Given an array of integers print the missing elements that lie in range 0-99. If there are more than one missing, collate them, otherwise just print the number.
Note that the input array may not be sorted and may contain numbers outside the range [0-99], but only this range is to be considered for printing missing elements.
Examples :
Input: {88, 105, 3, 2, 200, 0, 10}
Output: 1
4-9
11-87
89-99
Input: {9, 6, 900, 850, 5, 90, 100, 99}
Output: 0-4
7-8
10-89
91-98
Expected time complexity O(n), where n is the size of the input array.
The idea is to use a boolean array of size 100 to keep track of array elements that lie in range 0 to 99. We first traverse input array and mark such present elements in the boolean array. Once all present elements are marked, the boolean array is used to print missing elements.
Following is the implementation of above idea.
C++
// C++ program for print missing elements#include <bits/stdc++.h>#define LIMIT 100using namespace std;// A O(n) function to print missing elements in an arrayvoid printMissing(int arr[], int n){ // Initialize all number from 0 to 99 as NOT seen bool seen[LIMIT] = {false}; // Mark present elements in range [0-99] as seen for (int i=0; i<n; i++) if (arr[i] < LIMIT) seen[arr[i]] = true; // Print missing element int i = 0; while (i < LIMIT) { // If i is missing if (seen[i] == false) { // Find if there are more missing elements after i int j = i+1; while (j < LIMIT && seen[j] == false) j++; // Print missing single or range (i+1 == j)? cout << i : cout << "\n" << i << "-" << j-1; // Update u i = j; } else i++; }}// Driver programint main(){ int arr[] = {88, 105, 3, 2, 200, 0, 10}; int n = sizeof(arr)/sizeof(arr[0]); printMissing(arr, n); return 0;}// This code is contributed by shivanisinghss2110 |
C
// C program for print missing elements#include<stdio.h>#define LIMIT 100// A O(n) function to print missing elements in an arrayvoid printMissing(int arr[], int n){ // Initialize all number from 0 to 99 as NOT seen bool seen[LIMIT] = {false}; // Mark present elements in range [0-99] as seen for (int i=0; i<n; i++) if (arr[i] < LIMIT) seen[arr[i]] = true; // Print missing element int i = 0; while (i < LIMIT) { // If i is missing if (seen[i] == false) { // Find if there are more missing elements after i int j = i+1; while (j < LIMIT && seen[j] == false) j++; // Print missing single or range (i+1 == j)? printf("%dn", i): printf("%d-%dn", i, j-1); // Update u i = j; } else i++; }}// Driver programint main(){ int arr[] = {88, 105, 3, 2, 200, 0, 10}; int n = sizeof(arr)/sizeof(arr[0]); printMissing(arr, n); return 0;} |
Java
class PrintMissingElement { // A O(n) function to print missing elements in an array void printMissing(int arr[], int n) { int LIMIT = 100; boolean seen[] = new boolean[LIMIT]; // Initialize all number from 0 to 99 as NOT seen for (int i = 0; i < LIMIT; i++) seen[i] = false; // Mark present elements in range [0-99] as seen for (int i = 0; i < n; i++) { if (arr[i] < LIMIT) seen[arr[i]] = true; } // Print missing element int i = 0; while (i < LIMIT) { // If i is missing if (seen[i] == false) { // Find if there are more missing elements after i int j = i + 1; while (j < LIMIT && seen[j] == false) j++; // Print missing single or range int p = j-1; System.out.println(i+1==j ? i : i + "-" + p); // Update u i = j; } else i++; } } // Driver program to test above functions public static void main(String[] args) { PrintMissingElement missing = new PrintMissingElement(); int arr[] = {88, 105, 3, 2, 200, 0, 10}; int n = arr.length; missing.printMissing(arr, n); }} |
Python3
