Majority element in a circular array of 0’s and 1’s

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:

  1. Divide circular array into p subarrays each of size q.
  2. In each subarray, the number which is in majority will get stored into array B.
  3. 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:

  1. First of all, iterate over the circular array and count the total number of 1’s in each of p subarray(of size q).
  2. Store this number into another array (of size p).
  3. If in this case, 1 is in majority then, print yes.
  4. 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.
  5. 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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