the first array int aidx = ans.get(p - 1).getSecond().getSecond(); // Second array number int bx = ans.get(p).getSecond().getFirst(); // Element's index removed from the second array int bidx = ans.get(p).getSecond().getSecond(); System.out.println("Array " + ax + ", index " + aidx); System.out.println("Array " + bx + ", index " + bidx); flag = true; break; } } // If no pairs are found if (!flag) System.out.println("No special pair exists"); } public static void main(String[] args) { // k sets of arrays List<Integer>[] a = new List[]{ List.of(8, 1, 4, 7, 1), List.of(10, 10), List.of(1, 3, 4, 7, 3, 2, 2) }; int k = a.length; // Calling the function to print the pairs if any printPair(a, k); }} |
Python3
def print_pair(a): k = len(a) # stores the sum removing one element, # array number and index of removed element ans = [] # traverse in every array for i in range(k): # length of array l = len(a[i]) sum = 0 # compute total sum of array for j in range(l): sum = sum + a[i][j] # remove each element once and insert sum in # ans list along with index for j in range(l): ans.append((sum - a[i][j], (i + 1, j))) # sort the ans list so that # same sum values after removing # a single element comes together ans.sort() flag = False # iterate and check if any adjacent sum are equal for p in range(1, len(ans)): # check if the adjacent sum belong to different array # if the adjacent sum is equal if ans[p - 1][0] == ans[p][0] and ans[p - 1][1][0] != ans[p][1][0]: # first array number ax = ans[p - 1][1][0] # element's index removed from first array aidx = ans[p - 1][1][1] # second array number bx = ans[p][1][0] # element's index removed from second array bidx = ans[p][1][1] print(f"Array {ax}, index {aidx}") print(f"Array {bx}, index {bidx}") flag = True break # If no pairs are found if not flag: print("No special pair exists")# k sets of arraya = [ [8, 1, 4, 7, 1], [10, 10], [1, 3, 4, 7, 3, 2, 2]]# Calling Function to print the pairs if anyprint_pair(a) |
C#
using System;using System.Collections.Generic;using System.Linq;class Program{ // Function to print the pair of array and index // of element to be removed to make sum equal static void PrintPair(List<int>[] a, int k) { // stores the sum removing one element, // array number, and index of removed element List<Tuple<int, Tuple<int, int>>> ans = new List<Tuple<int, Tuple<int, int>>>(); // traverse in every array for (int i = 0; i < k; i++) { // length of array int l = a[i].Count; int sum = 0; // compute the total sum of array for (int j = 0; j < l; j++) { sum = sum + a[i][j]; } // remove each element once and insert sum in // ans list along with index for (int j = 0; j < l; j++) { ans.Add(new Tuple<int, Tuple<int, int>>(sum - a[i][j], new Tuple<int, int>(i + 1, j))); } } // sort the ans list so that // same sum values after removing // a single element come together ans = ans.OrderBy(t => t.Item1).ToList(); bool flag = false; // iterate and check if any adjacent sums are equal for (int p = 1; p < ans.Count; p++) { // check if the adjacent sum belongs to a different array // if the adjacent sum is equal if (ans[p - 1].Item1 == ans[p].Item1 && ans[p - 1].Item2.Item1 != ans[p].Item2.Item1) { // first array number int ax = ans[p - 1].Item2.Item1; // element's index removed from the first array int aidx = ans[p - 1].Item2.Item2; // second array number int bx = ans[p].Item2.Item1; // element's index removed from the second array int bidx = ans[p].Item2.Item2; Console.WriteLine($"Array {ax}, index {aidx}"); Console.WriteLine($"Array {bx}, index {bidx}"); flag = true; break; } } // If no pairs are found if (!flag) Console.WriteLine("No special pair exists"); } // Driver Code static void Main() { // k sets of array List<int>[] a = { new List<int> { 8, 1, 4, 7, 1 }, new List<int> { 10, 10 }, new List<int> { 1, 3, 4, 7, 3, 2, 2 } }; int k = a.Length; // Calling Function to print the pairs if any PrintPair(a, k); }} |
Javascript
function printPair(arrays) { const ans = []; // Traverse through each array for (let i = 0; i < arrays.length; i++) { const array = arrays[i]; const sum = array.reduce((acc, val) => acc + val, 0); // Remove each element once and insert sum in ans array along with index for (let j = 0; j < array.length; j++) { ans.push([sum - array[j], [i + 1, j]]); } } // Sort the ans array so that same sum values after removing a single element come together ans.sort((a, b) => a[0] - b[0]); let flag = false; // Iterate and check if any adjacent sums are equal for (let p = 1; p < ans.length; p++) { // Check if the adjacent sums belong to different arrays and if the sums are equal if (ans[p - 1][0] === ans[p][0] && ans[p - 1][1][0] !== ans[p][1][0]) { const ax = ans[p - 1][1][0]; const aidx = ans[p - 1][1][1]; const bx = ans[p][1][0]; const bidx = ans[p][1][1]; console.log(`Array ${ax}, index ${aidx}`); console.log(`Array ${bx}, index ${bidx}`); flag = true; break; } } // If no pairs are found if (!flag) { console.log("No special pair exists"); }}// Driver Codeconst arrays = [ [8, 1, 4, 7, 1], [10, 10], [1, 3, 4, 7, 3, 2, 2]];// Calling the function to print the pairs if anyprintPair(arrays);// code is contributed by shinjanpatra |
Output
Array 1, index 4 Array 3, index 5
Time Complexity : 
, or simply, Space Complexity : , or simply, 
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