# Find largest d in array such that a + b + c = d

Given a set S (all distinct elements) of integers, find the largest d such that a + b + c = d
where a, b, c, and d are distinct elements of S.

```Constraints:
1 ≤ number of elements in the set ≤ 1000
INT_MIN ≤ each element in the set ≤ INT_MAX
```

Examples :

Input : S[] = {2, 3, 5, 7, 12}
Output : 12
Explanation: 12 is the largest d which can be represented as 12 = 2 + 3 + 7

Input : S[] = {2, 16, 64, 256, 1024}
Output : No solution

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1(Brute Force)
We can solve this problem using simple brute force approach which is not very efficient as |S| can be as large as 1000. We’ll sort the set of elements and start by finding the largest d by equating it with the sum of all possible combinations of a, b and c.
Below is the implementation of above idea :

## C++

 `// CPP Program to find the largest d ` `// such that d = a + b + c ` `#include ` `using` `namespace` `std; ` ` `  `int` `findLargestd(``int` `S[], ``int` `n) ` `{ ` `    ``bool` `found = ``false``; ` ` `  `    ``// sort the array in ` `    ``// ascending order ` `    ``sort(S, S + n); ` ` `  `    ``// iterating from backwards to  ` `    ``// find the required largest d ` `    ``for` `(``int` `i = n - 1; i >= 0; i--)  ` `    ``{ ` `        ``for` `(``int` `j = 0; j < n; j++)  ` `        ``{ ` ` `  `            ``// since all four a, b, c,  ` `            ``// d should be distinct ` `            ``if` `(i == j) ` `                ``continue``; ` ` `  `            ``for` `(``int` `k = j + 1; k < n; k++)  ` `            ``{ ` `                ``if` `(i == k) ` `                    ``continue``; ` ` `  `                ``for` `(``int` `l = k + 1; l < n; l++)  ` `                ``{ ` `                    ``if` `(i == l) ` `                        ``continue``; ` ` `  `                    ``// if the current combination   ` `                    ``// of j, k, l in the set is  ` `                    ``// equal to S[i] return this  ` `                    ``// value as this would be the  ` `                    ``// largest d since we are ` `                    ``//  iterating in descending order  ` `                    ``if` `(S[i] == S[j] + S[k] + S[l]) ` `                    ``{ ` `                        ``found = ``true``; ` `                        ``return` `S[i]; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``if` `(found == ``false``) ` `        ``return` `INT_MIN; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Set of distinct Integers ` `    ``int` `S[] = { 2, 3, 5, 7, 12 }; ` `    ``int` `n = ``sizeof``(S) / ``sizeof``(S[0]); ` ` `  `    ``int` `ans = findLargestd(S, n); ` `    ``if` `(ans == INT_MIN) ` `        ``cout << ``"No Solution"` `<< endl; ` `    ``else` `        ``cout << ``"Largest d such that a + b + "` `             ``<< ``"c = d is "` `<< ans << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java Program to find the largest ` `// such that d = a + b + c ` `import` `java.io.*; ` `import` `java.util.Arrays; ` ` `  `class` `GFG  ` `{  ` `     `  `// function to find largest d  ` `static` `int` `findLargestd(``int` `[]S, ``int` `n) ` `{ ` `    ``boolean` `found = ``false``; ` ` `  `    ``// sort the array in ` `    ``// ascending order ` `    ``Arrays.sort(S); ` ` `  `    ``// iterating from backwards to ` `    ``// find the required largest d ` `    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)  ` `    ``{ ` `        ``for` `(``int` `j = ``0``; j < n; j++) ` `        ``{ ` ` `  `            ``// since all four a, b, c,  ` `            ``// d should be distinct ` `            ``if` `(i == j) ` `                ``continue``; ` ` `  `            ``for` `(``int` `k = j + ``1``; k < n; k++)  ` `            ``{ ` `                ``if` `(i == k) ` `                    ``continue``; ` ` `  `                ``for` `(``int` `l = k + ``1``; l < n; l++)  ` `                ``{ ` `                    ``if` `(i == l) ` `                        ``continue``; ` ` `  `                    ``// if the current combination   ` `                    ``// of j, k, l in the set is  ` `                    ``// equal to S[i] return this  ` `                    ``// value as this would be the  ` `                    ``// largest d since we are   ` `                    ``// iterating in descending order  ` `                    ``if` `(S[i] == S[j] + S[k] + S[l]) ` `                    ``{ ` `                        ``found = ``true``; ` `                        ``return` `S[i]; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``if` `(found == ``false``) ` `        ``return` `Integer.MAX_VALUE; ` ` `  `    ``return` `-``1``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String []args)  ` `{  ` `    ``// Set of distinct Integers ` `    ``int` `[]S = ``new` `int``[]{ ``2``, ``3``, ``5``, ``7``, ``12` `}; ` `    ``int` `n = S.length; ` ` `  `    ``int` `ans = findLargestd(S, n); ` `    ``if` `(ans == Integer.MAX_VALUE) ` `        ``System.out.println(``"No Solution"``); ` `    ``else` `        ``System.out.println(``"Largest d such that "` `+  ` `                         ``"a + "` `+ ``"b + c = d is "` `+  ` `                                             ``ans ); ` `         `  `}  ` `} ` ` `  `// This code is contributed by Sam007 `

