Given an array of binary integers, suppose these values are kept on the circumference of a circle at an equal distance. We need to tell whether is it possible to draw a regular polygon using only 1s as its vertices and if it is possible then print the maximum number of sides that regular polygon has.
Example:
Input : arr[] = [1, 1, 1, 0, 1, 0] Output : Polygon possible with side length 3 We can draw a regular triangle having 1s as its vertices as shown in below diagram (a). Input : arr[] = [1, 0, 1, 0, 1, 0, 1, 0, 1, 1] Output : Polygon possible with side length 5 We can draw a regular pentagon having 1s as its vertices as shown in below diagram (b).
We can solve this problem by getting a relation between the number of vertices a possible polygon can have and a total number of values in the array. Let a possible regular polygon in the circle has K vertices or K sides then it should satisfy two things to be the answer,
If the given array size is N, then K should divide N otherwise K vertices can’t divide N vertices in an equal manner to be a regular polygon.
Next thing is there should be one at every vertex of the chosen polygon.
After the above points, we can see that for solving this problem we need to iterate over divisors of N and then check whether every value of the array at chosen divisor’s distance is 1 or not. If it is 1 then we found our solution. We can iterate over all divisors in O(sqrt(N)) time just by iterating over from 1 to sqrt(N). you can read more about that here.
Implementation:
C++
// C++ program to find whether a regular polygon // is possible in circle with 1s as vertices #include <bits/stdc++.h> using namespace std; // method returns true if polygon is possible with // 'midpoints' number of midpoints bool checkPolygonWithMidpoints(int arr[], int N, int midpoints) { // loop for getting first vertex of polygon for (int j = 0; j < midpoints; j++) { int val = 1; // loop over array values at 'midpoints' distance for (int k = j; k < N; k += midpoints) { // and(&) all those values, if even one of // them is 0, val will be 0 val &= arr[k]; } /* if val is still 1 and (N/midpoints) or (number of vertices) are more than two (for a polygon minimum) print result and return true */ if (val && N/midpoints > 2) { cout << "Polygon possible with side length " <<(N/midpoints) << endl; return true; } } return false; } // method prints sides in the polygon or print not // possible in case of no possible polygon void isPolygonPossible(int arr[], int N) { // limit for iterating over divisors int limit = sqrt(N); for (int i = 1; i <= limit; i++) { // If i divides N then i and (N/i) will // be divisors if (N % i == 0) { // check polygon for both divisors if (checkPolygonWithMidpoints(arr, N, i) || checkPolygonWithMidpoints(arr, N, (N/i))) return; } } cout << "Not possiblen"; } // Driver code to test above methods int main() { int arr[] = {1, 0, 1, 0, 1, 0, 1, 0, 1, 1}; int N = sizeof(arr) / sizeof(arr[0]); isPolygonPossible(arr, N); return 0; } |
Java
// Java program to find whether a regular polygon // is possible in circle with 1s as vertices class Test { // method returns true if polygon is possible with // 'midpoints' number of midpoints static boolean checkPolygonWithMidpoints(int arr[], int N, int midpoints) { // loop for getting first vertex of polygon for (int j = 0; j < midpoints; j++) { int val = 1; // loop over array values at 'midpoints' distance for (int k = j; k < N; k += midpoints) { // and(&) all those values, if even one of // them is 0, val will be 0 val &= arr[k]; } /* if val is still 1 and (N/midpoints) or (number of vertices) are more than two (for a polygon minimum) print result and return true */ if (val != 0 && N/midpoints > 2) { System.out.println("Polygon possible with side length " + N/midpoints); return true; } } return false; } // method prints sides in the polygon or print not // possible in case of no possible polygon static void isPolygonPossible(int arr[], int N) { // limit for iterating over divisors int limit = (int)Math.sqrt(N); for (int i = 1; i <= limit; i++) { // If i divides N then i and (N/i) will // be divisors if (N % i == 0) { // check polygon for both divisors if (checkPolygonWithMidpoints(arr, N, i) || checkPolygonWithMidpoints(arr, N, (N/i))) return; } } System.out.println("Not possible"); } // Driver method public static void main(String args[]) { int arr[] = {1, 0, 1, 0, 1, 0, 1, 0, 1, 1}; isPolygonPossible(arr, arr.length); } } |
Python3
# Python3 program to find whether a # regular polygon is possible in circle # with 1s as vertices from math import sqrt # method returns true if polygon is # possible with 'midpoints' number # of midpoints def checkPolygonWithMidpoints(arr, N, midpoints) : # loop for getting first vertex of polygon for j in range(midpoints) : val = 1 # loop over array values at # 'midpoints' distance for k in range(j , N, midpoints) : # and(&) all those values, if even # one of them is 0, val will be 0 val &= arr[k] # if val is still 1 and (N/midpoints) or (number # of vertices) are more than two (for a polygon # minimum) print result and return true if (val and N // midpoints > 2) : print("Polygon possible with side length" , (N // midpoints)) return True return False # method prints sides in the polygon or print # not possible in case of no possible polygon def isPolygonPossible(arr, N) : # limit for iterating over divisors limit = sqrt(N) for i in range(1, int(limit) + 1) : # If i divides N then i and (N/i) # will be divisors if (N % i == 0) : # check polygon for both divisors if (checkPolygonWithMidpoints(arr, N, i) or checkPolygonWithMidpoints(arr, N, (N // i))): return print("Not possiblen") # Driver code if __name__ == "__main__" : arr = [1, 0, 1, 0, 1, 0, 1, 0, 1, 1] N = len(arr) isPolygonPossible(arr, N) # This code is contributed by Ryuga |
C#
