Maximum Frequency Number
Ninja is given an array of integers that contain numbers in random order. He needs to write a program to find and return the number which occurs the maximum times in the given input. He needs your help to solve this problem.

If two or more elements contend for the maximum frequency, return the element which occurs in the array first i.e. whose index is lowest.

For example,

For 'arr' = [ 1, 2, 3, 1, 2]. you need to return 1.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. The first line of each test case contains a single integer ‘N’ denoting the size of the array. The second line of each test case contains ‘N’ space-separated integers denoting the elements of the array.

Output Format:

For each case, we need to print an integer that has the maximum frequency. The output of each test case will be printed in a separate line.

Note:

You do not need to input or print anything, and it has already been taken care of. Just implement the given function.

Constraints:

1

Question

Maximum Frequency Number

Ninja is given an array of integers that contain numbers in random order. He needs to write a program to find and return the number which occurs the maximum times in the given input. He needs your help to solve this problem.

If two or more elements contend for the maximum frequency, return the element which occurs in the array first i.e. whose index is lowest.

For example,

For 'arr' = [ 1, 2, 3, 1, 2]. you need to return 1.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run.

The first line of each test case contains a single integer ‘N’ denoting the size of the array.

The second line of each test case contains ‘N’ space-separated integers denoting the elements of the array.

Output Format:

For each case, we need to print an integer that has the maximum frequency.

The output of each test case will be printed in a separate line.

Note:

You do not need to input or print anything, and it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 5
1 <= N <= 10000
-10 ^ 3 <= |arr| <= 10 ^ 3

Time Limit: 1 sec

Accepted Answer

Brute Force Approach

Here, we can simply run two loops. The outer loop picks all elements one by one and the inner loop finds the frequency of the picked element and compares it with the maximum present so far.

Algorithm:

Declare 4 variables as ‘maxFrequency’ , ‘currentFrequency’ , ‘maxElement’ , ‘currentElement’ and initialize them with 0
Run a loop from ‘i’ = ‘0’ to ‘N’
- Set ‘currentElement’ as arr[i] and ‘currentFrequency’ to 0
- Run a loop ‘j’ = ‘i’ to ‘N’
  - If ‘arr[j]’ is equal to currentElement
    - Increment currentFrequency by 1
- If our ‘currentFrequency’ is greater than ‘maxFrequency’
  - Update ‘maxFrequency’ and ‘maxElement' with 'currentFrequency' and ‘currentFrequency’ respectively.
Return 'maxElement'

Space Complexity: O(1)Explanation:

O(1)

We use only some variables that require constant space.

Time Complexity: O(n^2)Explanation:

O(N ^ 2) where N is the length of the array.

We iterate through the array to fixed our starting point and for every starting point, we make a loop that will count the frequency of the picked element. Therefore we are doing N + (N - 1) + (N - 2) + ..+ 1 = (N * ( N + 1)) / 2 iterations. So our Time Complexity is O(N ^ 2).

Java (SE 1.8)


/*
	Time Complexity: O(N ^ 2)
	Space Complexity: O(1)
	Where N is the total number of elements in the array.
*/
public class Solution {

	public static int maxFrequencyNumber(int n, int[] arr) {

		// Declare 4 variables as maxFrequency,currentFrequency,
		// maxElement,currentElement and initialize them with 0
		int maxFrequency = 0;
		int maxElement = 0;
		int currentElement = 0;
		int currentFrequency = 0;

		for (int i = 0; i < n; ++i) {

			// Set currentElement as arr[i] and currentFrequency to 0
			currentElement = arr[i];
			currentFrequency = 0;

			// Loop to calculate the frequency of the currentElement
			for (int j = i; j < n; ++j) {
				if (currentElement == arr[j]) {
					currentFrequency++;
				}
			}

			// If our currentFrequency is greater than maxFrequency
			if (currentFrequency > maxFrequency) {

