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 visualized 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++
// 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
// 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.. |
Python3
# 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#
// 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.. |
Javascript
<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> |
Output
1 3-6 9-10
Time complexity: O(n log n)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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