README_EN.md

1228. Missing Number In Arithmetic Progression

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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

Python3

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

class Solution:
    def missingNumber(self, arr: List[int]) -> int:
        n = len(arr)
        d = (arr[-1] - arr[0]) // n
        for i in range(1, n):
            if arr[i] != arr[i - 1] + d:
                return arr[i - 1] + d
        return arr[0]

Java

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(int[] arr) {
        int n = arr.length;
        int d = (arr[n - 1] - arr[0]) / n;
        for (int i = 1; i < n; ++i) {
            if (arr[i] != arr[i - 1] + d) {
                return arr[i - 1] + d;
            }
        }
        return arr[0];
    }
}

C++

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 {
public:
    int missingNumber(vector<int>& arr) {
        int n = arr.size();
        int d = (arr[n - 1] - arr[0]) / n;
        for (int i = 1; i < n; ++i)
            if (arr[i] != arr[i - 1] + d) return arr[i - 1] + d;
        return arr[0];
    }
};

Go

func missingNumber(arr []int) int {
	n := len(arr)
	x := (arr[0] + arr[n-1]) * (n + 1) / 2
	y := 0
	for _, v := range arr {
		y += v
	}
	return x - y
}

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]
}

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