				// Update maxFrequency and maxElement with currentFrequency and currentFrequency
				// respectively.
				maxFrequency = currentFrequency;
				maxElement = currentElement;
			}
		}

		return maxElement;
	}

}

C++ (g++ 5.4)

/*
    Time Complexity: O(N ^ 2)
    Space Complexity: O(1)

    Where N is the total number of elements in the array.
*/


int maximumFrequency(vector &arr, int n)
{
	// Declare 4 variables as ‘maxFrequency’, ‘currentFrequency’, ‘maxElement’, ‘currentElement’ and initialize them with 0
	int maxFrequency = 0;
	int maxElement = 0;
	int currentElement = 0;
	int currentFrequency = 0;


	for (int i = 0; i < n; ++i)
	{
		// Set currentElement as arr[i] and currentFrequency to 0
		currentElement = arr[i];
		currentFrequency = 0;

		// Loop to calculate the frequency of the currentElement
		for (int j = i; j < n; ++j)
		{
			if(currentElement == arr[j])
			{
				currentFrequency++;
			}
		}

		// If our currentFrequency is greater than maxFrequency
		if(currentFrequency > maxFrequency)
		{

			// Update maxFrequency and maxElement with currentFrequency and currentFrequency respectively.
			maxFrequency = currentFrequency;
			maxElement = currentElement;
		}
	}
	
	return maxElement;
}

Python (3.5)

'''

    Time Complexity: O(N ^ 2)

    Space Complexity: O(1)



    Where N is the total number of elements in the array.

'''





def maximumFrequency(arr, n):

	

	# Declare 4 variables as 'maxFrequency', 'currentFrequency', 'maxElement', 'currentElement' and initialize them with 0

	maxFrequency = 0

	maxElement = 0

	currentElement = 0

	currentFrequency = 0



	for i in range(n):

		

		# Set currentElement as arr[i] and currentFrequency to 0

		currentElement = arr[i]

		currentFrequency = 0

		

		# Loop to calculate the frequency of the currentElement



		for j in range(i, n):

			if(currentElement == arr[j]):

				currentFrequency += 1

				

		# If our currentFrequency is greater than maxFrequency

		if(currentFrequency > maxFrequency):

			

			# Update maxFrequency and 'maxElement' with 'currentFrequency' and 'currentFrequency' respectively.

			maxFrequency = currentFrequency

			maxElement = currentElement

			

	return maxElement

Accepted Answer

Greedy Approach

We can use hashing and count the frequency of all the elements and simultaneously find the maximum frequency. Now we can iterate through the array and find the first element with the maximum frequency and return that element.

Algorithm:

Declare a unorder map
Declare a variable ‘maxFrequency' and initialize it with ‘0’
Run a loop from ‘i’ = ‘0’ to ‘N’
- Update the frequency of current element arr[i]
- Update the ‘maxFrequency’ till the current element.
Declare a variable ‘maxElement’ and initialize it with ‘0’
Run a loop from ‘i’ = ‘0’ to ‘N’
- If the frequency of our current element is equal to the ‘maxFrequency’ computed
  - Update the ‘maxElement’ with the current element and break the loop
Return the 'maxElement'

Space Complexity: O(n)Explanation:

O(N) where N is the length of the array.

Using hashing, we count and store the frequency for all the unique elements. So, in the worst case, if we have all the unique elements, then we will require O(N) space.

Time Complexity: O(n)Explanation:

O(N) where N is the length of the array.