## Python3

 `# Python Program to find the largest  ` `# d such that d = a + b + c ` ` `  `def` `findLargestd(S, n) : ` `    ``found ``=` `False` ` `  `    ``# sort the array in ascending order ` `    ``S.sort() ` ` `  `    ``# iterating from backwards to  ` `    ``# find the required largest d ` `    ``for` `i ``in` `range``(n``-``1``, ``-``1``, ``-``1``) : ` `        ``for` `j ``in` `range``(``0``, n) : ` ` `  `            ``# since all four a, b, c,  ` `            ``# d should be distinct ` `            ``if` `(i ``=``=` `j) : ` `                ``continue` ` `  `            ``for` `k ``in` `range``(j ``+` `1``, n) : ` `                ``if` `(i ``=``=` `k) : ` `                    ``continue` ` `  `                ``for` `l ``in` `range``(k``+``1``, n) : ` `                    ``if` `(i ``=``=` `l) : ` `                        ``continue` ` `  `                    ``# if the current combination  ` `                    ``# of j, k, l in the set is  ` `                    ``# equal to S[i] return this  ` `                    ``# value as this would be the  ` `                    ``# largest d since we are  ` `                    ``# iterating in descending order ` `                    ``if` `(S[i] ``=``=` `S[j] ``+` `S[k] ``+` `S[l]) : ` `                        ``found ``=` `True` `                        ``return` `S[i]                  ` ` `  `    ``if` `(found ``=``=` `False``) : ` `        ``return` `-``1` ` `  `# Driver Code ` ` `  `# Set of distinct Integers ` `S ``=` `[ ``2``, ``3``, ``5``, ``7``, ``12` `] ` `n ``=` `len``(S) ` ` `  `ans ``=` `findLargestd(S, n) ` `if` `(ans ``=``=` `-``1``) : ` `    ``print` `(``"No Solution"``)  ` `else` `: ` `    ``print` `(``"Largest d such that a + b +"` `,  ` `        ``"c = d is"` `,ans) ` ` `  `# This code is contributed by Manish Shaw ` `# (manishshaw1) `

