If your strings only contain ASCII characters [0, 127] you can "compress" the string using a custom 6 or 7-bit code page.
You can do this several ways, but I think one of the simpler methods is to define an array holding all allowed characters - a LUT, lookup-table if you like, then use its index value as the encoded value. You would of course have to manually mask and shift the encoded value into a typed array.
If your LUT looked like this:
var lut = " abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789.,:;!(){}";
you would in this case deal with a LUT of length 71 which means we would need to use a 7-bit range or [0, 127] (if length were 64 we could've reduced the it to 6-bit [0, 63] values).
Then you would take each characters in the string and convert to index values (you would normally do all the following steps in a single operation but I have separated them for simplicity):
var lut = " abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789.,:;!(){}";
var str = "I like bananas !";
var page = [];
Array.prototype.forEach.call(str, function(ch) {
var i = lut.indexOf(ch);
if (i < 0) throw "Invalid character - can't encode";
page.push(i);
});
console.log("Intermediate page:", page);
You can always tweak the LUT so that the most used characters are in the beginning, then support variable encoding bit-range, find max value and use that to determine what range you want to encode in. You can add an initial bit as a flag as to which range the encoding uses (for example bit 0 set if 6-bit fits, otherwise use 7-bit range).
Now that you know the indices we can start to encode the binary output itself using a 7-bit approach. Since JavaScript only support byte values, i.e. 8-bit width, we have to do all the split, shift and merge operations manually.
This means we need to keep track of remainder and position on a bit-level.
Say first index value was the following 7-bit value (full 7-bit range for readability - all in pseudo format):
&b01111111
The first step would be to shift it over to bit position 0 and keep track of a remainder:
&b01111111 << 1
Resulting in:
&b11111110
^
new bit position: 7
new remainder : 1
Then the next index value, for example:
&b01010101
would be encoded like this - first convert to 7-bit value in its own byte representation:
&b01010101 << 1 => &b10101010
Then get the reminder part first. To obtain this will shift everything right-wise using 8-bit minus the current remainder (within modulo of 8):
remainderValue = &b10101010 >>> (8 - remainder)
leaving us with the following representation:
&b00000001
(Note that we use triple >>>
to shift right to avoid issues with sign.)
Next step now is to merge this value with our previous value that has already been encoded and stored into our destination byte array - for this we'll use an OR operation:
Index 0 New value Result in index 0 (index of dst. array)
&b11111110 | &b00000001 => &b11111111
then go to next index in our destination array and store the rest of the current value, then update the remainder and position.
The "leftover" of the byte is calculated like this using the original (after shifting it) 7-bit byte value:
leftover = &b10101010 << remainder => &b01010100
which we now put into the next position:
Index 0 Index 1 (destination array index, not page index)
&b11111111 01010100
^
new bit position: 14
new remainder : 2
And so on with the remaining index values. See this answer for actual code on how you can do this in JavaScript - the code in this answer doesn't deal with string encoding per-se, but it shows how you can shift byte buffers bit-wise which is essentially the same you need for this task.
To calculate the remainder step, use 8-bits minus your custom bit-range:
step = 8 - newRange (here 7) => 1
This will also be the start remainder. For each character, you'll add the step to remainder after it has been processed, but remember to use modulo 8 (byte width) when you use it for shifting:
remainder += step;
numOfBitsToShift = remainder % 8;
Bit-position uses of course the bit-range, in this case 7:
bitPosition += 7;
Then to find which indices you're dealing with you divide the bitPosition on 8, if any decimal you have to deal with two indexes (old and new), if no decimal the current position represents new index only (only shift is needed for current index value).
You can also use modulo and when modulo of remainder = step you know you that you are dealing with a single index in the destination.
To calculate the final length you would use the bit-length and length of string, then ceil the result so that all characters will fit into a 8-byte byte array which is the only array we can get in JavaScript:
dstLength = Math.ceil(7 * str.length / 8);
To decode you just reverse all the steps.
An alternative, if you use long strings or have to move forward fast, is to use an established compressor such as zlib which has a very compact header as well as good performance in JavaScript in the case of the linked solution. This will also deal with "patterns" in the string to further optimize the resulting size.
Disclaimer: as this is mostly a theoretical answer there might be some errors. Feel free to comment if any are found. Refer to linked answer for actual code example.