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 ?
CodePudding user response:
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 < nums.length; 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.
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
.
CodePudding user response:
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);
}