Find all ranges of consecutive numbers from Array

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.

Below is the implementation of the above approach:

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// C++ program to find the ranges of
// consecutive numbers from array
#include<bits/stdc++.h>
using namespace std;
 
// Function to find consecutive ranges
vector<string> consecutiveRanges(int a[], int n)
{
    int length = 1;
    vector<string> 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<string> 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
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// Java program to find the ranges of
// consecutive numbers from array
 
import java.util.*;
 
class GFG {
 
    // Function to find consecutive ranges
    static List<String>
    consecutiveRanges(int[] a)
    {
        int length = 1;
        List<String> list
            = new ArrayList<String>();
 
        // 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));
    }
}
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# 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
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// 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<String> consecutiveRanges(int[] a)
{
    int length = 1;
    List<String> list = new List<String>();
 
    // 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<String> 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
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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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