Amdocs Coding Question | Return Map | Count Frequency of each word in String

Hello friends, I hope you all are doing great. In this post today we will learn to solve the  coding questions which is being asked by the Amdocs company recently. If you are preparing for selection in any company then this post will help you definitely.

Lets Start

Return Map or HashMap

In this task you will get an array of string that contains words, you are required to return a map which will contain each unique word as a key, and the number of occurrences of each word as a value.\n\nFor example:

Given the following array of string:

"hello", "world", "this", "is", "this", "hello"

you will return:

{'hello': 2, 'world': 1, 'this': 2, 'is': 1}

Example :-

• Input:
• "hello", "world", "this", "is", "this", "hello"
• Output:
• {'hello': 2, 'world': 1, 'this': 2, 'is': 1}

Explanation:-

Frequency of hello is 2, world is 1 , this is 2  etc

Code in Python

```# Code in Python
class TestImpl:
def hashMap(self, listString):
mydict = {}
for i in listString:
if (mydict._contains_(i)):
mydict[i] = mydict[i] + 1;
else:
mydict[i] = 1;
print(mydict)

obj = TestImpl()
obj.hashMap(["hello", "world", "this", "is", "this", "hello"])

"""
{'hello': 2, 'world': 1, 'this': 2, 'is': 1}
"""
```

Code in Java
```import java.util.HashMap;
import java.util.Scanner;

class Range {

static HashMap ReturnHashMap(String ar[]) {
HashMap<String, Integer> map = new HashMap<>();
for (int i = 0; i < ar.length; i++) {
if (!map.containsKey(ar[i])) {
for (int j = 0; j < ar.length; j++) {
if (ar[i] == ar[j]) {
if (map.containsKey(ar[i])) {
map.put(ar[i], map.get(ar[i]) + 1);
} else {
map.put(ar[i], 1);
}
}
}
}
}
return map;
}

public static void main(String[] args) {
String arr[] = {"hello", "world", "this", "is", "this", "hello"};
HashMap<String, Integer> map = new HashMap<>();
map = ReturnHashMap(arr);
System.out.println(map);
}
}
/*
OUTPUT
{world=1, this=2, is=1, hello=2}
*/
```

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