# 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 is92, 93, 94 Input  : arr[] = {1, 5, 92, 4, 78, 6, 7};Output : 4 The largest set of consecutive elements is4, 5, 6, 7`

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`

## Javascript

 ``

Output
```3
```

Time complexity : O(n)
Space complexity: O(n)

Another approach: The idea is to sort the array. We will traverse through the array and check if the difference between the current element and the previous element is one or not. If the difference is one we will increment the count of the length of the current sequence. Otherwise, we will check if the count of the length of our current subsequence is greater than the length of our previously counted sequence. If it is, we will update our answer and then we will update the count to one to start counting the length of another sequence. By repeating this for all elements, we can find the lengths of all consecutive sequences in the array. Finally, we return the length of the largest sequence.

## C++

 `// CPP program to find largest consecutive numbers present``// in arr.``#include ``using` `namespace` `std;` `int` `findLongestConseqSubseq(``int` `arr[], ``int` `n)``{ ``  ` `  ``// The longest sequence in an empty array is, of``    ``// course, 0, so we can simply return that.``    ``if` `(n == 0) {``        ``return` `0;``    ``}` `    ``// We will arrange array elements in ascending order``    ``// using sort function.``    ``sort(arr, arr + n);` `    ``// check each possible sequence from the start then``    ``// update optimal length``    ``int` `ans = 1;``    ``int` `count = 1;``    ``for` `(``int` `i = 1; i < n; i++)``    ``{``      ` `        ``// For handling duplicate elements``        ``if` `(arr[i] != arr[i - 1])``        ``{``          ` `            ``// if difference between current element and``            ``// previous element is 1 then we want to``            ``// update our current sequence count``            ``if` `(arr[i] - arr[i - 1] == 1) {``                ``count += 1;``            ``}``          ` `            ``// otherwise, we will update our count to``            ``// zero to check for other sequences. before``            ``// updating count value we have to check if``            ``// current sequence length is more than our``            ``// ans. if count > ans then we want to``            ``// update our ans.``            ``else` `{``                ``ans = max(ans, count);``                ``count = 1;``            ``}``        ``}``    ``}``  ` `    ``// To handle the case in which last element is``    ``// present in longest sequence.``    ``return` `max(ans, count);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 94, 93, 1000, 5, 92, 78 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``  ` `    ``// Function call``    ``cout << findLongestConseqSubseq(arr, n) << endl;``    ``return` `0;``}` `// This code is contributed by Naveen Shah.`

## Java

 `// Java program to find largest consecutive numbers present``// in arr.``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `  ``static` `int` `findLongestConseqSubseq(``int``[] arr, ``int` `n)``  ``{` `    ``// The longest sequence in an empty array is, of``    ``// course, 0, so we can simply return that.``    ``if` `(n == ``0``) {``      ``return` `0``;``    ``}` `    ``// We will arrange array elements in ascending order``    ``// using sort function.``    ``Arrays.sort(arr);` `    ``// check each possible sequence from the start then``    ``// update optimal length``    ``int` `ans = ``1``;``    ``int` `count = ``1``;``    ``for` `(``int` `i = ``1``; i < n; i++) {``      ``// For handling duplicate elements``      ``if` `(arr[i] != arr[i - ``1``]) {``        ``// if difference between current element and``        ``// previous element is 1 then we want to``        ``// update our current sequence count``        ``if` `(arr[i] - arr[i - ``1``] == ``1``) {``          ``count += ``1``;``        ``}``        ``// otherwise, we will update our count to``        ``// zero to check for other sequences. before``        ``// updating count value we have to check if``        ``// current sequence length is more than our``        ``// ans. if count > ans then we want to``        ``// update our ans.``        ``else` `{``          ``ans = Math.max(ans, count);``          ``count = ``1``;``        ``}``      ``}``    ``}` `    ``// To handle the case in which last element is``    ``// present in longest sequence.``    ``return` `Math.max(ans, count);``  ``}` `  ``public` `static` `void` `main(String[] args)``  ``{``    ``int``[] arr = { ``1``, ``94``, ``93``, ``1000``, ``5``, ``92``, ``78` `};``    ``int` `n = arr.length;` `    ``// Function call``    ``System.out.print(findLongestConseqSubseq(arr, n));``  ``}``}` `// This code is contributed by lokeshmvs21.`

