Maximum Frequency Number Problem Statement

Given an array of integers with numbers in random order, write a program to find and return the number which appears the most frequently in the array.

If multiple elements have the same maximum frequency, return the element that appears first in the array (i.e., the element with the lowest index).

Input

```The first line contains an integer 'T' which denotes the number of test cases. For each test case, the input consists of: - An integer 'N' on the first line denoting the size of the array. - 'N' space-separated integers on the second line denoting the elements of the array.```

Output

```For each test case, print the integer that has the maximum frequency on a new line.```

Example

```Input arr: [1, 2, 3, 1, 2] Output: 1```

Constraints

1

Question

Maximum Frequency Number Problem Statement

Given an array of integers with numbers in random order, write a program to find and return the number which appears the most frequently in the array.

If multiple elements have the same maximum frequency, return the element that appears first in the array (i.e., the element with the lowest index).

Input

```The first line contains an integer 'T' which denotes the number of test cases. For each test case, the input consists of: - An integer 'N' on the first line denoting the size of the array. - 'N' space-separated integers on the second line denoting the elements of the array.```

Output

```For each test case, print the integer that has the maximum frequency on a new line.```

Example

```Input arr: [1, 2, 3, 1, 2] Output: 1```

Constraints

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

Note

```You are not required to handle input or output operations. Simply implement the required function.```

Accepted Answer

Find the number with the maximum frequency in an array of integers.

Create a hashmap to store the frequency of each element in the array.
Iterate through the array to update the frequency count.
Find the element with the maximum frequency and return it.
If multiple elements have the same maximum frequency, return the one with the lowest index.

Accepted Answer

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