Given an array of integers, Check if there exist four elements at different indexes in the array whose sum is equal to a given value k. For example, if the given array is {1 5 1 0 6 0} and k = 7, then your function should print “YES” as (1+5+1+0=7).
Examples:
Input : arr[] = {1 5 1 0 6 0}
k = 7
Output : YES
Input : arr[] = {38 7 44 42 28 16 10 37
33 2 38 29 26 8 25}
k = 22
Output : NOWe have discussed different solutions in below two sets. Find four elements that sum to a given value | Set 1 (n^3 solution) Find four elements that sum to a given value | Set 2 ( O(n^2Logn) Solution) In this post, an optimized solution is discussed that works in O(n2) on average. The idea is to create a hashmap to store pair sums.
Loop i = 0 to n-1 :
Loop j = i + 1 to n-1
calculate sum = arr[i] + arr[j]
If (k-sum) exist in hash
a) Check in hash table for all
pairs of indexes which form
(k-sum).
b) If there is any pair with no
common indexes.
return true
Else update hash table
EndLoop;
EndLoop;Implementation:
CPP
// C++ program to find if there exist 4 elements// with given sum#include <bits/stdc++.h>using namespace std;
// function to check if there exist four// elements whose sum is equal to kbool findfour(int arr[], int n, int k)
{ // map to store sum and indexes for
// a pair sum
unordered_map<int, vector<pair<int, int> > > hash;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// calculate the sum of each pair
int sum = arr[i] + arr[j];
// if k-sum exist in map
if (hash.find(k - sum) != hash.end()) {
auto num = hash.find(k - sum);
vector<pair<int, int> > v = num->second;
// check for index coincidence as if
// there is a common that means all
// the four numbers are not from
// different indexes and one of the
// index is repeated
for (int k = 0; k < num->second.size();
k++) {
pair<int, int> it = v[k];
// if all indexes are different then
// it means four number exist
// set the flag and break the loop
if (it.first != i && it.first != j
&& it.second != i && it.second != j)
return true;
}
}
// store the sum and index pair in hashmap
hash[sum].push_back(make_pair(i, j));
}
}
hash.clear();
return false;
}// Driver codeint main()
{ int k = 7;
int arr[] = { 1, 5, 1, 0, 6, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
if (findfour(arr, n, k))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
} |
Java
import java.util.HashMap;
import java.util.Map;
import java.util.Vector;
public class Gfg {
public static boolean findfour(int[] arr, int n, int k)
{
Map<Integer, Vector<Pair> > hash = new HashMap<>();
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// calculate the sum of each pair
int sum = arr[i] + arr[j];
// if k-sum exists in map
if (hash.containsKey(k - sum)) {
Vector<Pair> v = hash.get(k - sum);
for (int kk = 0; kk < v.size(); kk++) {
Pair it = v.get(kk);
if (it.first != i && it.first != j
&& it.second != i
&& it.second != j) {
return true;
}
}
}
Vector<Pair> vec = new Vector<>();
vec.add(new Pair(i, j));
hash.put(sum, vec);
}
}
hash.clear();
return false;
}
public static void main(String[] args)
{
int k = 7;
int[] arr = { 1, 5, 1, 0, 6, 0 };
int n = arr.length;
if (findfour(arr, n, k)) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}class Pair {
public int first;
public int second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
} |
Python3
# function to check if there exist four# elements whose sum is equal to kdef findfour(arr, n, k):
# dictionary to store sum and indexes for
# a pair sum
hash = {}
for i in range(n):
for j in range(i + 1, n):
# calculate the sum of each pair
s = arr[i] + arr[j]
# if k-sum exist in dictionary
if k-s in hash:
# check for index coincidence as if
# there is a common that means all
# the four numbers are not from
# different indexes and one of the
# index is repeated
for pair in hash[k-s]:
if pair[0] != i and pair[0] != j and pair[1] != i and pair[1] != j:
return True
# store the sum and index pair in dictionary
if s in hash:
hash[s].append((i, j))
else:
hash[s] = [(i, j)]
return False
# Driver codek = 7
arr = [1, 5, 1, 0, 6, 0]
n = len(arr)
if findfour(arr, n, k):
print("YES")
else:
print("NO")
# This code is contributed by divya_p123.
|
C#
using System;
using System.Collections.Generic;
class Gfg {
public static bool findfour(int[] arr, int n, int k)
{
Dictionary<int, List<Tuple<int, int> > > hash
= new Dictionary<int,
List<Tuple<int, int> > >();
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// calculate the sum of each pair
int sum = arr[i] + arr[j];
// if k-sum exists in dictionary
if (hash.ContainsKey(k - sum)) {
List<Tuple<int, int> > v
= hash[k - sum];
for (int kk = 0; kk < v.Count; kk++) {
Tuple<int, int> it = v[kk];
if (it.Item1 != i && it.Item1 != j
&& it.Item2 != i
&& it.Item2 != j) {
return true;
}
}
}
if (!hash.ContainsKey(sum)) {
hash.Add(sum,
new List<Tuple<int, int> >());
}
hash[sum].Add(Tuple.Create(i, j));
}
}
hash.Clear();
return false;
}
public static void Main(string[] args)
{
int k = 7;
int[] arr = { 1, 5, 1, 0, 6, 0 };
int n = arr.Length;
if (findfour(arr, n, k)) {
Console.WriteLine("YES");
}
else {
Console.WriteLine("NO");
}
}
}class Pair {
public int first;
public int second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
} |
Javascript
// JavaScript Code implementation.const findFour = (arr, n, k) => { let hash = {};
for (let i = 0; i < n; i++)
{
for (let j = i + 1; j < n; j++)
{
// calculate the sum of each pair
let sum = arr[i] + arr[j];
// if k-sum exists in map
if (hash[k - sum]) {
let v = hash[k - sum];
for (let kk = 0; kk < v.length; kk++) {
let it = v[kk];
if (it[0] !== i && it[0] !== j && it[1] !== i && it[1] !== j) {
return true;
}
}
}
if (!hash[sum]) {
hash[sum] = [];
}
hash[sum].push([i, j]);
}
}
hash = {};
return false;
};let k = 7;let arr = [1, 5, 1, 0, 6, 0];let n = arr.length;if (findFour(arr, n, k)) {
console.log("YES");
} else {
console.log("NO");
}// This code is contributed by lokeshmvs21. |
Output
YES
Time Complexity: O(n^2)
Auxiliary Space: O(n) for hashmap
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