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GoogleInterviewCodingChallenge3.java
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GoogleInterviewCodingChallenge3.java
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public class GoogleInterviewCodingChallenge3 {
public static void main(String[] args) {
/*
Given an array nums containing n distinct numbers in the range [0,n], retrun the only number in the range
thas is missing from the array.
follow up : Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity ?
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
*/
int[] nums = {3,0,1};
int[] nums1 = {0,1};
int[] nums2 = {9,6,4,2,3,5,7,0,1};
System.out.println(missingNumber(nums));
System.out.println(missingNumber(nums1));
System.out.println(missingNumber(nums2));
}
static int missingNumber(int[] nums){
int res = 0;
for(int i = 0; i<nums.length;i++){
res ^= nums[i];
}
for(int i = 1; i<= nums.length;i++){
res ^= i;
}
return res;
}
}