# Find all ranges of consecutive numbers from Array

• Difficulty Level : Hard
• Last Updated : 03 Jun, 2021

Given a sorted array arr[] consisting of N integers without any duplicates, the task is to find the ranges of consecutive numbers from that array.
Examples:

Input: arr[] = {1, 2, 3, 6, 7}
Output: 1->3, 6->7
Explanation:
There are two ranges of consecutive number from that array.
Range 1 = 1 -> 3
Range 2 = 6 -> 7
Input: arr[] = {-1, 0, 1, 2, 5, 6, 8}
Output: -1->2, 5->6, 8
Explanation:
There are three ranges of consecutive number from that array.
Range 1 = -1 -> 2
Range 2 = 5 -> 6
Range 3 = 8

Approach: The idea is to traverse the array from the initial position and for every element in the array, check the difference between the current element and the previous element.

• If the difference between the current element and the previous element is 1 then we just increment the length variable. We use the length variable to build the range “A -> B”. Since only the range is required, we don’t need to store all the elements between A and B. We just only need to know the length of this range.
• If the difference between the current element and the previous element is doesn’t equal to 1, we build the range between the first element of the range and the current previous element as the last range.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the ranges of``// consecutive numbers from array``#include``using` `namespace` `std;` `// Function to find consecutive ranges``vector consecutiveRanges(``int` `a[], ``int` `n)``{``    ``int` `length = 1;``    ``vector list;``    ` `    ``// If the array is empty,``    ``// return the list``    ``if` `(n == 0)``    ``{``        ``return` `list;``    ``}``    ` `    ``// Traverse the array from first position``    ``for``(``int` `i = 1; i <= n; i++)``    ``{``    ` `        ``// Check the difference between the``        ``// current and the previous elements``        ``// If the difference doesn't equal to 1``        ``// just increment the length variable.``        ``if` `(i == n || a[i] - a[i - 1] != 1)``        ``{``    ` `            ``// If the range contains``            ``// only one element.``            ``// add it into the list.``            ``if` `(length == 1)``            ``{``                ``list.push_back(to_string(a[i - length]));``            ``}``            ``else``            ``{``    ` `                ``// Build the range between the first``                ``// element of the range and the``                ``// current previous element as the``                ``// last range.``                ``string temp = to_string(a[i - length]) +``                            ``" -> "` `+ to_string(a[i - 1]);``                ``list.push_back(temp);``            ``}``    ` `            ``// After finding the first range``            ``// initialize the length by 1 to``            ``// build the next range.``            ``length = 1;``        ``}``        ``else``        ``{``            ``length++;``        ``}``    ``}``    ``return` `list;``}` `// Driver Code.``int` `main()``{` `    ``// Test Case 1:``    ``int` `arr1[] = { 1, 2, 3, 6, 7 };``    ``int` `n = ``sizeof``(arr1) / ``sizeof``(arr1[0]);``    ` `    ``vector ans = consecutiveRanges(arr1, n);``    ``cout << ``"["``;``    ``for``(``int` `i = 0; i < ans.size(); i++)``    ``{``        ``if``(i == ans.size() - 1)``            ``cout << ans[i] << ``"]"` `<< endl;``        ``else``            ``cout << ans[i] << ``", "``;``    ``}``    ` `    ``// Test Case 2:``    ``int` `arr2[] = { -1, 0, 1, 2, 5, 6, 8 };``    ``n = ``sizeof``(arr2) / ``sizeof``(arr2[0]);``    ``ans = consecutiveRanges(arr2, n);``    ` `    ``cout << ``"["``;``    ``for``(``int` `i = 0; i < ans.size(); i++)``    ``{``        ``if``(i == ans.size() - 1)``            ``cout << ans[i] << ``"]"` `<< endl;``        ``else``            ``cout << ans[i] << ``", "``;``    ``}``    ` `    ``// Test Case 3:``    ``int` `arr3[] = { -1, 3, 4, 5, 20, 21, 25 };``    ``n = ``sizeof``(arr3) / ``sizeof``(arr3[0]);``    ``ans = consecutiveRanges(arr3, n);``    ` `    ``cout << ``"["``;``    ``for``(``int` `i = 0; i < ans.size(); i++)``    ``{``        ``if``(i == ans.size() - 1)``            ``cout << ans[i] << ``"]"` `<< endl;``        ``else``            ``cout << ans[i] << ``", "``;``    ``}``}` `// This code is contributed by Surendra_Gangwar`

