Check if elements of an array can be arranged in a Circle with consecutive difference as 1
Given an array of 
numbers. The task is to check if it is possible to arrange all the numbers in a circle so that any two neighboring numbers differ exactly by 1. Print “YES” if it is possible to get such arrangement and “NO” otherwise.
Examples:
Input: arr[] = {1, 2, 3, 2}
Output: YES
The circle formed is:
1
2 2
3
Input: arr[] = {3, 5, 8, 4, 7, 6, 4, 7}
Output: NO
Below is the step by step algorithm to solve this problem:
- First insert all the elements in a multiset.
- Remove the first element of the set and store it in a curr variable.
- Traverse until the size of multiset reduced to 0.
- Remove elements that are 1 greater or 1 smaller than the curr value.
- If there is a value with difference more than 1 then “no circle possible”.
- Check if it’s initial and final values of curr variable are same, print “YES” if it is, otherwise print “NO”.
Below is the implementation of above approach:
CPP
// C++ program to check if elements of array// can be arranged in Circle with consecutive// difference as 1#include <bits/stdc++.h>using namespace std;// Function to check if elements of array can// be arranged in Circle with consecutive// difference as 1int circlePossible(int arr[], int n){ multiset<int> s; // Initialize the multiset with array // elements for (int i = 0; i < n; i++) s.insert(arr[i]); // Get a pointer to first element int cur = *s.begin(); // Store the first element in a temp variable int start = cur; // Remove the first element s.erase(s.begin()); // Traverse until multiset is non-empty while (s.size()) { // Elements which are 1 greater than the // current element, remove their first occurrence // and increment curr if (s.find(cur + 1) != s.end()) s.erase(s.find(++cur)); // Elements which are 1 less than the // current element, remove their first occurrence // and decrement curr else if (s.find(cur - 1) != s.end()) s.erase(s.find(--cur)); // If the set is non-empty and contains element // which differs by curr from more than 1 // then circle is not possible return else { cout << "NO"; return 0; } } // Finally, check if curr and first differs by 1 if (abs(cur - start) == 1) cout << "YES"; else cout << "NO"; return 0;}// Driver codeint main(){ int arr[] = { 1, 1, 2, 2, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); circlePossible(arr, n); return 0;} |
Output:
YES
.png)