As we are traveling the entire array only twice, so time complexity is O(N)

C++ (g++ 5.4)

/*
    Time Complexity: O(N)
    Space Complexity: O(N)

    Where N is the total number of elements in the array.
*/


int maximumFrequency(vector &arr, int n)
{
	// Declare a unorder map
	unordered_map frequency;

	// Declare variables maxFrequency and maxElement and initialize them with ‘0’
	int maxFrequency = 0;
	int maxElement = 0;
	for (int i = 0; i < n; ++i)
	{
		// Update the frequency of current element arr[i]
		frequency[arr[i]]++;

		// Update the maxFrequency till the current element
		maxFrequency = max( maxFrequency, frequency[arr[i]]);
	} 
	
	for (int i = 0; i < n; ++i)
	{
		// If the frequency of our current element is equal to the maxFrequency computed
		if(maxFrequency == frequency[arr[i]])
		{
			// Update the maxElement with the current element and break the loop
			maxElement = arr[i];
			break;
		}
	}
	
	return maxElement;
}

Python (3.5)

'''

    Time Complexity: O(N)

    Space Complexity: O(N)



    Where N is the total number of elements in the array.

'''





def maximumFrequency(arr, n):

	

	# Declare a hash map to store frequency.

	frequency = dict()

	

	# Declare variables maxFrequency and maxElement and initialize them with 0.

	

	maxFrequency = 0

	

	maxElement = 0



	for i in range(n):

		

		# Update the frequency of current element arr[i]



		if(arr[i] in frequency):

			frequency[arr[i]] += 1

		else:

			frequency[arr[i]] = 1

			

		# Update the maxFrequency till the current element

		maxFrequency = max( maxFrequency, frequency[arr[i]])



	for i in range(n):

		

		# If the frequency of our current element is equal to the maxFrequency computed

		if(maxFrequency == frequency[arr[i]]):



			# Update the maxElement with the current element and break the loop

			maxElement = arr[i]

			break

		

	return maxElement

Java (SE 1.8)


/*
	Time Complexity: O(N)
	Space Complexity: O(N)
	Where N is the total number of elements in the array.
*/

import java.util.HashMap;

public class Solution {

	public static int maxFrequencyNumber(int n, int[] arr) {

		HashMap hm = new HashMap();
		for (int i : arr) {

			hm.put(i, hm.getOrDefault(i, 0) + 1);
		}
		int max = 0, ans = Integer.MIN_VALUE;
		for (int i : arr) {
			if (hm.get(i) > max) {
				max = hm.get(i);
				ans = i;
			}
		}
		return ans;
	}
}

Accepted Answer

int maxcount = 0;

int element_having_max_freq = 0;

for (int i = 0; i < n; i++) {

int count = 0;

for (int j = 0; j < n; j++) {

if (arr[i] == arr[j]) {

count++;

}

if (count > maxcount) {

maxcount = count;

element_having_max_freq = arr[i];

}

return element_having_max_freq;

Accepted Answer

int maxcount = 0;

int element_having_max_freq = 0;

for (int i = 0; i < n; i++) {

int count = 0;

for (int j = 0; j < n; j++) {

if (arr[i] == arr[j]) {

count++;

}

if (count > maxcount) {

maxcount = count;

element_having_max_freq = arr[i];

}

return element_having_max_freq;

Accepted Answer

int maxcount = 0;

int element_having_max_freq = 0;

for (int i = 0; i < n; i++) {

int count = 0;

for (int j = 0; j < n; j++) {

if (arr[i] == arr[j]) {

count++;

}

if (count > maxcount) {

maxcount = count;

element_having_max_freq = arr[i];

}

return element_having_max_freq;

Accepted Answer

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. The first line of each test case contains a single integer ‘N’ denoting the size of the array. The second line of each test case contains ‘N’ space-separated integers denoting the elements of the array.

Output Format:

For each case, we need to print an integer that has the maximum frequency. The output of each test case will be printed in a separate line.

Note:

You do not need to input or print anything, and it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 5 1 <= N <= 100

Ninja is given an array of integers that contain numbers in random order. He needs to write a program to find and return the number which occurs the maximum times in the given input. He needs your help to solve this problem.

If two or more elements contend for the maximum frequency, return the element which occurs in the array first i.e. whose index is lowest.

For example,

Input Format:

Output Format:

Note:

Constraints:

Top Walmart SDE-2 interview questions & answers

Popular interview questions of SDE-2