## Python3

 `# Python3 program to find largest consecutive``# numbers present in arr.`  `def` `findLongestConseqSubseq(arr, n):``    ``#The longest sequence in an empty array is, of course, 0, so we can simply return that. ``    ``if` `n``=``=``0``:``       ``return` `0``      ` `    ``#We will arrange array elements in ascending order using sort function.``    ``arr.sort()` `    ``# check each possible sequence from the start``    ``# then update optimal length``    ``ans ``=` `1``    ``count ``=` `1``    ``for` `i ``in` `range``(``1``, n):``      ``#For handling duplicate elements``      ``if` `arr[i]!``=``arr[i``-``1``]:` `        ``# if difference between current element and previous element is 1``        ``# then we want to update our current sequence count` `        ``if` `arr[i]``-``arr[i``-``1``] ``=``=` `1``:``            ``count ``+``=` `1` `        ``# otherwise, we will update our count to zero to check for other sequences.``        ``# before updating count value we have to check if current sequence length is more than our ans.``        ``# if count > ans then we want to update our ans.` `        ``else``:``            ``ans ``=` `max``(ans, count)``            ``count ``=` `1` `    ``return` `max``(ans,count) ``#To handle the case in which last element is present in longest sequence.`  `# 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 sanjanasikarwar24`

## C#

 `// C# program to find largest consecutive numbers present``// in arr.``using` `System;``using` `System.Collections;` `public` `class` `GFG {` `  ``static` `int` `findLongestConseqSubseq(``int``[] arr, ``int` `n)``  ``{` `    ``// The longest sequence in an empty array is, of``    ``// course, 0, so we can simply return that.``    ``if` `(n == 0) {``      ``return` `0;``    ``}` `    ``// We will arrange array elements in ascending order``    ``// using sort function.``    ``Array.Sort(arr);` `    ``// check each possible sequence from the start then``    ``// update optimal length``    ``int` `ans = 1;``    ``int` `count = 1;``    ``for` `(``int` `i = 1; i < n; i++) {``      ``// For handling duplicate elements``      ``if` `(arr[i] != arr[i - 1]) {``        ``// if difference between current element and``        ``// previous element is 1 then we want to``        ``// update our current sequence count``        ``if` `(arr[i] - arr[i - 1] == 1) {``          ``count += 1;``        ``}``        ``// otherwise, we will update our count to``        ``// zero to check for other sequences. before``        ``// updating count value we have to check if``        ``// current sequence length is more than our``        ``// ans. if count > ans then we want to``        ``// update our ans.``        ``else` `{``          ``ans = Math.Max(ans, count);``          ``count = 1;``        ``}``      ``}``    ``}` `    ``// To handle the case in which last element is``    ``// present in longest sequence.``    ``return` `Math.Max(ans, count);``  ``}` `  ``static` `public` `void` `Main()``  ``{` `    ``// Code``    ``int``[] arr = { 1, 94, 93, 1000, 5, 92, 78 };``    ``int` `n = arr.Length;` `    ``// Function call``    ``Console.Write(findLongestConseqSubseq(arr, n));``  ``}``}` `// This code is contributed by lokeshmvs21.`

## Javascript

 `function` `findLongestConseqSubseq(arr, n) {``  ``// The longest sequence in an empty array is, of course, 0, so we can simply return that.``  ``if` `(n == 0) ``return` `0;` `  ``// We will arrange array elements in ascending order using sort function.``  ``arr.sort((a, b) => a - b);` `  ``// check each possible sequence from the start``  ``// then update optimal length``  ``let ans = 1,``    ``count = 1;``  ``for` `(let i = 1; i < n; i++) {``    ``// For handling duplicate elements``    ``if` `(arr[i] != arr[i - 1]) {``      ``// if difference between current element and previous element is 1``      ``// then we want to update our current sequence count``      ``if` `(arr[i] - arr[i - 1] == 1) {``        ``count++;``      ``}``      ``// otherwise, we will update our count to zero to check for other sequences.``      ``// before updating count value we have to check if current sequence length is more than our ans.``      ``// if count > ans then we want to update our ans.``      ``else` `{``        ``ans = Math.max(ans, count);``        ``count = 1;``      ``}``    ``}``  ``}` `  ``// To handle the case in which last element is present in longest sequence.``  ``return` `Math.max(ans, count);``}` `// Driver code``let arr = [1, 94, 93, 1000, 5, 92, 78];``let n = arr.length;``console.log(findLongestConseqSubseq(arr, n));``//This code is contributed by sanjanasikarwar24`

Output
```3
```

Time complexity : O(nlogn)

Space complexity: O(1)

