if(bMind.charAt(i) = ("a"))

There are several problems here: single = is an assignment operator. That is why you get a message that left side is not a variable. Change the single = to a double == to make a comparison.

Second: In most other languages, having double quotes makes a string (a pointer to a sequence of characters). Having single quotes makes a primitive char. Result: You are comparing a primitive char to a pointer. You should probably do well by changing to this:

if (bMind.charAt(i) == 'a')

It should solve your problem.

Edit: I wrote a much more elaborate answer for you, but it got lost when I accidentally pressed tab and then backspace, causing my browser to go back to the previous page. Yay, Chrome! I always wondered why you have assigned backspace to "go back". So done this a lot of times already... Now I am kind of discouraged. My point was very quickly explained elegant like this:

int characterCount[256];

//Remember to clear the array...

//Then you have some string "myString" or anything you name it

//Then the counting of occurrences is as simple as this:
for (int i = 0; i < length; i++) {
  characterCount[myString.charAt(i)]++;
}

To print the statistics:

for (int i = 'a'; i <= 'z'; i++) {
  std::cout << (char)i << " = " << characterCount[i];
}

Oh, it's Java! Well, the idea is still there! ;) I am mostly into C++/ObjC for the time being, but just see this as pseudocode. It's been years since I've touched Java, but I remember something like "System.out.writeLn" or something.