Hello coders, Welcome back to codingbroz! Today you are going to learn an interesting array interview problem i.e. Find the missing number between 0 to n range.
Problem Statement :
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.
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
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.
Example 4:
Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.
Now, lets see the solution of Find the missing number between 0 to N
Program in Java to find the missing number between 0 to N :
class Solution {
public int missingNumber(int[] nums) {
int sum = 0;
int n = nums.length;
for (int num : nums) {
sum += num;
}
int arrSum = (((n + 1)) * n) / 2;
if (arrSum == sum) return 0;
else return arrSum - sum;
}
}
This is How you can Find missing number between 0 to N range. To learn and solve more interview problems read more problems on www.codingbroz.com, Thanks for reading !!!