Check if elements of given array can be rearranged such that (arr[i] + i*K) % N = i for all values of i in range [0, N-1]

• Last Updated : 05 Oct, 2021

Given an array arr[] consisting of N positive integers and an integer K, the task is to check if the array elements can be rearranged such that (arr[i] + i*K) % N = i for all values of i in the range [0, N-1].

Examples:

Input: arr[] = {2, 1, 0}, K = 5
Output: Yes
Explanation: The given array can be rearranged to {0, 2, 1}. Hence the values after updation becomes {(0 + 0*5) % 3, (2 + 1*5) % 3, (1 + 2*5) % 3} => {0%3, 7%3, 11%3} => {0, 1, 2} having all elements equal to their indices in the array.

Input: arr[] = {1, 1, 1, 1, 1}, K = 5
Output: No

Naive Approach: The given problem can be solved by generating all the possible permutations of the given array arr[] and check if there exists any such permutation that satisfies the given criteria.

Time Complexity: O(N*N!)
Auxiliary Space: O(N)

Efficient Approach: The above approach can also be optimized with the help of the Set data structure using Recursion. Below are a few observations to solve the given problem:

• The fact that each array element arr[i] is updated as (arr[i] + i*K) % N. So, the value arr[i] % N and i*K % N can be calculated independently.
• If a multiset A contains all the values of arr[i] % N and multiset B contains all the values of i*K % N for all values of i in the range [0, N-1],  generate all possible combinations of elements in A and B and store (A[i] + B[i]) % N in a set. If the size of the resulting set is N, it is possible to rearrange the array in the required way.

Using the above observations, the given problem can be solved by the following steps:

• Create a multiset A containing all the values of arr[i] % N for all values of i in the range [0, N-1].
• Similarly, create a multiset B contains all the values of i*K % N for all values of i in the range [0, N-1].
• Create a recursive function to iterate over all pairs of integers in A and B, add their sum modulo N in set C and recursively call for the remaining elements.
• If at any point, the size of the set C = N, return true else return false.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include using namespace std;   // Function to check if it is possible // to generate all numbers in range // [0, N-1] using the sum of elements // in the multiset A and B mod N bool isPossible(multiset A,                 multiset B,                 set C, int N) {     // If no more pair of elements     // can be selected     if (A.size() == 0 || B.size() == 0) {           // If the number of elements         // in C = N, then return true         if (C.size() == N) {             return true;         }           // Otherwise return false         else {             return false;         }     }       // Stores the value of final answer     bool ans = false;       // Iterate through all the pairs in     // the given multiset A and B     for (auto x : A) {         for (auto y : B) {               // Stores the set A without x             multiset _A = A;             _A.erase(_A.find(x));               // Stores the set B without y             multiset _B = B;             _B.erase(_B.find(y));               // Stores the set C with (x+y)%N             set _C = C;             _C.insert((x + y) % N);               // Recursive call             ans = (ans                    || isPossible(                           _A, _B, _C, N));         }     }       // Return Answer     return ans; }   // Function to check if it is possible // to rearrange array elements such that // (arr[i] + i*K) % N = i void rearrangeArray(     int arr[], int N, int K) {     // Stores the values of arr[] modulo N     multiset A;     for (int i = 0; i < N; i++) {         A.insert(arr[i] % N);     }       // Stores all the values of     // i*K modulo N     multiset B;     for (int i = 0; i < N; i++) {         B.insert((i * K) % N);     }       set C;       // Print Answer     if (isPossible(A, B, C, N)) {         cout << "YES";     }     else {         cout << "NO";     } }   // Driver Code int main() {     int arr[] = { 1, 2, 0 };     int K = 5;     int N = sizeof(arr) / sizeof(arr);       rearrangeArray(arr, N, K);       return 0; }

Python3

 # Python3 program for the above approach   # Function to check if it is possible # to generate all numbers in range # [0, N-1] using the sum of elements #+ in the multiset A and B mod N def isPossible(A, B, C, N):          # If no more pair of elements     # can be selected     if(len(A) == 0 or len(B) == 0):                  # If the number of elements         # in C = N, then return true         if(len(C) == N):             return True                     # Otherwise return false         else:             return False                 # Stores the value of final answer     ans = False     for x in A:                 # Iterate through all the pairs in         # the given multiset A and B         for y in B:                        # Stores the set A without x             _A = A             _A.remove(x)                           # Stores the set A without y             _B = B             _B.remove(y)                            # Stores the set A without x+y%N             _C = C             _C.add((x+y) % N)                           # Recursive call             ans = (ans or isPossible(_A, _B, _C, N))                   return ans   # Function to check if it is possible # to rearrange array elements such that # (arr[i] + i*K) % N = i def rearrangeArray(arr, N, K):        # Stores the values of arr[] modulo N     A = []     for i in range(N):         A.append(arr[i] % N)     A.sort()            # Stores all the values of     # i*K modulo N     B = []     for i in range(N):         B.append((i*K) % N)     B.sort()     C = set()           # Print Answer     if isPossible(A, B, C, N):         print("YES")     else:         print("NO")   # Driver code arr = [1, 2, 0] K = 5 N = len(arr) rearrangeArray(arr, N, K)   # This code is contributed by parthmanchanda81

Output:

YES

Time Complexity: O(N*2N)
Auxiliary Space: O(1)

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