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
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)