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# Modify sequence of first N natural numbers to a given array by replacing pairs with their GCD

• Last Updated : 22 Mar, 2023

Given an integer N and an array arr[], the task is to check if a sequence of first N natural numbers, i.e. {1, 2, 3, .. N} can be made equal to arr[] by choosing any pair (i, j) from the sequence and replacing both i and j by GCD of i and j. If possible, then print “Yes”. Otherwise, print “No”.

Examples:

Input: N = 4, arr[] = {1, 2, 3, 2}
Output: Yes
Explanation: For the pair (2, 4) in the sequence {1, 2, 3, 4}, GCD(2, 4) = 2. Now, the sequence modifies to {1, 2, 3, 2}, which is same as arr[].

Input: N = 3, arr[] = {1, 2, 2}
Output: No

Approach:

We can start by iterating over all pairs of numbers (i, j) such that 1 ≤ i < j ≤ N and finding their GCD using the Euclidean algorithm. Then, if we find a pair (i, j) such that GCD(i, j) is equal to arr[i-1] and arr[j-1], we can replace both i and j with GCD(i, j) and continue with the modified sequence. If we are able to modify the sequence to be equal to arr[], then we can print “Yes”. Otherwise, we can print “No”.

Algorithm:

• Initialize a sequence of first N natural numbers, i.e. {1, 2, 3, .. N}.
• Iterate over all pairs of numbers (i, j) such that 1 ≤ i < j ≤ N.
• Compute the GCD of i and j using the Euclidean algorithm.
• If GCD(i, j) is equal to arr[i-1] and arr[j-1], replace both i and j with GCD(i, j) in the sequence.
• If the modified sequence is equal to arr[], print “Yes” and return.
• If no such pair (i, j) is found, print “No” and return.

Implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Function to compute the GCD using the Euclidean algorithm``int` `gcd(``int` `a, ``int` `b) {``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to check if a sequence of first N natural numbers can be made equal to arr[]``void` `checkSequence(``int` `N, ``int` `arr[]) {``    ``// Initialize the sequence of first N natural numbers``    ``vector<``int``> seq(N);``    ``for` `(``int` `i = 0; i < N; i++)``        ``seq[i] = i+1;` `    ``// Iterate over all pairs of numbers (i, j) such that 1 ≤ i < j ≤ N``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = i+1; j < N; j++) {``            ``// Compute the GCD of i and j``            ``int` `g = gcd(seq[i], seq[j]);``            ``// If GCD(i, j) is equal to arr[i-1] and arr[j-1], replace both i and j with GCD(i, j) in the sequence``            ``if` `(g == arr[i] && g == arr[j]) {``                ``seq[i] = seq[j] = g;``                ``break``;``            ``}``        ``}``        ``// If the modified sequence is equal to arr[], print "Yes" and return``        ``if` `(seq == vector<``int``>(arr, arr+N)) {``            ``cout << ``"Yes\n"``;``            ``return``;``        ``}``    ``}``    ``// If no such pair (i, j) is found, print "No" and return``    ``cout << ``"No\n"``;``}` `// Driver code``int` `main() {``    ``int` `N = 4;``    ``int` `arr[] = {1, 2, 3, 2};``    ``checkSequence(N, arr);` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {``    ``// Function to compute the GCD using the Euclidean``    ``// algorithm``    ``public` `static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == ``0``)``            ``return` `b;``        ``return` `gcd(b % a, a);``    ``}` `    ``// Function to check if a sequence of first N natural``    ``// numbers can be made equal to arr[]``    ``public` `static` `void` `checkSequence(``int` `N, ``int``[] arr)``    ``{``        ``// Initialize the sequence of first N natural``        ``// numbers``        ``List seq = ``new` `ArrayList<>();``        ``for` `(``int` `i = ``0``; i < N; i++)``            ``seq.add(i + ``1``);` `        ``// Iterate over all pairs of numbers (i, j) such``        ``// that 1 ≤ i < j ≤ N``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``for` `(``int` `j = i + ``1``; j < N; j++) {``                ``// Compute the GCD of i and j``                ``int` `g = gcd(seq.get(i), seq.get(j));``                ``// If GCD(i, j) is equal to arr[i-1] and``                ``// arr[j-1], replace both i and j with``                ``// GCD(i, j) in the sequence``                ``if` `(g == arr[i] && g == arr[j]) {``                    ``seq.set(i, g);``                    ``seq.set(j, g);``                    ``break``;``                ``}``            ``}``            ``// If the modified sequence is equal to arr[],``            ``// print "Yes" and return``            ``if` `(seq.equals(Arrays.asList(``                    ``Arrays.stream(arr).boxed().toArray(``                        ``Integer[] ::``new``)))) {``                ``System.out.println(``"Yes"``);``                ``return``;``            ``}``        ``}``        ``// If no such pair (i, j) is found, print "No" and``        ``// return``        ``System.out.println(``"No"``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `N = ``4``;``        ``int``[] arr = { ``1``, ``2``, ``3``, ``2` `};``        ``checkSequence(N, arr);``    ``}``}``// Contributed by sdeadityasharma`

