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# Maximum Frequency Number

Last Updated: 30 Nov, 2020
Difficulty: Easy

## PROBLEM STATEMENT

#### 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
`````` ## Approach 1

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'