Leetcode：计算 O(n) 时间复杂度的元素

Question

Given an integer array arr, count element x such that x + 1 is also in arr.If there're duplicates in arr, count them separately.

Example 1: Input: arr = [1,2,3] Output: 2 Explanation: 1 and 2 are counted cause 2 and 3 are in arr.

Example 2: Input: arr = [1,1,2] Output: 2 Explanation: 1 counted twice cause 2 is in arr.

Example 3: Input: arr = [1,1,3,3,5,5,7,7] Output: 0 Explanation: No numbers are counted, cause there's no 2, 4, 6, or 8 in arr.

Example 4: Input: arr = [1,1,2,2] Output: 2 Explanation: Two 1s are counted cause 2 is in arr.

MyCode: (Does not work for Example 2 only(for now))

int count = 0;
HashSet set = new HashSet();
for(int i=0;i<arr.length;i++) {
    if(set.contains(arr[i]-1) || set.contains(arr[i]+1)) {
        count++;
        set.add(arr[i]);
    }else {
        set.add(arr[i]);
    }
}
return count;

Answer 1

Why are you counting arr[i]-1 ?. The condition set.contains(arr[i]+1) is enough to count as per the definition.

I guess you are trying to achieve this in a single pass. This does not work because, you have to count the duplicate occurences as well.

Example 2 would work if the input was in a different order. ex: [2,1,1] . your solution is more like counting the arr[i]+1 s than counting such numbers itself.

Here's my code with 2 pass:

class Solution {
    public int countElements(int[] arr) {
        Set<Integer> lookup = new HashSet<>();
        for(Integer i: arr){
            lookup.add(i);
        }
        
        int count =0;
        
        for(Integer i:arr){
            if(lookup.contains(i+1)){
                count++;
            }
        }
        return count;
    }
}

1. Add all elements to a set. This has complexity of O(n) and helps with O(1) lookup.
2. Iterate through the array and see if that element is present in the set. This has complexity of O(n) * O(1) viz O(n)

Although we are iterating the array twice, the time complexity will be O(n).

Answer 2

Swift Solution O(n)

class Solution {
    func countElements(_ arr: [Int]) -> Int {
        var arrSet = Set(arr)
        var result = 0

        for num in arr {
            if arrSet.contains(num+1) {
                result += 1
            }
        }
        return result
    }
}

Answer 3

simple solution

class Solution:
    def countElements(self, arr: List[int]) -> int:
        c=0
        for i in arr:
         if(i+1 in arr):
                c+=1
        if(c>0):
            return c
        else:
            return 0```