Biorąc pod uwagę tablicę N liczby całkowite. Zadanie polega na sprawdzeniu, czy ze wszystkich podanych elementów można utworzyć ciąg arytmetyczny. Jeśli to możliwe, wydrukuj „Tak”, w przeciwnym razie wydrukuj „Nie”.
Przykłady:
Wejście : tablica [] = {0 12 4 8}
Wyjście : Tak
Zmień układ podanej tablicy na {0 4 8 12}, co tworzy postęp arytmetyczny.
Wejście : tablica [] = {12 40 11 20}
Wyjście : NIE
Korzystanie z sortowania — czas O(n log n).
Pomysł polega na posortowaniu podanej tablicy. Po sortowaniu sprawdź, czy różnice pomiędzy kolejnymi elementami są takie same, czy nie. Jeśli wszystkie różnice są takie same, możliwy jest postęp arytmetyczny. Proszę zapoznać się - Program do sprawdzania postępu arytmetycznego do wdrożenia tego podejścia.
Korzystanie z sortowania przez zliczanie - O(n) czasu i O(n) przestrzeni
Możemy zmniejszyć przestrzeń wymaganą w metodzie 3, jeśli podaną tablicę można modyfikować.
- Znajdź najmniejsze i drugie najmniejsze elementy.
- Znajdź d = drugi_najmniejszy - najmniejszy
- Odejmij najmniejszy element od wszystkich elementów.
- Teraz, jeśli dana tablica reprezentuje AP, wszystkie elementy powinny mieć postać i*d, gdzie i zmienia się od 0 do n-1.
- Jeden po drugim podziel wszystkie zredukowane elementy za pomocą d. Jeśli jakikolwiek element nie jest podzielny przez d, zwróć wartość false.
- Teraz, jeśli tablica reprezentuje AP, musi to być permutacja liczb od 0 do n-1. Możemy to łatwo sprawdzić za pomocą sortowania przez zliczanie.
Poniżej implementacja tej metody:
C++// C++ program to check if a given array // can form arithmetic progression #include using namespace std; // Checking if array is permutation // of 0 to n-1 using counting sort bool countingsort(int arr[] int n) { int count[n] = { 0 }; // Counting the frequency for (int i = 0; i < n; i++) { count[arr[i]]++; } // Check if each frequency is 1 only for (int i = 0; i <= n-1; i++) { if (count[i] != 1) return false; } return true; } // Returns true if a permutation of arr[0..n-1] // can form arithmetic progression bool checkIsAP(int arr[] int n) { int smallest = INT_MAX second_smallest = INT_MAX; for (int i = 0; i < n; i++) { // Find the smallest and // update second smallest if (arr[i] < smallest) { second_smallest = smallest; smallest = arr[i]; } // Find second smallest else if (arr[i] != smallest && arr[i] < second_smallest) second_smallest = arr[i]; } // Find the difference between smallest and second // smallest int diff = second_smallest - smallest; for (int i = 0; i < n; i++) { arr[i]=arr[i]-smallest; } for(int i=0;i<n;i++) { if(arr[i]%diff!=0) { return false; } else { arr[i]=arr[i]/diff; } } // If array represents AP it must be a // permutation of numbers from 0 to n-1. // Check this using counting sort. if(countingsort(arrn)) return true; else return false; } // Driven Program int main() { int arr[] = { 20 15 5 0 10 }; int n = sizeof(arr) / sizeof(arr[0]); (checkIsAP(arr n)) ? (cout << 'Yes' << endl) : (cout << 'No' << endl); return 0; // This code is contributed by Pushpesh Raj }
Java // Java program to check if a given array // can form arithmetic progression import java.io.*; class GFG { // Checking if array is permutation // of 0 to n-1 using counting sort static boolean countingsort(int arr[] int n) { int[] count = new int[n]; for(int i = 0; i < n; i++) count[i] = 0; // Counting the frequency for (int i = 0; i < n; i++) { count[arr[i]]++; } // Check if each frequency is 1 only for (int i = 0; i <= n-1; i++) { if (count[i] != 1) return false; } return true; } // Returns true if a permutation of arr[0..n-1] // can form arithmetic progression static boolean checkIsAP(int arr[] int n) { int smallest = Integer.MAX_VALUE second_smallest = Integer.MAX_VALUE ; for (int i = 0; i < n; i++) { // Find the smallest and // update second smallest if (arr[i] < smallest) { second_smallest = smallest; smallest = arr[i]; } // Find second smallest else if (arr[i] != smallest && arr[i] < second_smallest) second_smallest = arr[i]; } // Find the difference between smallest and second // smallest int diff = second_smallest - smallest; for (int i = 0; i < n; i++) { arr[i] = arr[i] - smallest; } for(int i = 0; i < n; i++) { if(arr[i] % diff != 0) { return false; } else { arr[i] = arr[i]/diff; } } // If array represents AP it must be a // permutation of numbers from 0 to n-1. // Check this using counting sort. if(countingsort(arrn)) return true; else return false; } // Driven Program public static void main (String[] args) { int arr[] = { 20 15 5 0 10 }; int n = arr.length; if(checkIsAP(arr n)) System.out.println('Yes'); else System.out.println('No'); } } // This code is contributed by Utkarsh
