Horje
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++

<?php
//PHP  implementation of above approach
 
// Function to check if 1 is the majority
// element or not
 
function  majority($a, $p, $q, $size)
{
    // assuming starting and ending index of 1st subarray
     $start = 0; $ends = $q;
 
    // to store majority of p
    $arr=array(0);
 
    // subarrays each of size q ;
    $k = 0;
 
    // Loop to calculate total number
    // of 1's in subarray which will get
    // stored in array arr[]
    while ($k < $p) {
        $one = 0;
        for ($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
     $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
        $dist_one = 0;
 
        // Check if 1 is in majority in
        // this subarray
        for ($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;
            echo  "Yes" ,"\n";
 
            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;
 
         $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 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) {
        echo  "No" ,"\n";
    }
}
 
// Driver code
    $p = 3; $q = 3;
    $n = $p * $q;
 
     $a = array( 0, 0, 1, 1, 0, 1, 1, 0, 0 );
 
    // circular array of given size
    majority($a, $p, $q, $n);
 
 
 
#This Code is Contributed by ajit
?>

Javascript

<script>
 
    // Javascript implementation
    // of above approach
     
    // Function to check if 1 is the
    // majority element or not
    function majority(a, p, q, size)
    {
        // assuming starting and ending
        // index of 1st subarray
        let start = 0, ends = q;
 
        // to store majority of p
        let arr = new Array(p);
        arr.fill(0);
 
        // subarrays each of size q ;
        let k = 0;
 
        // Loop to calculate total number
        // of 1's in subarray which will
        // get stored in array arr[]
        while (k < p)
        {
            let one = 0;
            for (let 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
        let 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
            let dist_one = 0;
 
            // Check if 1 is in majority in
            // this subarray
            for (let i = 0; i < p; i++)
            {
                if (arr[i] > parseInt(q / 2, 10))
                {
                    dist_one++;
                }
            }
 
            // If 1 is in majority return
            if (dist_one > parseInt(p / 2, 10))
            {
                found = true;
                document.write("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;
            }
 
            let 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 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)
        {
            document.write("No");
        }
    }
     
    let p = 3, q = 3;
    let n = p * q;
   
    let a = [0, 0, 1, 1, 0, 1, 1, 0, 0];
   
    // circular array of given size
    majority(a, p, q, n);
     
</script>

Output

Yes

Complexity Analysis:

  • Time Complexity: O(p*q)
  • Auxiliary Space: O(1)

 


Reffered: https://www.geeksforgeeks.org

Arrays

Related
Count elements in the given range which have maximum number of divisors Count elements in the given range which have maximum number of divisors
Elements that occurred only once in the array Elements that occurred only once in the array
Longest subarray having average greater than or equal to x | Set-2 Longest subarray having average greater than or equal to x | Set-2
Sorting a dynamic 2-dimensional array of Strings Sorting a dynamic 2-dimensional array of Strings
Find the largest area rectangular sub-matrix whose sum is equal to k Find the largest area rectangular sub-matrix whose sum is equal to k

Type:
 Geek
Category:
 Coding
Sub Category:
 Tutorial
Uploaded by:
 Admin
Views:
 10



 

Horje