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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.


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


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:



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  • sdiz

    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)