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.

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)*.

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.

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.

## C++

// C++ program to find the maximum repeating number #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; }

## Java

// Java program to find the maximum repeating number import java.io.*; class MaxRepeating { // Returns maximum repeating element in arr[0..n-1]. // The array elements are in range from 0 to k-1 static 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 function to check for above function*/ public static void main (String[] args) { int arr[] = {2, 3, 3, 5, 3, 4, 1, 7}; int n = arr.length; int k=8; System.out.println("Maximum repeating element is: " + maxRepeating(arr,n,k)); } } /* This code is contributed by Devesh Agrawal */

## Python3

# Python program to find the maximum repeating number # Returns maximum repeating element in arr[0..n-1]. # The array elements are in range from 0 to k-1 def maxRepeating(arr, n, k): # Iterate though input array, for every element # arr[i], increment arr[arr[i]%k] by k for i in range(0, n): arr[arr[i]%k] += k # Find index of the maximum repeating element max = arr[0] result = 0 for i in range(1, n): if arr[i] > max: max = arr[i] result = i # Uncomment this code to get the original array back #for i in range(0, n): # arr[i] = arr[i]%k # Return index of the maximum element return result # Driver program to test above function arr = [2, 3, 3, 5, 3, 4, 1, 7] n = len(arr) k = 8 print("The maximum repeating number is",maxRepeating(arr, n, k)) # This code is contributed by # Smitha Dinesh Semwal

## C#

//C# program to find the maximum repeating // number using System; class GFG { // Returns maximum repeating element // in arr[0..n-1]. // The array elements are in range // from 0 to k-1 static 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; } } // Return index of the // maximum element return result; } /*Driver function to check for above function*/ public static void Main () { int []arr = {2, 3, 3, 5, 3, 4, 1, 7}; int n = arr.Length; int k=8; Console.Write("Maximum repeating " + "element is: " + maxRepeating(arr,n,k)); } } // This code is contributed by Sam007.

Output:

The maximum repeating number is 3

**Exercise:**

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.