// C# program to find whether // a regular polygon is possible // in circle with 1s as vertices using System; class GFG { // method returns true if // polygon is possible // with 'midpoints' number // of midpoints static bool checkPolygonWithMidpoints(int []arr, int N, int midpoints) { // loop for getting first // vertex of polygon for (int j = 0; j < midpoints; j++) { int val = 1; // loop over array values // at 'midpoints' distance for (int k = j; k < N; k += midpoints) { // and(&) all those values, // if even one of them is 0, // val will be 0 val &= arr[k]; } /* if val is still 1 and (N/midpoints) or (number of vertices) are more than two (for a polygon minimum) print result and return true */ if (val != 0 && N / midpoints > 2) { Console.WriteLine("Polygon possible with " + "side length " + N / midpoints); return true; } } return false; } // method prints sides in the // polygon or print not possible // in case of no possible polygon static void isPolygonPossible(int []arr, int N) { // limit for iterating // over divisors int limit = (int)Math.Sqrt(N); for (int i = 1; i <= limit; i++) { // If i divides N then i // and (N/i) will be divisors if (N % i == 0) { // check polygon for // both divisors if (checkPolygonWithMidpoints(arr, N, i) || checkPolygonWithMidpoints(arr, N, (N / i))) return; } } Console.WriteLine("Not possible"); } // Driver Code static public void Main () { int []arr = {1, 0, 1, 0, 1, 0, 1, 0, 1, 1}; isPolygonPossible(arr, arr.Length); } } // This code is contributed by jit_t |
PHP
<?php // PHP program to find whether // a regular polygon is possible // in circle with 1s as vertices // method returns true if polygon // is possible with 'midpoints' // number of midpoints function checkPolygonWithMidpoints($arr, $N, $midpoints) { // loop for getting first // vertex of polygon for ($j = 0; $j < $midpoints; $j++) { $val = 1; // loop over array values // at 'midpoints' distance for ($k = $j; $k < $N; $k += $midpoints) { // and(&) all those values, // if even one of them is 0, // val will be 0 $val &= $arr[$k]; } /* if val is still 1 and (N/midpoints) or (number of vertices) are more than two (for a polygon minimum) print result and return true */ if ($val && $N / $midpoints > 2) { echo "Polygon possible with side length " , ($N / $midpoints) , "\n"; return true; } } return false; } // method prints sides in // the polygon or print not // possible in case of no // possible polygon function isPolygonPossible($arr, $N) { // limit for iterating // over divisors $limit = sqrt($N); for ($i = 1; $i <= $limit; $i++) { // If i divides N then // i and (N/i) will be // divisors if ($N % $i == 0) { // check polygon for // both divisors if (checkPolygonWithMidpoints($arr, $N, $i) || checkPolygonWithMidpoints($arr, $N, ($N / $i))) return; } } echo "Not possiblen"; } // Driver Code $arr = array(1, 0, 1, 0, 1, 0, 1, 0, 1, 1); $N = sizeof($arr); isPolygonPossible($arr, $N); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript program to // find whether a regular polygon // is possible in circle with 1s as vertices // method returns true if polygon is possible with // 'midpoints' number of midpoints function checkPolygonWithMidpoints(arr, N, midpoints) { // loop for getting first vertex of polygon for (let j = 0; j < midpoints; j++) { let val = 1; // loop over array values at // 'midpoints' distance for (let k = j; k < N; k += midpoints) { // and(&) all those values, // if even one of // them is 0, val will be 0 val &= arr[k]; } /* if val is still 1 and (N/midpoints) or (number of vertices) are more than two (for a polygon minimum) print result and return true */ if (val && parseInt(N/midpoints) > 2) { document.write( "Polygon possible with side length " + parseInt(N/midpoints) + "<br>" ); return true; } } return false; } // method prints sides in the polygon or print not // possible in case of no possible polygon function isPolygonPossible(arr, N) { // limit for iterating over divisors let limit = Math.sqrt(N); for (let i = 1; i <= limit; i++) { // If i divides N then i and (N/i) will // be divisors if (N % i == 0) { // check polygon for both divisors if (checkPolygonWithMidpoints(arr, N, i) || checkPolygonWithMidpoints(arr, N, parseInt(N/i))) return; } } document.write("Not possible"); } // Driver code to test above methods let arr = [1, 0, 1, 0, 1, 0, 1, 0, 1, 1]; let N = arr.length; isPolygonPossible(arr, N); </script> |
Output
Polygon possible with side length 5
Time Complexity: O(sqrt(N)2)
Auxiliary Space: O(1)
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