Codingbat challenge: sameEnds Stream API Solution

Question

Given the task sameEnds from CodingBat:

Return true if the group of N numbers at the start and end of the array are the same. For example, with {5, 6, 45, 99, 13, 5, 6}, the ends are the same for n=0 and n=2, and false for n=1 and n=3. You may assume that n is in the range 0..nums.length inclusive.
sameEnds([5, 6, 45, 99, 13, 5, 6], 1) → false
sameEnds([5, 6, 45, 99, 13, 5, 6], 2) → true
sameEnds([5, 6, 45, 99, 13, 5, 6], 3) → false

My solution to this problem passes the vast majority of the tests, but not all of them:

public boolean sameEnds(int[] nums, int len) {
  
  if (nums.length >= len * 2) {
    for (int i = 0, j = nums.length - 1 ; i < len && len > 0; i++, j--) {
       if (nums[i] != nums[j]) {
         return false;
       }
    }
  }
  
  return true;
}

My questions are the following:

What can be done in order to fix my solution?
Is it possible to solve this task using Stream API ?

The spec doesn't prohibit len from being greater than nums.length / 2. The start and end can overlap. — John Kugelman
– John Kugelman, Commented Jun 9, 2022 at 20:12

Alexander Ivanchenko · Accepted Answer · 2022-06-10 10:09:00Z

You can use allMatch() operation in order to implement it with streams.

This solution passes all test cases on CodingBat:

public boolean sameEnds(int[] nums, int len) {
    return java.util.stream.IntStream.range(0, len)
        .allMatch(n -> nums[n] == nums[nums.length - (len - n)]);
}

A fix to your imperative solution might look like this:

public boolean sameEnds(int[] nums, int len) {
    for (int i = 0, j = nums.length - 1 - (len - 1); i < len; i++, j++) {
        if (nums[i] != nums[j]) {
            return false;
        }
    }
    return true;
}

I removed the wrapping if condition because when nums.length >= len * 2 is evaluated to false it means that subarrays that need to be compared overlap, but it doesn't automatically mean that these subarrays are equal.

Condition len > 0 is redundant because it is guaranteed to be in the range [0,nums.length].

Variable j that denotes position in the tail subarray has been initialized to nums.length - 1 - (len - 1) - last valid index minus subarray's length. And the so-called increment statement of the for loop was changed from j-- to j++.

Eritrean · Accepted Answer · 2022-06-09 20:44:31Z

1

You can use the built in method Arrays.equals if you are using Java 9 or higher:

public boolean sameEnds(int[] nums, int len) {
    return Arrays.equals(nums, 0, len, nums, nums.length - len, nums.length);
}

edited Jun 9, 2022 at 20:44

answered Jun 9, 2022 at 20:37

Eritrean

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