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Formatted question description: https://leetcode.ca/all/1228.html

# 1228. Missing Number In Arithmetic Progression

Easy

## 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.

Then, a value from arr was removed that was not the first or last value in the array.

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] <= 10^5

## Solution

For the original arithmetic progression (no value removed), all pairs of adjacent numbers in the array have the same difference.

After a value is removed, then one pair of adjacent numbers in the array will have a difference that has a greater absolute value.

If a difference with a greater absolute value is found, then find the two numbers with such the difference with a greater absolute value, and the removed value is the average value of the two numbers.

• class Solution {
public int missingNumber(int[] arr) {
int length = arr.length;
int prevDif = arr - arr;
for (int i = 2; i < length; i++) {
int curDif = arr[i] - arr[i - 1];
if (Math.abs(curDif) < Math.abs(prevDif))
return arr[i - 1] - curDif;
else if (Math.abs(curDif) > Math.abs(prevDif))
return arr[i - 1] + prevDif;
prevDif = curDif;
}
return arr[length - 1] + prevDif;
}
}

• // OJ: https://leetcode.com/problems/missing-number-in-arithmetic-progression/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int missingNumber(vector<int>& A) {
int N = A.size(), d = (A.back() - A) / N;
if (d == 0) return A;
for (int i = 1; i < N; ++i) {
if (A[i] != A[i - 1] + d) return A[i - 1] + d;
}
return -1;
}
};

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