Given a circular array containing only 0’s and 1’s, of size n where n = p*q (p and q are both odd integers). The task is to check if there is a way such that 1 will be in majority after applying the following operations:

- Divide circular array into p subarrays each of size q.
- In each subarray, the number which is in majority will get stored into array B.
- Now, 1 will said to be in majority if it is in majority into array B.

**Note:** A number is in majority in an array if it occurs more than half times of the size of an array.

**Examples:**

Input:p = 3, q = 3, array[] = {0, 0, 1, 1, 0, 1, 1, 0, 0}

Output:Yes

Assume index of the array from 1 to N, Since the array is circular so index N and 1 will be adjacent.

Divide this circular array into subarray in this way :-

{2, 3, 4}, {5, 6, 7} and {8, 9, 1}. [These are the index of elements]

In {2, 3, 4}, 1 is in majority,

in {5, 6, 7}, again 1 is in majority and

In {8, 9, 1}, 0 is in majority.

Now insert 1, 1, 0 into array B so array B = {1, 1, 0}

In array B, 1 is the majority element so print Yes.

Input:p = 3, q = 3, array[] = {1, 0, 0, 1, 1, 0, 1, 0, 0}

Output:No

No matter how you divide this circular subarray,

1 will not be in majority. Hence, the answer is No.

**Approach:**

- First of all, iterate over the circular array and count the total number of 1’s in each of p subarray(of size q).
- Store this number into another array (of size p).
- If in this case, 1 is in majority then, print yes.
- Else take another set by moving the previous set’s index 1 unit either increasing or decreasing it, and keep track of only those indices which are new in given set and update number of 1’s in the array.
- Repeat the 2nd and 3rd step.

We will repeat q times if there is no majority of 1 found until that, the answer would be NO, else ( like in the previous example case ) the answer would be 1.

Since at the maximum, we can repeat this process q times and each time we are tracking only two elements in each of p subarray.

**Explanation:**

In given example-1, we will divide circular subarray as [indices] => {1, 2, 3}, {4, 5, 6}, {7, 8, 9} and store number of 1’s in another array which are [1, 2, 1] in subarrays respectively.

Taking another set in example 1 by increasing 1 unit so set will be {2, 3, 4}, {5, 6, 7}, {8, 9, 1} now in set 1 only change is inclusion of element 4 and removal of element 1, we will only track these so updated number of 1’s will be 2, 2, 0.

Below is the implementation of above approach:

## C++

// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // Function to check if 1 is the majority // element or not void majority(bool a[], int p, int q, int size) { // assuming starting and ending index of 1st subarray int start = 0, ends = q; // to store majority of p int arr[p]; // subarrays each of size q ; int k = 0; // Loop to calculate total number // of 1's in subarray which will get // stored in array arr[] while (k < p) { int one = 0; for (int j = start; j < ends; j++) { if (a[j] == 1) { one++; } } // starting index of next subarray start = ends; // ending index of next subarray ends = ends + q; // storing 1's arr[k] = one; k++; } start = 0; ends = q; // variable to keep a check // if 1 is in majority or not bool found = 0; // In this case, we are repeating // the task of calculating // total number of 1's backward while (ends > 0) { // to store the total number of 1's int dist_one = 0; // Check if 1 is in majority in // this subarray for (int i = 0; i < p; i++) if (arr[i] > q / 2) dist_one++; // If 1 is in majority return if (dist_one > p / 2) { found = 1; cout << "Yes" << endl; return; } // shifting starting index of // subarray by 1 unit leftwards start--; // shifting ending index of // subarray by 1 unit leftwards ends--; // to ensure it is a valid index // ( array is circular) -1 index means // last index of a circular array if (start < 0) start = size + start; int st = start, en = ends, l = 0; // now to track changes occur // due to shifting of the subarray while (en < size) { if (a[st % size] != a[en % size]) { // st refers to starting index of // new subarray and en refers to // last element of of same subarray // but in previous iteration if (a[st % size] == 1) arr[l]++; else arr[l]--; } l++; // now repeating the same // for other subarrays too st = (st + q); en = en + q; } } if (found == 0) { cout << "No" << endl; } } // Driver code int main() { int p = 3, q = 3; int n = p * q; bool a[] = { 0, 0, 1, 1, 0, 1, 1, 0, 0 }; // circular array of given size majority(a, p, q, n); return 0; }

## Java

// Java implementation of above approach import java.util.*; import java.lang.*; import java.io.*; class GFG { // Function to check if 1 is the majority // element or not static void majority(int a[], int p, int q, int size) { // assuming starting and ending index of 1st subarray int start = 0, ends = q; // to store majority of p int []arr = new int[p]; // subarrays each of size q ; int k = 0; // Loop to calculate total number // of 1's in subarray which will get // stored in array arr[] while (k < p) { int one = 0; for (int j = start; j < ends; j++) { if (a[j] == 1) { one++; } } // starting index of next subarray start = ends; // ending index of next subarray ends = ends + q; // storing 1's arr[k] = one; k++; } start = 0; ends = q; // variable to keep a check // if 1 is in majority or not boolean found = false; // In this case, we are repeating // the task of calculating // total number of 1's backward while (ends > 0) { // to store the total number of 1's int dist_one = 0; // Check if 1 is in majority in // this subarray for (int i = 0; i < p; i++) if (arr[i] > q / 2) dist_one++; // If 1 is in majority return if (dist_one > p / 2) { found = true; System.out.println( "Yes" ); return; } // shifting starting index of // subarray by 1 unit leftwards start--; // shifting ending index of // subarray by 1 unit leftwards ends--; // to ensure it is a valid index // ( array is circular) -1 index means // last index of a circular array if (start < 0) start = size + start; int st = start, en = ends,l = 0; // now to track changes occur // due to shifting of the subarray while (en < size) { if (a[st % size] != a[en % size]) { // st refers to starting index of // new subarray and en refers to // last element of of same subarray // but in previous iteration if (a[st % size] == 1) arr[l]++; else arr[l]--; } l++; // now repeating the same // for other subarrays too st = (st + q); en = en + q; } } if (found == false ) { System.out.println( "No" ); } } // Driver code public static void main(String args[]) { int p = 3, q = 3; int n = p * q; int a[] = { 0, 0, 1, 1, 0, 1, 1, 0, 0 }; // circular array of given size majority(a, p, q, n); } } // This code is Contributed by tufan_gupta2000

**Output:**

Yes

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