Write a function , char* findx (const char* s, const char* x), that* finds the first occurrence of the C-style string x in s. This is using C++Write a function , char* findx (const char* s, const char* x), that Solution Here is the program and please let me know if any errors occurs char* findx (const char* s, const char* x) { assert(s); assert(x); size_t len_s = m_strlen(s); size_t len_x = m_strlen(x); assert(len_s >= len_x); char* p_to_match = nullptr; for (size_t i = 0; i < len_s; ++i) { if (*(s + i) == *x) { p_to_match = const_cast(s + i); if (len_x == 1) return p_to_match; const char* next_s = (s + i + 1); const char* first_x = (x + 1); for (size_t j = 0; j < len_x - 1; ++x) { if (*(next_s + j) != *(first_x + j)) break; if (j == len_x - 2) return p_to_match; } } } return nullptr;.

1. Write a method rarest that accepts a map whose keys are strings and whose values are integers
as a parameter and returns the integer value that occurs the fewest times in the map. If there is a
tie, return the smaller integer value. If the map is empty, throw an exception. For example,
suppose the map contains mappings from students' names (strings) to their ages (integers). Your
method would return the least frequently occurring age. Consider a map variable m containing
the following key/value pairs: {Alyssa=22, Char=25, Dan=25, Jeff=20, Kasey=20, Kim=20,
Mognan=25, Ryan=25, Stef=22} Three people are age 20 (Jeff, Kasey, and Kim), two people
are age 22 (Alyssa and Stef), and four people are age 25 (Char, Dan, Mogran, and Ryan). So a
call of rarest (m) returns 22 because only two people are that age. If there is a tie (two or more
rarest ages that occur the same number of times), return the youngest age among them. For
example, if we added another pair of Kelly=22 to the map above, there would now be a tie of
three people of age 20 (Jeff, Kasey, Kim) and three people of age 22 (Alyssa, Kelly, Stef). So a
call of rarest(m) would now return 20 because 20 is the smaller of the rarest values. Type your
solution here: public static int rarest(Map m) {}
Solution
Code:
public static int rarest(Map m) {
int age = 0;
if (m.size() == 0)
m = null;
Collection c = m.values();
Object values[] = c.toArray();
Set set = new TreeSet();
for (Object a : values) {
set.add(a);
}
Map m2 = new HashMap();
for (Object o : set) {
m2.put(o, 0);
}
for (Object a : values) {
for (Object b : set) {
if (a.equals(b)) {
age = m2.get(b);

2. age++;
m2.put(b, age);
}
}
}
int n = 0;
for (Object o : set) {
if (n == 0 || m2.get(o) < n) {
n = m2.get(o);
age = (int) o;
} else {
if (m2.get(o) == n) {
if ((int) o < age) {
age = (int) o;
}
}
}
}
return age;
}