Skip to content

1228. Missing Number In Arithmetic Progression 👍

  • Time: $O(\log n)$
  • Space: $O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
class Solution {
 public:
  int missingNumber(vector<int>& arr) {
    const int n = arr.size();
    const int delta = (arr.back() - arr.front()) / n;
    int l = 0;
    int r = n - 1;

    while (l < r) {
      const int m = (l + r) / 2;
      if (arr[m] == arr[0] + m * delta)
        l = m + 1;
      else
        r = m;
    }

    return arr[0] + l * delta;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
class Solution {
  public int missingNumber(int[] arr) {
    final int n = arr.length;
    final int delta = (arr[n - 1] - arr[0]) / n;
    int l = 0;
    int r = n - 1;

    while (l < r) {
      final int m = (l + r) / 2;
      if (arr[m] == arr[0] + m * delta)
        l = m + 1;
      else
        r = m;
    }

    return arr[0] + l * delta;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
class Solution:
  def missingNumber(self, arr: List[int]) -> int:
    n = len(arr)
    delta = (arr[-1] - arr[0]) // n
    l = 0
    r = n - 1

    while l < r:
      m = (l + r) // 2
      if arr[m] == arr[0] + m * delta:
        l = m + 1
      else:
        r = m

    return arr[0] + l * delta