Python # Python program to check if a given array # can form arithmetic progression import sys # Checking if array is permutation # of 0 to n-1 using counting sort def countingsort( arr n): count = [0]*n; # Counting the frequency for i in range(0 n): count[arr[i]] += 1; # Check if each frequency is 1 only for i in range(0 n - 1): if (count[i] != 1): return False; return True; # Returns true if a permutation of arr[0..n-1] # can form arithmetic progression def checkIsAP( arr n): smallest = sys.maxsize; second_smallest = sys.maxsize; for i in range(0n): # Find the smallest and # update second smallest if (arr[i] < smallest) : second_smallest = smallest; smallest = arr[i]; # Find second smallest elif (arr[i] != smallest and arr[i] < second_smallest): second_smallest = arr[i]; # Find the difference between smallest and second # smallest diff = second_smallest - smallest; for i in range(0n): arr[i]=arr[i]-smallest; for i in range(0n): if(arr[i]%diff!=0): return False; else: arr[i]=(int)(arr[i]/diff); # If array represents AP it must be a # permutation of numbers from 0 to n-1. # Check this using counting sort. if(countingsort(arrn)): return True; else: return False; # Driven Program arr = [ 20 15 5 0 10 ]; n = len(arr); if(checkIsAP(arr n)): print('Yes'); else: print('NO'); # This code is contributed by ratiagrawal.
C# using System; class GFG { // Checking if array is permutation // of 0 to n-1 using counting sort static bool CountingSort(int[] arr int n) { // Counting the frequency int[] count = new int[n]; for (int i = 0; i < n; i++) { count[arr[i]]++; } // Check if each frequency is 1 only for (int i = 0; i <= n - 1; i++) { if (count[i] != 1) { return false; } } return true; }// Returns true if a permutation of arr[0..n-1] // can form arithmetic progression static bool CheckIsAP(int[] arr int n) {// Find the smallest and // update second smallest int smallest = int.MaxValue; int secondSmallest = int.MaxValue; for (int i = 0; i < n; i++) { if (arr[i] < smallest) { secondSmallest = smallest; smallest = arr[i]; } else if (arr[i] != smallest && arr[i] < secondSmallest) { secondSmallest = arr[i]; } } int diff = secondSmallest - smallest; for (int i = 0; i < n; i++) { arr[i] = arr[i] - smallest; } for (int i = 0; i < n; i++) { if (arr[i] % diff != 0) { return false; } else { arr[i] = arr[i] / diff; } } // If array represents AP it must be a // permutation of numbers from 0 to n-1. // Check this using counting sort. if (CountingSort(arr n)) { return true; } else { return false; } } // Driven Program static void Main(string[] args) { int[] arr = new int[] { 20 15 5 0 10 }; int n = arr.Length; Console.WriteLine(CheckIsAP(arr n) ? 'Yes' : 'No'); } }
JavaScript // Javascript program to check if a given array // can form arithmetic progression // Checking if array is permutation // of 0 to n-1 using counting sort function countingsort( arr n) { let count=new Array(n).fill(0); // Counting the frequency for (let i = 0; i < n; i++) { count[arr[i]]++; } // Check if each frequency is 1 only for (let i = 0; i <= n-1; i++) { if (count[i] != 1) return false; } return true; } // Returns true if a permutation of arr[0..n-1] // can form arithmetic progression function checkIsAP( arr n) { let smallest = Number.MAX_SAFE_INTEGER second_smallest = Number.MAX_SAFE_INTEGER; for (let i = 0; i < n; i++) { // Find the smallest and // update second smallest if (arr[i] < smallest) { second_smallest = smallest; smallest = arr[i]; } // Find second smallest else if (arr[i] != smallest && arr[i] < second_smallest) second_smallest = arr[i]; } // Find the difference between smallest and second // smallest let diff = second_smallest - smallest; for (let i = 0; i < n; i++) { arr[i]=arr[i]-smallest; } for(let i=0;i<n;i++) { if(arr[i]%diff!=0) { return false; } else { arr[i]=arr[i]/diff; } } // If array represents AP it must be a // permutation of numbers from 0 to n-1. // Check this using counting sort. if(countingsort(arrn)) return true; else return false; } // Driven Program let arr = [20 15 5 0 10 ]; let n = arr.length; (checkIsAP(arr n)) ? (console.log('Yesn')) : (console.log('Non')); // // This code was contributed by poojaagrawal2.
