# Print missing elements that lie in range 0 – 99

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.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 ` `#define LIMIT 100 ` `using` `namespace` `std; ` ` `  `// A O(n) function to print missing elements in an array ` `void` `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

## C

 `// C program for print missing elements ` `#include ` `#define LIMIT 100 ` ` `  `// A O(n) function to print missing elements in an array ` `void` `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

## 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); ` `    ``} ` `} `

## PHP

 ` `

Output :

```1
4-9
11-87
89-99```

Time complexity of the above program is O(n).

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Improved By : jit_t, shivanisinghss2110

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