Given an array arr[] consisting of N integers, the task is to check if any permutation of the array elements exists where the sum of every pair of adjacent elements is not divisible by 3. If it is possible, then print “Yes”. Otherwise, print “No”.
Examples:
Input: arr[] = {1, 2, 3, 3}
Output: Yes
Explanation:
Since there exist at least 1 combination {3, 2, 3, 1} where sum of every adjacent pairs is not divisible by 3.Input: arr[] = {3, 6, 1, 9}
Output: No
Naive Approach: The simplest approach is to generate all permutations of the given array and check if there exists an arrangement in which the sum of no two adjacent elements is divisible by 3. If it is found to be true, then print “Yes”. Otherwise, print “No”.
Time Complexity: O(N!)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to observe that the only possible remainders for all the array elements i.e., {0, 1, 2}. To segregate these three numbers in such a way that the sum of two adjacent elements is not divisible by 3. Follow the steps below:
- Count all the numbers into three parts having remainder 0, 1, and 2. Let the count be a, b, and c respectively.
- Now arrange the numbers having remainder as 0 with the numbers having remainder as 1 or 2 such that their sum will not be divisible by 3. Below are the conditions where this condition can be true:
- If a ? 1 and a ? b + c + 1
- If a and b both are equals to 0 and c > 0
- If a and c both are equals to 0 and b > 0
- If there is no way to arrange all the numbers in the above way then there is no permutation such that their sum of adjacent elements is not divisible by 3. Therefore, print “No”.
- If the condition in Step 2 is found to be true, then print “Yes”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to checks if any permutation// of the array exists whose sum of// adjacent pairs is not divisible by 3void factorsOf3(int arr[], int N){ int a = 0, b = 0, c = 0; for (int i = 0; i < N; i++) { // Count remainder 0 if (arr[i] % 3 == 0) a++; // Count remainder 1 else if (arr[i] % 3 == 1) b++; // Count remainder 2 else if (arr[i] % 3 == 2) c++; } // Condition for valid arrangements if (a >= 1 && a <= b + c + 1) cout << "Yes" << endl; else if (a == 0 && b == 0 && c > 0) cout << "Yes" << endl; else if (a == 0 && c == 0 && b > 0) cout << "Yes" << endl; else cout << "No" << endl;}// Driver Codeint main(){ // Given array arr[] int arr[] = { 1, 2, 3, 3 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call factorsOf3(arr, N); return 0;} |
Java
// Java program for // the above approachclass GFG{// Function to checks if any permutation// of the array exists whose sum of// adjacent pairs is not divisible by 3static void factorsOf3(int arr[], int N){ int a = 0, b = 0, c = 0; for (int i = 0; i < N; i++) { // Count remainder 0 if (arr[i] % 3 == 0) a++; // Count remainder 1 else if (arr[i] % 3 == 1) b++; // Count remainder 2 else if (arr[i] % 3 == 2) c++; } // Condition for valid arrangements if (a >= 1 && a <= b + c + 1) System.out.print("Yes" + "\n"); else if (a == 0 && b == 0 && c > 0) System.out.print("Yes" + "\n"); else if (a == 0 && c == 0 && b > 0) System.out.print("Yes" + "\n"); else System.out.print("No" + "\n");}// Driver Codepublic static void main(String[] args){ // Given array arr[] int arr[] = {1, 2, 3, 3}; int N = arr.length; // Function Call factorsOf3(arr, N);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program for the above approach# Function to checks if any permutation# of the array exists whose sum of# adjacent pairs is not divisible by 3def factorsOf3(arr, N): a = 0 b = 0 c = 0 for i in range(N): # Count remainder 0 if (arr[i] % 3 == 0): a += 1 # Count remainder 1 elif (arr[i] % 3 == 1): b += 1 # Count remainder 2 elif (arr[i] % 3 == 2): c += 1 # Condition for valid arrangements if (a >= 1 and a <= b + c + 1): print("Yes") elif (a == 0 and b == 0 and c > 0): print("Yes") elif (a == 0 and c == 0 and b > 0): print("Yes") else: print("No")# Driver Code# Given array arr[]arr = [ 1, 2, 3, 3 ]N = len(arr)# Function callfactorsOf3(arr, N)# This code is contributed by Shivam Singh |
C#
// C# program for // the above approachusing System;class GFG{// Function to checks if any // permutation of the array // exists whose sum of// adjacent pairs is not // divisible by 3static void factorsOf3(int []arr, int N){ int a = 0, b = 0, c = 0; for (int i = 0; i < N; i++) { // Count remainder 0 if (arr[i] % 3 == 0) a++; // Count remainder 1 else if (arr[i] % 3 == 1) b++; // Count remainder 2 else if (arr[i] % 3 == 2) c++; } // Condition for valid arrangements if (a >= 1 && a <= b + c + 1) Console.Write("Yes" + "\n"); else if (a == 0 && b == 0 && c > 0) Console.Write("Yes" + "\n"); else if (a == 0 && c == 0 && b > 0) Console.Write("Yes" + "\n"); else Console.Write("No" + "\n");}// Driver Codepublic static void Main(String[] args){ // Given array []arr int []arr = {1, 2, 3, 3}; int N = arr.Length; // Function Call factorsOf3(arr, N);}}// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript program for the above approach// Function to checks if any permutation// of the array exists whose sum of// adjacent pairs is not divisible by 3function factorsOf3(arr, N){ let a = 0, b = 0, c = 0; for(let i = 0; i < N; i++) { // Count remainder 0 if (arr[i] % 3 == 0) a++; // Count remainder 1 else if (arr[i] % 3 == 1) b++; // Count remainder 2 else if (arr[i] % 3 == 2) c++; } // Condition for valid arrangements if (a >= 1 && a <= b + c + 1) document.write("Yes" + "<br/>"); else if (a == 0 && b == 0 && c > 0) document.write("Yes" + "<br/>"); else if (a == 0 && c == 0 && b > 0) document.write("Yes" + "<br/>"); else document.write("No" + "<br/>");}// Driver Code// Given array arr[]let arr = [ 1, 2, 3, 3 ];let N = arr.length;// Function CallfactorsOf3(arr, N);// This code is contributed by chinmoy1997pal </script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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