Wyjście
Yes
Złożoność czasu - O(n)
Przestrzeń pomocnicza - O(n)
Haszowanie z pojedynczym przebiegiem — czas O(n) i spacja O(n).
Podstawową ideą jest znalezienie wspólnej różnicy AP poprzez znalezienie maksymalnego i minimalnego elementu tablicy. Następnie zacznij od wartości maksymalnej i kontynuuj zmniejszanie wartości o wspólną różnicę, sprawdzając, czy ta nowa wartość jest obecna w mapie mieszającej, czy nie. Jeśli w którymkolwiek momencie wartości nie ma w zestawie skrótów, przerwij pętlę. Idealna sytuacja po przerwaniu pętli jest taka, że wszystkie n elementów zostało uwzględnionych i jeśli tak, zwróć wartość true, w przeciwnym razie zwróć wartość false.
C++// C++ program for above approach #include using namespace std; bool checkIsAP(int arr[] int n) { unordered_set<int> st; int maxi = INT_MIN; int mini = INT_MAX; for (int i=0;i<n;i++) { maxi = max(arr[i] maxi); mini = min(arr[i] mini); st.insert(arr[i]); } // FINDING THE COMMON DIFFERENCE int diff = (maxi - mini) / (n - 1); int count = 0; // CHECK TERMS OF AP PRESENT IN THE HASHSET while (st.find(maxi)!=st.end()) { count++; maxi = maxi - diff; } if (count == n) return true; return false; } // Driver Code int main() { int arr[] = { 0 12 4 8 }; int n = 4; cout << boolalpha << checkIsAP(arr n); return 0; } // This code is contributed by Rohit Pradhan
Java /*package whatever //do not write package name here */ import java.io.*; import java.util.*; class GFG { public static void main(String[] args) { int[] arr = { 0 12 4 8 }; int n = arr.length; System.out.println(checkIsAP(arr n)); } static boolean checkIsAP(int arr[] int n) { HashSet<Integer> set = new HashSet<Integer>(); int max = Integer.MIN_VALUE; int min = Integer.MAX_VALUE; for (int i : arr) { max = Math.max(i max); min = Math.min(i min); set.add(i); } // FINDING THE COMMON DIFFERENCE int diff = (max - min) / (n - 1); int count = 0; // CHECK IF TERMS OF AP PRESENT IN THE HASHSET while (set.contains(max)) { count++; max = max - diff; } if (count == arr.length) return true; return false; } }
Python import sys def checkIsAP(arr n): Set = set() Max = -sys.maxsize - 1 Min = sys.maxsize for i in arr: Max = max(i Max) Min = min(i Min) Set.add(i) # FINDING THE COMMON DIFFERENCE diff = (Max - Min) // (n - 1) count = 0 # CHECK IF TERMS OF AP PRESENT IN THE HASHSET while (Max in Set): count += 1 Max = Max - diff if (count == len(arr)): return True return False # driver code arr = [ 0 12 4 8 ] n = len(arr) print(checkIsAP(arr n)) # This code is contributed by shinjanpatra
C# using System; using System.Collections.Generic; public class GFG { // C# program for above approach static bool checkIsAP(int[] arr int n) { HashSet<int> st = new HashSet<int>(); int maxi = int.MinValue; int mini = int.MaxValue; for (int i = 0; i < n; i++) { maxi = Math.Max(arr[i] maxi); mini = Math.Min(arr[i] mini); st.Add(arr[i]); } // FINDING THE COMMON DIFFERENCE int diff = (maxi - mini) / (n - 1); int count = 0; // CHECK IF TERMS OF AP PRESENT IN THE HASHSET while (st.Contains(maxi)) { count++; maxi = maxi - diff; } if (count == n) { return true; } return false; } // Driver Code internal static void Main() { int[] arr = { 0 12 4 8 }; int n = 4; Console.Write(checkIsAP(arr n)); } // This code is contributed by Aarti_Rathi }
JavaScript function checkIsAP(arr n){ set = new Set() let Max = Number.MIN_VALUE let Min = Number.MAX_VALUE for(let i of arr){ Max = Math.max(i Max) Min = Math.min(i Min) set.add(i) } // FINDING THE COMMON DIFFERENCE let diff = Math.floor((Max - Min) / (n - 1)) let count = 0 // CHECK IF TERMS OF AP PRESENT IN THE HASHSET while (set.has(Max)){ count += 1 Max = Max - diff } if (count == arr.length) return true return false } // driver code let arr = [ 0 12 4 8 ] let n = arr.length console.log(checkIsAP(arr n))
Wyjście
trueUtwórz quiz