Another approach: The idea is to use set. We traverse through the array and for every element, we check if it is the starting element of its sequence( no element whose value is less than the current element by one is present in the set ). If yes then by incrementing its value we search for other valid elements that could be present in the set and increment the length of the sequence accordingly. By repeating this for all elements, we can find the lengths of all consecutive sequences in the array. Finally, we return the length of the largest sequence

## C++

 `#include ``#include ``using` `namespace` `std;` `int` `findLongestConseqSubseq(``int` `arr[], ``int` `n)``{``    ``// We insert all the array elements into set.``    ``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];``            ``while` `(S.find(j) != S.end())``                ``j++;` `            ``// update optimal length if this length``            ``// is more``            ``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``(arr[0]);``    ``cout << findLongestConseqSubseq(arr, n) << endl;``    ``return` `0;``}``// This code is contributed by sanjanasikarwar24`

## Java

 `import` `java.util.HashSet;``import` `java.util.Set;` `public` `class` `Main {``  ``public` `static` `int` `findLongestConseqSubseq(``int``[] arr, ``int` `n) ``  ``{` `    ``// We insert all the array elements into set.``    ``Set 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] - ``1``))``      ``{` `        ``// Then check for next elements in the``        ``// sequence``        ``int` `j = arr[i];``        ``while` `(S.contains(j))``          ``j++;` `        ``// update optimal length if this length``        ``// is more``        ``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 is contributed by sanjanasikarwar24`

## Python3

 `# Python3 program to find largest consecutive``# numbers present in arr.``def` `findLongestConseqSubseq(arr, n):``    ``'''We insert all the array elements into set.'''` `    ``S ``=` `set``(arr)` `    ``# check each possible sequence from the start``    ``# then update optimal length``    ``ans ``=` `0``    ``for` `e ``in` `arr:` `        ``# i contains current element of array``        ``i ``=` `e``        ``# count represents the length of current sequence``        ``count ``=` `1` `        ``# if current element is the starting``        ``# element of a sequence``        ``if` `i``-``1` `not` `in` `S:``            ``# Then check for next elements in the``            ``# sequence``            ``while` `i``+``1` `in` `S:``                ``# increment the value of array element``                ``# and repeat search in the set``                ``i ``+``=` `1``                ``count ``+``=` `1` `            ``# Update optimal length if this length``            ``# is more.``            ``ans ``=` `max``(ans, count)``    ``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 sanjanasikarwar24`

## C#

 `using` `System;``using` `System.Collections.Generic;` `public` `class` `MainClass {``  ``public` `static` `int` `findLongestConseqSubseq(``int``[] arr, ``int` `n) {``    ``// We insert all the array elements into set.``    ``HashSet<``int``> S = ``new` `HashSet<``int``>();``    ``foreach` `(``int` `x ``in` `arr)``      ``S.Add(x);` `    ``// check each possible sequence from the start``    ``// then update optimal length``    ``int` `ans = 0;``    ``foreach` `(``int` `x ``in` `arr) {``      ``// if current element is the starting``      ``// element of a sequence``      ``if` `(!S.Contains(x - 1)) {``        ``// Then check for next elements in the``        ``// sequence``        ``int` `j = x;``        ``while` `(S.Contains(j)) j++;` `        ``// update optimal length if this length``        ``// is more``        ``ans = Math.Max(ans, j - x);``      ``}``    ``}``    ``return` `ans;``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main() {``    ``int``[] arr = { 1, 94, 93, 1000, 5, 92, 78 };``    ``int` `n = arr.Length;``    ``Console.WriteLine(findLongestConseqSubseq(arr, n));``  ``}``}``//This code is contributed by sanjanasikarwar24`

## Javascript

 `function` `findLongestConseqSubseq(arr, n)``{` `  ``// We insert all the array elements into set.``  ``let S = ``new` `Set(arr);` `  ``// check each possible sequence from the start``  ``// then update optimal length``  ``let ans = 0;``  ``for` `(let i = 0; i < n; i++)``  ``{``  ` `    ``// if current element is the starting``    ``// element of a sequence``    ``if` `(!S.has(arr[i] - 1))``    ``{``    ` `      ``// Then check for next elements in the``      ``// sequence``      ``let j = arr[i];``      ``while` `(S.has(j)) j++;` `      ``// update optimal length if this length``      ``// is more``      ``ans = Math.max(ans, j - arr[i]);``    ``}``  ``}``  ``return` `ans;``}` `// Driver code``let arr = [1, 94, 93, 1000, 5, 92, 78];``let n = arr.length;``console.log(findLongestConseqSubseq(arr, n));` `// This code is contributed by sanjanasikarwar24`

Output
```3
```

Time complexity: O(nlogn)

Space complexity: O(n)

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