268. Missing Number(Solution || Leetcode easy || Java)

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.

Constraints:

• n == nums.length
• 1 <= n <= 104
• 0 <= nums[i] <= n
• 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 (1):

class Solution {
public int missingNumber(int[] nums) {
int i = 0;

//first we will sort our array using cyclic sort
while(i < nums.length){
if(nums[i] != i && nums[i] < nums.length){
int temp = nums[i];
nums[i] = nums[nums[i]];
nums[temp] = temp;
}else{
i++;
}
}

//if the element is not found return that element
for(int j = 0; j < nums.length; j++){
if(nums[j] != j){
return j;
}
}

//if nothing is missing return length of array
return nums.length;
}
}

SOLUTION(2):

class Solution {
public int missingNumber(int[] nums) {
//as the range of array’s elements is 0 to n then, we will first find the sum of n natural numbers using the formula n*(n+1)/2. In the last we subtract our array’s elements sum from the expexted sum and that will give us our answer

int sum = 0;
int expectedSum = nums.length*(nums.length + 1)/2;

for(int i = 0; i < nums.length; i++){
sum += nums[i];
}

return expectedSum — sum;
}
}

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