Compress the array into Ranges - GeeksforGeeks

18 Nov, 2021
Given an array of integers of size N, The task is to print the consecutive integers as a range.Examples: 
Input : N = 7, arr=[7, 8, 9, 15, 16, 20, 25] Output : 7-9 15-16 20 25 Consecutive elements present are[ {7, 8, 9}, {15, 16}, {20}, {25} ] Hence output the result as 7-9 15-16 20 25Input : N = 6, arr=[1, 2, 3, 4, 5, 6] Output : 1-6  
 
Approach: The problem can be easily visualised as a variation of run length encoding problem.  
First sort the array.
Then, start a while loop for traversing the array to check the consecutive elements. The ending of the consecutive numbers will be denoted by j-1 and start by i at any particular instance.
Increment i by 1 if it do not falls in while loop otherwise increment it by j+1 so that it jumps to the next ith element which is out of current range.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ program to compress the array ranges#include <bits/stdc++.h>using namespace std; // Function to compress the array rangesvoid compressArr(int arr[], int n){    int i = 0, j = 0;    sort(arr, arr + n);    while (i < n) {         // start iteration from the        // ith array element        j = i;         // loop until arr[i+1] == arr[i]        // and increment j        while ((j + 1 < n) &&                 (arr[j + 1] == arr[j] + 1)) {            j++;        }         // if the program do not enter into        // the above while loop this means that        // (i+1)th element is not consecutive        // to i th element        if (i == j) {            cout << arr[i] << " ";             // increment i for next iteration            i++;        }        else {            // print the consecutive range found            cout << arr[i] << "-" << arr[j] << " ";             // move i jump directly to j+1            i = j + 1;        }    }} // Driver codeint main(){     int n = 7;    int arr[n] = { 1, 3, 4, 5, 6, 9, 10 };     compressArr(arr, n);}
// Java program to compress the array rangesimport java.util.Arrays; class GFG{ // Function to compress the array rangesstatic void compressArr(int arr[], int n){    int i = 0, j = 0;    Arrays.sort(arr);    while (i < n)    {         // start iteration from the        // ith array element        j = i;         // loop until arr[i+1] == arr[i]        // and increment j        while ((j + 1 < n) &&                (arr[j + 1] == arr[j] + 1))        {            j++;        }         // if the program do not enter into        // the above while loop this means that        // (i+1)th element is not consecutive        // to i th element        if (i == j)        {            System.out.print( arr[i] + " ");             // increment i for next iteration            i++;        }        else        {            // print the consecutive range found            System.out.print( arr[i] + "-" + arr[j] + " ");             // move i jump directly to j+1            i = j + 1;        }    }}     // Driver code    public static void main (String[] args)    {        int n = 7;        int arr[] = { 1, 3, 4, 5, 6, 9, 10 };         compressArr(arr, n);    }} // This code is contributed by anuj_67..
# Python program to compress the array ranges # Function to compress the array rangesdef compressArr(arr, n):    i = 0;    j = 0;    arr.sort();    while (i < n):         # start iteration from the        # ith array element        j = i;         # loop until arr[i+1] == arr[i]        # and increment j        while ((j + 1 < n) and                (arr[j + 1] == arr[j] + 1)):            j += 1;         # if the program do not enter into        # the above while loop this means that        # (i+1)th element is not consecutive        # to i th element        if (i == j):            print(arr[i], end=" ");             # increment i for next iteration            i+=1;        else:            # print the consecutive range found            print(arr[i], "-", arr[j], end=" ");             # move i jump directly to j+1            i = j + 1; # Driver coden = 7;arr = [ 1, 3, 4, 5, 6, 9, 10 ];compressArr(arr, n); # This code is contributed by PrinciRaj1992
// C# program to compress the array rangesusing System; class GFG{ // Function to compress the array rangesstatic void compressArr(int []arr, int n){    int i = 0, j = 0;    Array.Sort(arr);    while (i < n)    {         // start iteration from the        // ith array element        j = i;         // loop until arr[i+1] == arr[i]        // and increment j        while ((j + 1 < n) &&                (arr[j + 1] == arr[j] + 1))        {            j++;        }         // if the program do not enter into        // the above while loop this means that        // (i+1)th element is not consecutive        // to i th element        if (i == j)        {            Console.Write( arr[i] + " ");             // increment i for next iteration            i++;        }        else        {            // print the consecutive range found            Console.Write( arr[i] + "-" + arr[j] + " ");             // move i jump directly to j+1            i = j + 1;        }    }} // Driver codepublic static void Main (){    int n = 7;    int []arr = { 1, 3, 4, 5, 6, 9, 10 };     compressArr(arr, n);}} // This code is contributed by anuj_67..
<script>    // Javascript program to compress the array ranges         // Function to compress the array ranges    function compressArr(arr, n)    {        let i = 0, j = 0;        arr.sort(function(a, b){return a - b});        while (i < n)        {             // start iteration from the            // ith array element            j = i;             // loop until arr[i+1] == arr[i]            // and increment j            while ((j + 1 < n) && (arr[j + 1] == arr[j] + 1))            {                j++;            }             // if the program do not enter into            // the above while loop this means that            // (i+1)th element is not consecutive            // to i th element            if (i == j)            {                document.write( arr[i] + " ");                 // increment i for next iteration                i++;            }            else            {                // print the consecutive range found                document.write( arr[i] + "-" + arr[j] + " ");                 // move i jump directly to j+1                i = j + 1;            }        }    }         let n = 7;    let arr = [ 1, 3, 4, 5, 6, 9, 10 ];       compressArr(arr, n); </script>
1 3-6 9-10
 
Time complexity: O(nlogn) 
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