## C#

 `// C# Program to find the largest ` `// such that d = a + b + c ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `// function to find largest d  ` `static` `int` `findLargestd(``int` `[]S,  ` `                        ``int` `n) ` `{ ` `    ``bool` `found = ``false``; ` ` `  `    ``// sort the array  ` `    ``// in ascending order ` `    ``Array.Sort(S); ` ` `  `    ``// iterating from backwards to ` `    ``// find the required largest d ` `    ``for` `(``int` `i = n - 1; i >= 0; i--)  ` `    ``{ ` `        ``for` `(``int` `j = 0; j < n; j++) ` `        ``{ ` ` `  `            ``// since all four a, b, c,  ` `            ``// d should be distinct ` `            ``if` `(i == j) ` `                ``continue``; ` ` `  `            ``for` `(``int` `k = j + 1; k < n; k++)  ` `            ``{ ` `                ``if` `(i == k) ` `                    ``continue``; ` ` `  `                ``for` `(``int` `l = k + 1; l < n; l++)  ` `                ``{ ` `                    ``if` `(i == l) ` `                        ``continue``; ` ` `  `                    ``// if the current combination  ` `                    ``// of j, k, l in the set is  ` `                    ``// equal to S[i] return this  ` `                    ``// value as this would be the  ` `                    ``// largest dsince we are   ` `                    ``// iterating in descending order  ` `                    ``if` `(S[i] == S[j] + S[k] + S[l]) ` `                    ``{ ` `                        ``found = ``true``; ` `                        ``return` `S[i]; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``if` `(found == ``false``) ` `        ``return` `int``.MaxValue; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main()  ` `{  ` `    ``// Set of distinct Integers ` `    ``int` `[]S = ``new` `int``[]{ 2, 3, 5, 7, 12 }; ` `    ``int` `n = S.Length; ` ` `  `    ``int` `ans = findLargestd(S, n); ` `    ``if` `(ans == ``int``.MaxValue) ` `        ``Console.WriteLine( ``"No Solution"``); ` `    ``else` `        ``Console.Write(``"Largest d such that a + "` `+ ` `                          ``"b + c = d is "` `+ ans ); ` `         `  `}  ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 `= 0; ``\$i``--)  ` `    ``{ ` `        ``for` `( ``\$j` `= 0; ``\$j` `< ``\$n``; ``\$j``++)  ` `        ``{ ` ` `  `            ``// since all four a, b, c,   ` `            ``// d should be distinct ` `            ``if` `(``\$i` `== ``\$j``) ` `                ``continue``; ` ` `  `            ``for` `( ``\$k` `= ``\$j` `+ 1; ``\$k` `< ``\$n``; ``\$k``++)  ` `            ``{ ` `                ``if` `(``\$i` `== ``\$k``) ` `                    ``continue``; ` ` `  `                ``for` `( ``\$l` `= ``\$k` `+ 1; ``\$l` `< ``\$n``; ``\$l``++)  ` `                ``{ ` `                    ``if` `(``\$i` `== ``\$l``) ` `                        ``continue``; ` ` `  `                    ``// if the current combination  ` `                    ``// of j, k, l in the set is  ` `                    ``// equal to S[i] return this  ` `                    ``// value as this would be the  ` `                    ``// largest d since we are   ` `                    ``// iterating in descending order ` `                    ``if` `(``\$S``[``\$i``] == ``\$S``[``\$j``] + ``\$S``[``\$k``] + ``\$S``[``\$l``])  ` `                    ``{ ` `                        ``\$found` `= true; ` `                        ``return` `\$S``[``\$i``]; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``if` `(``\$found` `== false) ` `        ``return` `PHP_INT_MIN; ` `} ` ` `  `// Driver Code ` ` `  `// Set of distinct Integers ` `\$S` `= ``array``( 2, 3, 5, 7, 12 ); ` `\$n` `= ``count``(``\$S``); ` ` `  `\$ans` `= findLargestd(``\$S``, ``\$n``); ` `if` `(``\$ans` `== PHP_INT_MIN) ` `    ``echo` `"No Solution"` `; ` `else` `    ``echo` `"Largest d such that a + b + "` `,  ` `         ``"c = d is "` `, ``\$ans` `; ` ` `  `// This code is contributed by anuj_67. ` `?> `

Output :

```Largest d such that a + b + c = d is 12
```

This brute force solution has a time complexity of O((size of Set)4).

Method 2(Efficient Approach – Using Hashing)
The above problem statement (a + b + c = d) can be restated as finding a, b, c, d such that a + b = d – c. So this problem can be efficiently solved using hashing.

1. Store sums of all pairs (a + b) in a hash table
2. Traverse through all pairs (c, d) again and search for (d – c) in the hash table.
3. If a pair is found with the required sum, then make sure that all elements are distinct array elements and an element is not considered more than once.

Below is the implementation in C++.

## C++

 `// A hashing based CPP program to find largest d ` `// such that a + b + c = d. ` `#include ` `using` `namespace` `std; ` ` `  `// The function finds four elements with given sum X ` `int` `findFourElements(``int` `arr[], ``int` `n) ` `{ ` `    ``// Store sums  (a+b) of all pairs (a,b) in a  ` `    ``// hash table ` `    ``unordered_map<``int``, pair<``int``, ``int``> > mp; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``mp[arr[i] + arr[j]] = { i, j }; ` ` `  `    ``// Traverse through all pairs and find (d -c) ` `    ``// is present in hash table ` `    ``int` `d = INT_MIN; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``int` `abs_diff = ``abs``(arr[i] - arr[j]); ` ` `  `            ``// If d - c is present in hash table, ` `            ``if` `(mp.find(abs_diff) != mp.end()) { ` ` `  `                ``// Making sure that all elements are ` `                ``// distinct array elements and an element ` `                ``// is not considered more than once. ` `                ``pair<``int``, ``int``> p = mp[abs_diff]; ` `                ``if` `(p.first != i && p.first != j &&  ` `                    ``p.second != i && p.second != j) ` `                    ``d = max(d, max(arr[i], arr[j])); ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `d; ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 3, 5, 7, 12 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `res = findFourElements(arr, n); ` `    ``if` `(res == INT_MIN) ` `        ``cout << ``"No Solution."``; ` `    ``else` `        ``cout << res; ` `    ``return` `0; ` `} `

Output:

```12
```

The overall time complexity for this efficient approach is O(N2) (where N is the size of the set).
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