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contain duplicate
How should I optimize this code sample,getting. Tlc import java.util.*; class Solution { public boolean containsDuplicate(int[] nums) { //tlc int s =0; boolean ans=true; int e = nums.length-1; if (nums.length <= 0) { return false; } while(s<e){ if(nums[s]!=nums[e]){ ans =false; s++; e--; } } return ans; } }
9 Answers
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Yes i got what are you saying... thanks đ
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Mustafa A, vikram Rathod here is a way to reduce the number of comparisons to nlog(n). Only compare from i to the end of the list. The items before i have been compared already.
int end = nums.length;
for (int i = 0; i < end-1; i++) {
for (int j = i+1; j < end; j++) {
if (nums[i] == nums[j]) {
return true; // end here
}
}
}
return false; // no duplicates
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First of all. A method that doesn't use object attributes should be static. You don't want to be forced to create an instance to use it.
If you find a duplicate, you don't want to continue. You should return immediately.
Otherwise, I would improve the readability of the code.
Like this:
int end = nums.length;
for (int i = 0; i < end; i++) {
for (int j = 0; j < end; j++)
{
if (i != j) { // ignore same element
if (nums[i] == nums[j]) {
return true; // end here
}
}
}
}
return false; // no duplicates
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I think this could use the n^2 time complexity
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Yeah. But time time complexity is about worst case. This could take anything from 2 (the first two are duplicates) to n^2 (last two are duplicates).
You don't need to force it. If there is one duplicate, then the list contains duplicates.
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But its still showing TLC
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vikram Rathod Try starting j with 1 instead. Otherwise it will compare the first element with itself for no reason.
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vikram Rathod if you need further reduction in complexity, consider using a hashMap. That would reduce it to 2n -- that is, 1n to test for the number being already in the map with the containsKey() method, and subsequently 1n to add it to the map with the put() method.
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Brian Yeah. That's true. Good point.