## Java

 `// Java program to find the ranges of``// consecutive numbers from array` `import` `java.util.*;` `class` `GFG {` `    ``// Function to find consecutive ranges``    ``static` `List``    ``consecutiveRanges(``int``[] a)``    ``{``        ``int` `length = ``1``;``        ``List list``            ``= ``new` `ArrayList();` `        ``// If the array is empty,``        ``// return the list``        ``if` `(a.length == ``0``) {``            ``return` `list;``        ``}` `        ``// Traverse the array from first position``        ``for` `(``int` `i = ``1``; i <= a.length; i++) {` `            ``// Check the difference between the``            ``// current and the previous elements``            ``// If the difference doesn't equal to 1``            ``// just increment the length variable.``            ``if` `(i == a.length``                ``|| a[i] - a[i - ``1``] != ``1``) {` `                ``// If the range contains``                ``// only one element.``                ``// add it into the list.``                ``if` `(length == ``1``) {``                    ``list.add(``                        ``String.valueOf(a[i - length]));``                ``}``                ``else` `{` `                    ``// Build the range between the first``                    ``// element of the range and the``                    ``// current previous element as the``                    ``// last range.``                    ``list.add(a[i - length]``                             ``+ ``" -> "` `+ a[i - ``1``]);``                ``}` `                ``// After finding the first range``                ``// initialize the length by 1 to``                ``// build the next range.``                ``length = ``1``;``            ``}``            ``else` `{``                ``length++;``            ``}``        ``}` `        ``return` `list;``    ``}` `    ``// Driver Code.``    ``public` `static` `void` `main(String args[])``    ``{` `        ``// Test Case 1:``        ``int``[] arr1 = { ``1``, ``2``, ``3``, ``6``, ``7` `};``        ``System.out.print(consecutiveRanges(arr1));``        ``System.out.println();` `        ``// Test Case 2:``        ``int``[] arr2 = { -``1``, ``0``, ``1``, ``2``, ``5``, ``6``, ``8` `};``        ``System.out.print(consecutiveRanges(arr2));``        ``System.out.println();` `        ``// Test Case 3:``        ``int``[] arr3 = { -``1``, ``3``, ``4``, ``5``, ``20``, ``21``, ``25` `};``        ``System.out.print(consecutiveRanges(arr3));``    ``}``}`

