 # Check if Array Contains All Elements Of Some Given Range

Objective: Given an array of unsorted numbers, check if it contains all elements of some given range.

Examples:

```int[] arrA = { 11, 17, 13, 19, 15, 16, 12, 14 };
Range : 12-15
Output: True
```

Approach:

Naive Approach: Sorting . Time Complexity – O(nlogn).

Better Approach: Time Complexity – O(n).

• Find the range = y-x;
• Do the linear scan of array.
• Check if element falls within the range of x and y, (arrA[i]>=x && arrA[i]<=y)
• If Yes then calculate z = arrA[i]-x;
• Make the arrA[z] element as negative.
• Once the linear scan is done, just check all the elements in arrA[] from 0 to range are negative, if yes them array contains all the numbers of the given range, return true, else false.
• See the Picture below for better explanation

Complete Code:

Output:

```True
```

### 1 thought on “Check if Array Contains All Elements Of Some Given Range”

1. If we can make the assumption that all integers are positive, this below code I believe is slightly simpler than your O(N) solution.

O(range_start-range_end) space
O(N) runtime

`def check_in_range(items, start, end): # define an array 4 slots wide visited = [False]*(end-start+1)`

` for item in items:`

` # check for only items between our range if item >= start and item <= end: # if we find an item IE; 12 we need to mark it # as visited in our array, so 12-12 = 0, so it # goes it slot 0, 13 goes in slot 1, 13-12 1 etc... visited[item-start] = True`

` # return True if all were visited return all(visited)`