1228. Missing Number In Arithmetic Progression

Description

In some array arr, the values were in arithmetic progression: the values arr[i + 1] - arr[i] are all equal for every 0 <= i < arr.length - 1.

A value from arr was removed that was not the first or last value in the array.

Given arr, return the removed value.

Example 1:

Input: arr = [5,7,11,13]
Output: 9
Explanation: The previous array was [5,7,9,11,13].


Example 2:

Input: arr = [15,13,12]
Output: 14
Explanation: The previous array was [15,14,13,12].

Constraints:

• 3 <= arr.length <= 1000
• 0 <= arr[i] <= 105
• The given array is guaranteed to be a valid array.

Solutions

Solution 1: Arithmetic Series Sum Formula

The sum formula for an arithmetic series is $\frac{n(a_1 + a_n)}{2}$, where $n$ is the number of terms in the arithmetic series, $a_1$ is the first term of the arithmetic series, and $a_n$ is the last term of the arithmetic series.

Since the array given in the problem is an arithmetic series and is missing a number, the number of terms in the array is $n + 1$, the first term is $a_1$, and the last term is $a_n$, so the sum of the array is $\frac{n + 1}{2}(a_1 + a_n)$.

Therefore, the missing number is $\frac{n + 1}{2}(a_1 + a_n) - \sum_{i = 0}^n a_i$.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Where $n$ is the length of the array.

• class Solution {
public int missingNumber(int[] arr) {
int n = arr.length;
int x = (arr[0] + arr[n - 1]) * (n + 1) / 2;
int y = Arrays.stream(arr).sum();
return x - y;
}
}

• class Solution {
public:
int missingNumber(vector<int>& arr) {
int n = arr.size();
int x = (arr[0] + arr[n - 1]) * (n + 1) / 2;
int y = accumulate(arr.begin(), arr.end(), 0);
return x - y;
}
};

• class Solution:
def missingNumber(self, arr: List[int]) -> int:
return (arr[0] + arr[-1]) * (len(arr) + 1) // 2 - sum(arr)


• func missingNumber(arr []int) int {
n := len(arr)
d := (arr[n-1] - arr[0]) / n
for i := 1; i < n; i++ {
if arr[i] != arr[i-1]+d {
return arr[i-1] + d
}
}
return arr[0]
}