## Python3

 `# Python3 program to find``# the ranges of consecutive``# numbers from array` `# Function to find``# consecutive ranges``def` `consecutiveRanges(a, n):` `    ``length ``=` `1``    ``list` `=` `[]``    ` `    ``# If the array is empty,``    ``# return the list``    ``if` `(n ``=``=` `0``):``        ``return` `list``    ` `    ``# Traverse the array``    ``# from first position``    ``for` `i ``in` `range` `(``1``, n ``+` `1``):``    ` `        ``# Check the difference``        ``# between the current``        ``# and the previous elements``        ``# If the difference doesn't``        ``# equal to 1 just increment``        ``# the length variable.``        ``if` `(i ``=``=` `n ``or` `a[i] ``-``            ``a[i ``-` `1``] !``=` `1``):``       ` `            ``# If the range contains``            ``# only one element.``            ``# add it into the list.``            ``if` `(length ``=``=` `1``):``                ``list``.append(``str``(a[i ``-` `length]))``            ``else``:``    ` `                ``# Build the range between the first``                ``# element of the range and the``                ``# current previous element as the``                ``# last range.``                ``temp ``=` `(``str``(a[i ``-` `length]) ``+``                        ``" -> "` `+` `str``(a[i ``-` `1``]))``                ``list``.append(temp)``          ` `            ``# After finding the ``            ``# first range initialize``            ``# the length by 1 to``            ``# build the next range.``            ``length ``=` `1``       ` `        ``else``:``            ``length ``+``=` `1``    ``return` `list` `# Driver Code.``if` `__name__ ``=``=` `"__main__"``:` `    ``# Test Case 1:``    ``arr1 ``=` `[``1``, ``2``, ``3``, ``6``, ``7``]``    ``n ``=` `len``(arr1)``    ` `    ``ans ``=` `consecutiveRanges(arr1, n)``    ``print` `(``"["``, end ``=` `"")``    ``for` `i ``in` `range``(``len``(ans)):``    ` `        ``if``(i ``=``=` `len``(ans) ``-` `1``):``            ``print` `(ans[i], ``"]"``)``        ``else``:``            ``print` `(ans[i], end ``=` `", "``)``    ` `    ``# Test Case 2:``    ``arr2 ``=` `[``-``1``, ``0``, ``1``, ``2``, ``5``, ``6``, ``8``]``    ``n ``=` `len``(arr2)``    ``ans ``=` `consecutiveRanges(arr2, n)``    ` `    ``print` `(``"["``, end ``=` `"")``    ` `    ``for` `i ``in` `range` `(``len``(ans)): ``        ``if``(i ``=``=` `len``(ans) ``-` `1``):``            ``print` `(ans[i], ``"]"``)``        ``else``:``            ``print` `(ans[i], end ``=` `", "``)``  ` `    ``# Test Case 3:``    ``arr3 ``=` `[``-``1``, ``3``, ``4``, ``5``, ``20``, ``21``, ``25``]``    ``n ``=` `len``(arr3)``    ``ans ``=` `consecutiveRanges(arr3, n)``    ` `    ``print` `(``"["``, end ``=` `"")``    ` `    ``for` `i ``in` `range` `(``len``(ans)):   ``        ``if``(i ``=``=` `len``(ans) ``-` `1``):``            ``print` `(ans[i], ``"]"``)``        ``else``:``            ``print` `(ans[i], end ``=` `", "``)` `# This code is contributed by Chitranayal`

## C#

 `// C# program to find the ranges of``// consecutive numbers from array``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find consecutive ranges``static` `List consecutiveRanges(``int``[] a)``{``    ``int` `length = 1;``    ``List list = ``new` `List();` `    ``// If the array is empty,``    ``// return the list``    ``if` `(a.Length == 0)``    ``{``        ``return` `list;``    ``}` `    ``// Traverse the array from first position``    ``for``(``int` `i = 1; i <= a.Length; i++)``    ``{` `        ``// Check the difference between the``        ``// current and the previous elements``        ``// If the difference doesn't equal to 1``        ``// just increment the length variable.``        ``if` `(i == a.Length || a[i] - a[i - 1] != 1)``        ``{` `            ``// If the range contains``            ``// only one element.``            ``// add it into the list.``            ``if` `(length == 1)``            ``{``                ``list.Add(``                    ``String.Join(``""``, a[i - length]));``            ``}``            ``else``            ``{` `                ``// Build the range between the first``                ``// element of the range and the``                ``// current previous element as the``                ``// last range.``                ``list.Add(a[i - length] +``                ``" -> "` `+ a[i - 1]);``            ``}` `            ``// After finding the first range``            ``// initialize the length by 1 to``            ``// build the next range.``            ``length = 1;``        ``}``        ``else``        ``{``            ``length++;``        ``}``    ``}``    ``return` `list;``}` `static` `void` `print(List arr)``{``    ``Console.Write(``"["``);``    ``foreach``(String i ``in` `arr)``        ``Console.Write(i + ``", "``);``        ` `    ``Console.Write(``"]"``);``}` `// Driver Code.``public` `static` `void` `Main(String []args)``{` `    ``// Test Case 1:``    ``int``[] arr1 = { 1, 2, 3, 6, 7 };``    ``print(consecutiveRanges(arr1));``    ``Console.WriteLine();` `    ``// Test Case 2:``    ``int``[] arr2 = { -1, 0, 1, 2, 5, 6, 8 };``    ``print(consecutiveRanges(arr2));``    ``Console.WriteLine();` `    ``// Test Case 3:``    ``int``[] arr3 = { -1, 3, 4, 5, 20, 21, 25 };``    ``print(consecutiveRanges(arr3));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

```[1 -> 3, 6 -> 7]
[-1 -> 2, 5 -> 6, 8]
[-1, 3 -> 5, 20 -> 21, 25]```

Time Complexity: O(N), where N is the length of the array.

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