# Maximum consecutive numbers present in an array

Find the length of maximum number of consecutive numbers jumbled up in an array.

Examples:

```Input : arr[] = {1, 94, 93, 1000, 5, 92, 78};
Output : 3
The largest set of consecutive elements is
92, 93, 94

Input  : arr[] = {1, 5, 92, 4, 78, 6, 7};
Output : 4
The largest set of consecutive elements is
4, 5, 6, 7
```

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

The idea is to use hashing. We traverse through the array and for every element, we check if it is the starting element of its sequence. If yes then by incrementing its value we search the set and increment the length. By repeating this for all elements, we can find the lengths of all consecutive sets in array. Finally we return length of the largest set.

## C++

 `// CPP program to find largest consecutive numbers ` `// present in arr[]. ` `#include ` `using` `namespace` `std; ` ` `  `int` `findLongestConseqSubseq(``int` `arr[], ``int` `n) ` `{ ` `    ``/* We insert all the array elements into ` `       ``unordered set. */` `    ``unordered_set<``int``> S; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``S.insert(arr[i]); ` ` `  `    ``// check each possible sequence from the start ` `    ``// then update optimal length ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// if current element is the starting ` `        ``// element of a sequence ` `        ``if` `(S.find(arr[i] - 1) == S.end()) { ` ` `  `            ``// Then check for next elements in the ` `            ``// sequence ` `            ``int` `j = arr[i]; ` ` `  `            ``// increment the value of array element ` `            ``// and repeat search in the set ` `            ``while` `(S.find(j) != S.end()) ` `                ``j++;  ` ` `  `            ``// Update  optimal length if this length ` `            ``// is more. To get the length as it is  ` `            ``// incremented one by one ` `            ``ans = max(ans, j - arr[i]);  ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 94, 93, 1000, 5, 92, 78 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``cout << findLongestConseqSubseq(arr, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find largest consecutive  ` `// numbers present in arr[]. ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `findLongestConseqSubseq(``int` `arr[], ``int` `n) ` `{ ` `    ``/* We insert all the array elements into ` `    ``unordered set. */` `    ``HashSet S = ``new` `HashSet(); ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``S.add(arr[i]); ` ` `  `    ``// check each possible sequence from the start ` `    ``// then update optimal length ` `    ``int` `ans = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// if current element is the starting ` `        ``// element of a sequence ` `        ``if``(S.contains(arr[i]))  ` `        ``{ ` ` `  `            ``// Then check for next elements in the ` `            ``// sequence ` `            ``int` `j = arr[i]; ` ` `  `            ``// increment the value of array element ` `            ``// and repeat search in the set ` `            ``while` `(S.contains(j)) ` `                ``j++;  ` ` `  `            ``// Update optimal length if this length ` `            ``// is more. To get the length as it is  ` `            ``// incremented one by one ` `            ``ans = Math.max(ans, j - arr[i]);  ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = {``1``, ``94``, ``93``, ``1000``, ``5``, ``92``, ``78``}; ` `    ``int` `n = arr.length; ` `        ``System.out.println(findLongestConseqSubseq(arr, n)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python3

 `# Python3 program to find largest consecutive ` `# numbers present in arr. ` ` `  `def` `findLongestConseqSubseq(arr, n): ` `    ``'''We insert all the array elements into unordered set.'''` ` `  `    ``S ``=` `set``(); ` `    ``for` `i ``in` `range``(n): ` `        ``S.add(arr[i]); ` ` `  `    ``# check each possible sequence from the start ` `    ``# then update optimal length ` `    ``ans ``=` `0``; ` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# if current element is the starting ` `        ``# element of a sequence ` `        ``if` `S.__contains__(arr[i]): ` `             `  `            ``# Then check for next elements in the ` `            ``# sequence ` `            ``j ``=` `arr[i]; ` `             `  `            ``# increment the value of array element ` `            ``# and repeat search in the set ` `            ``while``(S.__contains__(j)): ` `                ``j ``+``=` `1``; ` ` `  `            ``# Update optimal length if this length ` `            ``# is more. To get the length as it is ` `            ``# incremented one by one ` `            ``ans ``=` `max``(ans, j ``-` `arr[i]); ` `    ``return` `ans; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[ ``1``, ``94``, ``93``, ``1000``, ``5``, ``92``, ``78` `]; ` `    ``n ``=` `len``(arr); ` `    ``print``(findLongestConseqSubseq(arr, n)); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# program to find largest consecutive  ` `// numbers present in arr[]. ` `using` `System; ` `using` `System.Collections.Generic; ``public` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `findLongestConseqSubseq(``int` `[]arr, ``int` `n) ` `{ ` `    ``/* We insert all the array elements into ` `    ``unordered set. */` `    ``HashSet<``int``> S = ``new` `HashSet<``int``>(); ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``S.Add(arr[i]); ` ` `  `    ``// check each possible sequence from the start ` `    ``// then update optimal length ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// if current element is the starting ` `        ``// element of a sequence ` `        ``if``(S.Contains(arr[i]))  ` `        ``{ ` ` `  `            ``// Then check for next elements in the ` `            ``// sequence ` `            ``int` `j = arr[i]; ` ` `  `            ``// increment the value of array element ` `            ``// and repeat search in the set ` `            ``while` `(S.Contains(j)) ` `                ``j++;  ` ` `  `            ``// Update optimal length if this length ` `            ``// is more. To get the length as it is  ` `            ``// incremented one by one ` `            ``ans = Math.Max(ans, j - arr[i]);  ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `[]arr = {1, 94, 93, 1000, 5, 92, 78}; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(findLongestConseqSubseq(arr, n)); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```3
```

Time complexity : O(n)

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