Given a permutation A of N integers, the task is to find another permutation B such that for every index i, A[i] + B[i] is a composite number. If no permutation is possible, print -1.
Examples:
Input: A[] = {2, 5, 1, 3, 4}
Output: 4 1 5 3 2
Explanation: A = {2 5 1 3 4} and B = {4 1 5 3 2}
- For i = 0, A[0] + B[0] = 2 + 4 = 6 which is not a prime number.
- For i = 1, A[1] + B[1] = 5 + 1 = 6 which is not a prime number.
- For i = 2, A[2] + B[2] = 1 + 5 = 6 which is not a prime number.
- For i = 3, A[3] + B[3] = 3 + 3 = 6 which is not a prime number.
- For i = 4, A[4] + B[4] = 4 + 2 = 6 which is not a prime number.
Input: A[] = {2, 1}
Output: -1
Explanation: There is no such permutation exist so print -1.
Approach: To solve the problem, follow the below idea:
We know that on adding two odd numbers or two even numbers, we always get an even number. So, for every index i, make B[i] = A[i] so that we will have even sum for all indices. Among all the even sums, no sum will be prime except 2 (A[i] = 1 and B[i] = 1). To handle this, we simply swap B[i] = 1 with the largest odd number in B[]. Also, if the size of permutation is less than 3, then it is impossible to construct a valid permutation B.
Step-by-step algorithm:
- Check if N < 3, return -1.
- Declare an array B of size N to store the permutation.
- Iterate over all the elements in A and initialize B[i] = A[i].
- Store index of 1 and largest odd number of B, as idx1 and idx2.
- Swap B[idx1] and B[idx2] and return B as the final answer.
Below is the implementation of the algorithm:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the valid permutation B
vector<int> solve(vector<int>& A, int N)
{
// vector to store array B
vector<int> B;
// If the permutation has less than three elements, then
// it is impossible to construct a valid permutation
if (N < 3)
return B;
// Variables to store the index of 1 and largest odd
// number
int idx1, idx2;
for (int i = 0; i < N; i++) {
B.push_back(A[i]);
// If B[i] = 1, then store its index in idx1
if (B[i] == 1) {
idx1 = i;
}
// If B[i] = largest odd number, store its index in
// idx2
else if ((N % 2 == 0 && B[i] == N - 1)
|| (N % 2 == 1 && B[i] == N)) {
idx2 = i;
}
}
// swap 1 and the largest odd number
swap(B[idx1], B[idx2]);
return B;
}
// Driver Code
int main()
{
// Sample Input
vector<int> A = { 2, 5, 1, 3, 4 };
int N = A.size();
vector<int> B = solve(A, N);
// If size of B is 0, then print -1
if (B.size() == 0)
cout << -1 << "\n";
else {
for (int i = 0; i < N; i++)
cout << B[i] << " ";
cout << "\n";
}
return 0;
}
Java
import java.util.ArrayList;
public class ValidPermutation {
// Function to find the valid permutation B
static ArrayList<Integer> solve(ArrayList<Integer> A, int N) {
// ArrayList to store array B
ArrayList<Integer> B = new ArrayList<>();
// If the permutation has less than three elements, then
// it is impossible to construct a valid permutation
if (N < 3)
return B;
// Variables to store the index of 1 and largest odd number
int idx1 = -1, idx2 = -1;
for (int i = 0; i < N; i++) {
B.add(A.get(i));
// If B[i] = 1, then store its index in idx1
if (B.get(i) == 1) {
idx1 = i;
}
// If B[i] = largest odd number, store its index in idx2
else if ((N % 2 == 0 && B.get(i) == N - 1)
|| (N % 2 == 1 && B.get(i) == N)) {
idx2 = i;
}
}
// swap 1 and the largest odd number
if (idx1 != -1 && idx2 != -1) {
int temp = B.get(idx1);
B.set(idx1, B.get(idx2));
B.set(idx2, temp);
}
return B;
}
// Driver Code
public static void main(String[] args) {
// Sample Input
ArrayList<Integer> A = new ArrayList<>();
A.add(2);
A.add(5);
A.add(1);
A.add(3);
A.add(4);
int N = A.size();
ArrayList<Integer> B = solve(A, N);
// If size of B is 0, then print -1
if (B.size() == 0)
System.out.println(-1);
else {
for (int i = 0; i < N; i++)
System.out.print(B.get(i) + " ");
System.out.println();
}
}
}
// This code is contributed by shivamgupta0987654321
Python3
def solve(A, N):
# If the permutation has less than three elements, then
# it is impossible to construct a valid permutation
if N < 3:
return []
B = A.copy()
# Variables to store the index of 1 and largest odd number
idx1, idx2 = None, None
for i in range(N):
# If B[i] = 1, then store its index in idx1
if B[i] == 1:
idx1 = i
# If B[i] = largest odd number, store its index in idx2
elif (N % 2 == 0 and B[i] == N - 1) or (N % 2 == 1 and B[i] == N):
idx2 = i
# swap 1 and the largest odd number
B[idx1], B[idx2] = B[idx2], B[idx1]
return B
# Driver Code
if __name__ == "__main__":
# Sample Input
A = [2, 5, 1, 3, 4]
N = len(A)
B = solve(A, N)
# If size of B is 0, then print -1
if len(B) == 0:
print(-1)
else:
print(' '.join(map(str, B)))
JavaScript
// Function to find the valid permutation B
function solve(A) {
// Vector to store array B
let B = [];
// If the permutation has less than three elements, then
// it is impossible to construct a valid permutation
if (A.length < 3)
return B;
// Variables to store the index of 1 and largest odd
// number
let idx1, idx2;
for (let i = 0; i < A.length; i++) {
B.push(A[i]);
// If B[i] = 1, then store its index in idx1
if (B[i] == 1) {
idx1 = i;
}
// If B[i] = largest odd number, store its index in
// idx2
else if ((A.length % 2 == 0 && B[i] == A.length - 1)
|| (A.length % 2 == 1 && B[i] == A.length)) {
idx2 = i;
}
}
// Swap 1 and the largest odd number
[B[idx1], B[idx2]] = [B[idx2], B[idx1]];
return B;
}
// Driver Code
// Sample Input
let A = [2, 5, 1, 3, 4];
let B = solve(A);
// If size of B is 0, then print -1
if (B.length === 0)
console.log(-1);
else {
console.log("Elements are: " + B.join(" "));
}
Output
2 1 5 3 4
Time Complexity: O(N), where N is the size of permutation.
Auxiliary Space: O(N).
Article Tags :
Recommended Articles