Genc Next

I solved the Problem using the concept of Dynamic Programming bottom up approach. In this i subdivided the problem into smaller problems and then used the result of smaller problems to calculate for large problem.
So calculating coins need for any small amount less than target and then using it further for larger subproblems.

int coinChange(vector const &arr, int sum)
{
int dp[sum + 1];
for (int i = 1; i <= sum; i++)
{
dp[i] = INT_MAX;
int result = INT_MAX;

for (int c: arr)
{
if (i - c >= 0) {
result = dp[i - c];
}

if (result != INT_MAX) {
dp[i] = min(dp[i], result + 1);
}
}
}

return dp[sum];
}

This was my code for the problem where i used a 1-d array to store the results of subproblems and used it to get the output.
In this i was calculating the coins for each amount less than target(sum) and then using it forward for further calculations and a last returning the result stored at last index of the array.

I solved the problem using a character array in which i stored all the characters from the string.
So i traversed the string and stored al the characters in the character array.
Then i reversed the array.
Now i again traversed the original string and checked if the particular element is character then replaced that with character from my reversed character array and for doing this i used a integer variable to keep pointing to my current index in the array.

string reverse(string s)
{
char temp[s.length()];
int x = 0;
for (int i = 0; i < s.length(); i++) {
if (s[i] >= 'a' && s[i] <= 'z'
|| s[i] >= 'A' && s[i] <= 'Z') {
temp[x] = s[i];
x++;
}
}

reverse(temp, temp + x);

x = 0;
for (int i = 0; i < s.length(); i++) {
if (s[i] >= 'a' && s[i] <= 'z'
|| s[i] >= 'A' && s[i] <= 'Z') {
s[i] = temp[x];
x++;
}
}
return s;
}

This was solution to the problem.

I started by taking two numbers as input from the user and then i asked user which operation he wanted to perform. then on the basis of user's choice i used Switch case statements to perform the operation and show the result on the console. I also used Exception handling in the division operation to check if DivisionByZero Exception occurs and handle it. ( the interviewer was impressed with this approach )

The idea is to first count number digits (or find order). Let the number of digits be n. For every digit r in input number x, compute rn. If sum of all such values is equal to n, then return true, else false.