1502.Can Make Arithmetic Progression From Sequence

tags: `Easy`,`Array`,`Sorting`

1502. Can Make Arithmetic Progression From Sequence

題目描述

A sequence of numbers is called an arithmetic progression if the difference between any two consecutive elements is the same.

Given an array of numbers arr, return true if the array can be rearranged to form an arithmetic progression. Otherwise, return false.

範例

Example 1:

Input: arr = [3,5,1]
Output: true
Explanation: We can reorder the elements as [1,3,5] or [5,3,1] with differences 2 and -2 respectively, between each consecutive elements.

Example 2:

Input: arr = [1,2,4]
Output: false
Explanation: There is no way to reorder the elements to obtain an arithmetic progression.

Constraints:

2 <= arr.length <= 1000
-10⁶ <= arr[i] <= 10⁶

解答

C#









public bool CanMakeArithmeticProgression(int[] arr) {
    Array.Sort(arr);
    for(int i = arr.Length - 1; i >= 2; i--){
        if(arr[i] - arr[i - 1] != arr[1] - arr[0]){
            return false;
        }
    }
    return true;
}

JimJun 6, 2023

C++


















class Solution {
public:
    bool canMakeArithmeticProgression(vector<int>& arr) {
        ios_base::sync_with_stdio(0), cin.tie(0);
        int n = arr.size();
        if (n == 2) {
            return true;
        }
        sort(arr.begin(), arr.end());
        int commonDiff = arr[0] - arr[1];
        for (int i = 1; i < n - 1; i ++) {
            if (arr[i] - arr[i + 1] != commonDiff) {
                return false;
            }
        }
        return true;
    }
};

Jerry Wu6 June, 2023

Javascript









function canMakeArithmeticProgression(arr) {
  arr.sort((a, b) => a - b);
  const r = arr[1] - arr[0];
  for (let i = 2; i < arr.length; i++) {
    if (arr[i] - arr[i - 1] !== r) return false;
  }

  return true;
}

Marsgoat6 June, 2023

看解答看到

O (n)

解，前面太直覺就用sort了，沒想到這題還可以這麼秀














function canMakeArithmeticProgression2(arr) {
  const min = Math.min(...arr);
  const max = Math.max(...arr);
  const n = arr.length;
  const r = (max - min) / (n - 1); // 找到公差
  const set = new Set(arr);
  if (r === 0) return true;
  if (r % 1 !== 0) return false;
  for (let i = 0; i < n; i++) {
    if (!set.has(min + i * r)) return false;
  }

  return true;
}

Marsgoat6 June, 2023

Python










class Solution:
    def canMakeArithmeticProgression(self, arr: List[int]) -> bool:
        arr = sorted(arr)
        r = arr[1] - arr[0]

        for i in range(2, len(arr)):
            if arr[i] - arr[i-1] != r:
                return False

        return True


















class Solution:
    def canMakeArithmeticProgression(self, arr: List[int]) -> bool:
        mn, mx = min(arr), max(arr)
        n = len(arr)
        if mx - mn == 0:
            return True
        if (mx - mn) % (n - 1):
            return False

        r = (mx - mn) // (n - 1)  # 算公差

        st = set()
        for num in arr:
            if (num - mn) % r:
                return False
            st.add(num)

        return len(st) == n

仿照上面公差的寫法
TC:

O (n)

SC:

O (n)

Ron ChenTue, June 7, 2023

Java













class Solution {
    public boolean canMakeArithmeticProgression(int[] arr) {
        Arrays.sort(arr);
        int r = arr[1] - arr[0];

        for(int i = 2; i < arr.length; i++) {
            if(arr[i] - arr[i-1] != r)
                return false;
        }

        return true;
    }
}























class Solution {
    public boolean canMakeArithmeticProgression(int[] arr) {
        int min = Arrays.stream(arr).min().getAsInt();
        int max = Arrays.stream(arr).max().getAsInt();
        int n = arr.length;

        if(max - min == 0)
            return true;
        if((max - min) % (n - 1) != 0)
            return false;

        int r = (max - min) / (n - 1);
        Set<Integer> set = new HashSet<>();

        for(int num : arr) {
            if((num - min) % r != 0)
                return false;
            set.add(num);
        }

        return set.size() == n;
    }
}