## Problem:

Given an array with values between 1 to n , find the missing number between 1 to n that is not present in the array

## Input:

[9,6,4,2,3,5,7,0,1]

## Output:

8

In the above array of length 9 we have all the numbers from 1 to 9 except 8 , so output is 8

Try out the solution here:

https://leetcode.com/problems/missing-number

## Solution:

### Hint:

The sum of all numbers upto a given number n can be calculated using the formula:

(n * (n+1)/2)

## Algorithm:

STEP1: Find the sum of all numbers upto the length of the given array using the formula (n*(n+1)/2) where n is the length of the array

STEP2: Find the sum of all numbers of the given array

STEP3: Find the difference between the sum calculated in step1 and that calculated in step2 , this gives the missing number

## Code:

```
class Solution {
public int missingNumber(int[] nums) {
int sumGivenNos = 0;
for(int i=0;i<nums.length;i++){
sumGivenNos += nums[i];
}
return ((nums.length * (nums.length +1))/2) - sumGivenNos;
}
}
```

Time complexity of above code is O(n) and space complexity is O(1)

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