Below is the one of the code using Xor
I've presented all the 5 techniques within the video description here
Question
Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.
Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
All the numbers of nums are unique.
Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
Solution
One of the approach is provided below
class Solution {
/**
x ^ 0 => x
x ^ x => 0
*/
public int missingNumber(int[] nums) {
final int n = nums.length;
// sum of n elements - sum of nums
return xorOverRange(n)
^ xorOverArray(nums);
}
int xorOverArray(int[] nums) {
int xor = 0;
for (int i = 0 ; i < nums.length; i++) {
xor ^= nums[i];
}
return xor;
}
int xorOverRange(int n) {
int xor = 0;
for (int i = 0 ; i <= n; i++) {
xor ^= i;
}
return xor;
}
}
For more different approach, checkout the video mentioned above.
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