# Python3 program for print missing elements# A O(n) function to print missing elements in an array def printMissing(arr, n) : LIMIT = 100 seen = [False]*LIMIT # Initialize all number from 0 to 99 as NOT seen for i in range(LIMIT) : seen[i] = False # Mark present elements in range [0-99] as seen for i in range(n) : if (arr[i] < LIMIT) : seen[arr[i]] = True # Print missing element i = 0 while (i < LIMIT) : # If i is missing if (seen[i] == False) : # Find if there are more missing elements after i j = i + 1 while (j < LIMIT and seen[j] == False) : j += 1 # Print missing single or range p = j - 1 if(i + 1 == j) : print(i) else : print(i, "-", p) # Update u i = j else : i += 1 # Driver codearr = [88, 105, 3, 2, 200, 0, 10] n = len(arr) printMissing(arr, n)# This code is contributed by divyesh072019. |
C#
using System;class GFG{ // A O(n) function to print missing elements in an array static void printMissing(int[] arr, int n) { int LIMIT = 100; bool[] seen = new bool[LIMIT]; int i; // Initialize all number from 0 to 99 as NOT seen for (i = 0; i < LIMIT; i++) seen[i] = false; // Mark present elements in range [0-99] as seen for (i = 0; i < n; i++) { if (arr[i] < LIMIT) seen[arr[i]] = true; } // Print missing element i = 0; while (i < LIMIT) { // If i is missing if (seen[i] == false) { // Find if there are more missing elements after i int j = i + 1; while (j < LIMIT && seen[j] == false) j++; // Print missing single or range int p = j - 1; if(i + 1 == j) { Console.WriteLine(i); } else { Console.WriteLine(i + "-" + p); } // Update u i = j; } else i++; } } // Driver code static void Main() { int[] arr = {88, 105, 3, 2, 200, 0, 10}; int n = arr.Length; printMissing(arr, n); }}// This code is contributed by divyeshrabadiya07. |
PHP
<?php// PHP program for print// missing elements$LIMIT= 100;// A O(n) function to print // missing elements in an arrayfunction printMissing($arr, $n){ global $LIMIT; // Initialize all number from // 0 to 99 as NOT seen $seen = (false); // Mark present elements in. // range [0-99] as seen for ($i = 0; $i < $n; $i++) if ($arr[$i] < $LIMIT) $seen[$arr[$i]] = true; // Print missing element $i = 0; while ($i < $LIMIT) { // If i is missing if ($seen[$i] == false) { // Find if there are more // missing elements after i $j = $i + 1; while ($j < $LIMIT && $seen[$j] == false) $j++; // Print missing // single or range if (($i + 1 == $j) == true) echo $i, "\n"; else echo $i , "-", $j - 1, "\n"; // Update u $i = $j; } else $i++; }}// Driver Code$arr = array (88, 105, 3, 2, 200, 0, 10);$n = sizeof($arr);printMissing($arr, $n);// This code is contributed by aj_36?> |
Javascript
<script>// Javascript program for print missing elements// A O(n) function to print missing // elements in an arrayfunction printMissing(arr, n) { let LIMIT = 100; let seen = new Array(LIMIT); let i; // Initialize all number from // 0 to 99 as NOT seen for(i = 0; i < LIMIT; i++) seen[i] = false; // Mark present elements in // range [0-99] as seen for(i = 0; i < n; i++) { if (arr[i] < LIMIT) seen[arr[i]] = true; } // Print missing element i = 0; while (i < LIMIT) { // If i is missing if (seen[i] == false) { // Find if there are more missing // elements after i let j = i + 1; while (j < LIMIT && seen[j] == false) j++; // Print missing single or range let p = j - 1; if (i + 1 == j) { document.write(i + "</br>"); } else { document.write(i + "-" + p + "</br>"); } // Update u i = j; } else i++; }}// Driver codelet arr = [88, 105, 3, 2, 200, 0, 10];let n = arr.length;printMissing(arr, n);// This code is contributed by suresh07</script> |
Output
1 4-9 11-87 89-99
Time complexity: O(n)
Auxiliary Space: O(1)
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