Output

`Yes`

Time Complexity: O(N * sqrt(N) * log(N))

Space Complexity: O(N)

Approach: The idea is based on the fact that the GCD of two numbers lies between 1 and the minimum of the two numbers. By definition of gcd, it’s the greatest number that divides both. Therefore, make the number at an index smaller if and only if there exists some number which is its factor. Hence, it can be concluded that for every ith index in the array, if the follow condition holds true, the array arr[] can be obtained from the sequence of first N natural numbers.

(i + 1) % arr[i] == 0

Follow the steps below to solve the problem:

• Traverse the array arr[] using variable i.
• For every ith index, check if (i + 1) % arr[i] is equal to 0 or not. If found to be false for any array element, print “No”.
• Otherwise, after complete traversal of the array, print “Yes”.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to check if array arr[]``// can be obtained from first N``// natural numbers or not``void` `isSequenceValid(vector<``int``>& B,``                     ``int` `N)``{``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``if` `((i + 1) % B[i] != 0) {``            ``cout << ``"No"``;``            ``return``;``        ``}``    ``}` `    ``cout << ``"Yes"``;``}` `// Driver Code``int` `main()``{``    ``int` `N = 4;``    ``vector<``int``> arr{ 1, 2, 3, 2 };` `    ``// Function Call``    ``isSequenceValid(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{``     ` `// Function to check if array arr[]``// can be obtained from first N``// natural numbers or not``static` `void` `isSequenceValid(``int``[] B,``                            ``int` `N)``{``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``if` `((i + ``1``) % B[i] != ``0``)``        ``{``            ``System.out.print(``"No"``);``            ``return``;``        ``}``    ``}``    ``System.out.print(``"Yes"``);``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``4``;``    ``int``[] arr = { ``1``, ``2``, ``3``, ``2` `};``    ` `    ``// Function Call``    ``isSequenceValid(arr, N);``}``}`` ` `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program for the above approach` `# Function to check if array arr[]``# can be obtained from first N``# natural numbers or not``def` `isSequenceValid(B, N):``    ` `    ``for` `i ``in` `range``(N):``        ``if` `((i ``+` `1``) ``%` `B[i] !``=` `0``):``            ``print``(``"No"``)``            ``return``        ` `    ``print``(``"Yes"``)``    ` `# Driver Code``N ``=` `4``arr ``=` `[ ``1``, ``2``, ``3``, ``2` `]`` ` `# Function Call``isSequenceValid(arr, N)` `# This code is contributed by susmitakundugoaldanga`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``      ` `// Function to check if array arr[]``// can be obtained from first N``// natural numbers or not``static` `void` `isSequenceValid(``int``[] B,``                            ``int` `N)``{``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``if` `((i + 1) % B[i] != 0)``        ``{``            ``Console.WriteLine(``"No"``);``            ``return``;``        ``}``    ``}``    ``Console.WriteLine(``"Yes"``);``}``  ` `// Driver code``public` `static` `void` `Main()``{``    ``int` `N = 4;``    ``int``[] arr = { 1, 2, 3, 2 };``     ` `    ``// Function Call``    ``isSequenceValid(arr, N);``}``}` `// This code is contributed by code_hunt`

## Javascript

 ``

Output:

`Yes`

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

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