Description#
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.
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
Solutions#
Solution 1: Sorting + Traversal#
We can first sort the array arr, then traverse the array, and check whether the difference between adjacent items is equal.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array arr.
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| class Solution:
def canMakeArithmeticProgression(self, arr: List[int]) -> bool:
arr.sort()
d = arr[1] - arr[0]
return all(b - a == d for a, b in pairwise(arr))
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| class Solution {
public boolean canMakeArithmeticProgression(int[] arr) {
Arrays.sort(arr);
int d = arr[1] - arr[0];
for (int i = 2; i < arr.length; ++i) {
if (arr[i] - arr[i - 1] != d) {
return false;
}
}
return true;
}
}
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| class Solution {
public:
bool canMakeArithmeticProgression(vector<int>& arr) {
sort(arr.begin(), arr.end());
int d = arr[1] - arr[0];
for (int i = 2; i < arr.size(); i++) {
if (arr[i] - arr[i - 1] != d) {
return false;
}
}
return true;
}
};
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| func canMakeArithmeticProgression(arr []int) bool {
sort.Ints(arr)
d := arr[1] - arr[0]
for i := 2; i < len(arr); i++ {
if arr[i]-arr[i-1] != d {
return false
}
}
return true
}
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| function canMakeArithmeticProgression(arr: number[]): boolean {
arr.sort((a, b) => a - b);
const n = arr.length;
for (let i = 2; i < n; i++) {
if (arr[i - 2] - arr[i - 1] !== arr[i - 1] - arr[i]) {
return false;
}
}
return true;
}
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| impl Solution {
pub fn can_make_arithmetic_progression(mut arr: Vec<i32>) -> bool {
arr.sort();
let n = arr.len();
for i in 2..n {
if arr[i - 2] - arr[i - 1] != arr[i - 1] - arr[i] {
return false;
}
}
true
}
}
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| /**
* @param {number[]} arr
* @return {boolean}
*/
var canMakeArithmeticProgression = function (arr) {
arr.sort((a, b) => a - b);
for (let i = 1; i < arr.length - 1; i++) {
if (arr[i] << 1 != arr[i - 1] + arr[i + 1]) {
return false;
}
}
return true;
};
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| int cmp(const void* a, const void* b) {
return *(int*) a - *(int*) b;
}
bool canMakeArithmeticProgression(int* arr, int arrSize) {
qsort(arr, arrSize, sizeof(int), cmp);
for (int i = 2; i < arrSize; i++) {
if (arr[i - 2] - arr[i - 1] != arr[i - 1] - arr[i]) {
return 0;
}
}
return 1;
}
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Solution 2: Hash Table + Mathematics#
We first find the minimum value $a$ and the maximum value $b$ in the array $arr$. If the array $arr$ can be rearranged into an arithmetic sequence, then the common difference $d = \frac{b - a}{n - 1}$ must be an integer.
We can use a hash table to record all elements in the array $arr$, then traverse $i \in [0, n)$, and check whether $a + d \times i$ is in the hash table. If not, it means that the array $arr$ cannot be rearranged into an arithmetic sequence, and we return false. Otherwise, after traversing the array, we return true.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array arr.
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| class Solution:
def canMakeArithmeticProgression(self, arr: List[int]) -> bool:
a = min(arr)
b = max(arr)
n = len(arr)
if (b - a) % (n - 1):
return False
d = (b - a) // (n - 1)
s = set(arr)
return all(a + d * i in s for i in range(n))
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| class Solution {
public boolean canMakeArithmeticProgression(int[] arr) {
int n = arr.length;
int a = arr[0], b = arr[0];
Set<Integer> s = new HashSet<>();
for (int x : arr) {
a = Math.min(a, x);
b = Math.max(b, x);
s.add(x);
}
if ((b - a) % (n - 1) != 0) {
return false;
}
int d = (b - a) / (n - 1);
for (int i = 0; i < n; ++i) {
if (!s.contains(a + d * i)) {
return false;
}
}
return true;
}
}
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| class Solution {
public:
bool canMakeArithmeticProgression(vector<int>& arr) {
auto [a, b] = minmax_element(arr.begin(), arr.end());
int n = arr.size();
if ((*b - *a) % (n - 1) != 0) {
return false;
}
int d = (*b - *a) / (n - 1);
unordered_set<int> s(arr.begin(), arr.end());
for (int i = 0; i < n; ++i) {
if (!s.count(*a + d * i)) {
return false;
}
}
return true;
}
};
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| func canMakeArithmeticProgression(arr []int) bool {
a, b := slices.Min(arr), slices.Max(arr)
n := len(arr)
if (b-a)%(n-1) != 0 {
return false
}
d := (b - a) / (n - 1)
s := map[int]bool{}
for _, x := range arr {
s[x] = true
}
for i := 0; i < n; i++ {
if !s[a+i*d] {
return false
}
}
return true
}
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| function canMakeArithmeticProgression(arr: number[]): boolean {
const n = arr.length;
const map = new Map<number, number>();
let min = Infinity;
let max = -Infinity;
for (const num of arr) {
map.set(num, (map.get(num) ?? 0) + 1);
min = Math.min(min, num);
max = Math.max(max, num);
}
if (max === min) {
return true;
}
if ((max - min) % (arr.length - 1)) {
return false;
}
const diff = (max - min) / (arr.length - 1);
for (let i = min; i <= max; i += diff) {
if (map.get(i) !== 1) {
return false;
}
}
return true;
}
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| use std::collections::HashMap;
impl Solution {
pub fn can_make_arithmetic_progression(arr: Vec<i32>) -> bool {
let n = arr.len() as i32;
let mut min = i32::MAX;
let mut max = i32::MIN;
let mut map = HashMap::new();
for &num in arr.iter() {
*map.entry(num).or_insert(0) += 1;
min = min.min(num);
max = max.max(num);
}
if min == max {
return true;
}
if (max - min) % (n - 1) != 0 {
return false;
}
let diff = (max - min) / (n - 1);
let mut k = min;
while k <= max {
if *map.get(&k).unwrap_or(&0) != 1 {
return false;
}
k += diff;
}
true
}
}
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