Saturday, January 18, 2014

Find the maximum repeating number in array where all the elements are in range 0 to k-1, k ∈ [0,N]

Given an array of size n, the array contains numbers in range from 0 to k-1 where k is a positive integer and k <= n. Find the maximum repeating number in this array. For example, let k be 10 the given array be arr[] = {1, 2, 2, 2, 0, 2, 0, 2, 3, 8, 0, 9, 2, 3}, the maximum repeating number would be 2. Expected time complexity is O(n) and extra space allowed is O(1). Modifications to array are allowed.

Solution 1 - Brute force
The naive approach is to run two loops, the outer loop picks an element one by one, the inner loop counts number of occurrences of the picked element. Finally return the element with maximum count. Time complexity of this approach is O(n^2).

Solution2 - Using element as index in auxiliary array for counting the frequency
A better approach is to create a count array of size k and initialize all elements of count[] as 0. Iterate through all elements of input array, and for every element arr[i], increment count[arr[i]]. Finally, iterate through count[] and return the index with maximum value. This approach takes O(n) time, but requires O(k) space.

Solution 3 - Using modulo operator
Following is the O(n) time and O(1) extra space approach. Let us understand the approach with a simple example where arr[] = {2, 3, 3, 5, 3, 4, 1, 7}, k = 8, n = 8 (number of elements in arr[]).
1) Iterate though input array arr[], for every element arr[i], increment arr[arr[i]%k] by k (arr[] becomes {2, 11, 11, 29, 11, 12, 1, 15 })
2) Find the maximum value in the modified array (maximum value is 29). Index of the maximum value is the maximum repeating element (index of 29 is 3).
3) If we want to get the original array back, we can iterate through the array one more time and do arr[i] = arr[i] % k where i varies from 0 to n-1.
How does the above algorithm work? Since we use arr[i]%k as index and add value k at the index arr[i]%k, the index which is equal to maximum repeating element will have the maximum value in the end. Note that k is added maximum number of times at the index equal to maximum repeating element and all array elements are smaller than k.
Following is C++ implementation of the above algorithm.
#include<iostream>
using namespace std;
 
// Returns maximum repeating element in arr[0..n-1].
// The array elements are in range from 0 to k-1
int maxRepeating(int* arr, int n, int k)
{
    // Iterate though input array, for every element
    // arr[i], increment arr[arr[i]%k] by k
    for (int i = 0; i< n; i++)
        arr[arr[i]%k] += k;
 
    // Find index of the maximum repeating element
    int max = arr[0], result = 0;
    for (int i = 1; i < n; i++)
    {
        if (arr[i] > max)
        {
            max = arr[i];
            result = i;
        }
    }
 
    /* Uncomment this code to get the original array back
       for (int i = 0; i< n; i++)
          arr[i] = arr[i]%k; */
 
    // Return index of the maximum element
    return result;
}
 
// Driver program to test above function
int main()
{
    int arr[] = {2, 3, 3, 5, 3, 4, 1, 7};
    int n = sizeof(arr)/sizeof(arr[0]);
    int k = 8;
 
    cout << "The maximum repeating number is " <<
         maxRepeating(arr, n, k) << endl;
 
    return 0;
}

Output : The max repeating number is 3

What else?
The above solution prints only one repeating element and doesn’t work if we want to print all maximum repeating elements. For example, if the input array is {2, 3, 2, 3}, the above solution will print only 3. What if we need to print both of 2 and 3 as both of them occur maximum number of times. Write a O(n) time and O(1) extra space function that prints all maximum repeating elements. (Hint: We can use maximum quotient arr[i]/n instead of maximum value in step 2).
Note that the above solutions may cause overflow if adding k repeatedly makes the value more than INT_MAX.
This article is compiled by Ashish Anand and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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