nested for loops

resolve nested for loops by preventing repeated iterations
published: (updated: )
by Harshvardhan J. Pandit
is part of: code optimisation
optimization programming

I saw a post on Facebook by a colleague from my lab about the runtime of her program being quite high.. I thought this can’t be right, imagining a large amount of data being crunched copiously over and over. Thinking that this would offer an interesting problem (and a way to procrastinate from work), I offered to help her. The next day we sat looking at the lines of code trying to identify what exactly was taking that much time, and where could we do things better. Scrolling through the various files and classes and methods all made up of Java, my eyes fell on a nesting of for-loops that went on and on and on. Years of stereotyped-advice rang out in my head as I recoiled looking at the indentation of each for-loop going on and over. I resisted the urge to point out that this is bad practice , thinking it would be condescending on my part, and more importantly - not necessarily good advice.

Understanding for loops

For loops have a purpose - to iterate a block of code over and over until the given condition no longer holds true. To that end, they are a fancy way of writing a while loop. Logically, they are equivalent -

for (int i=0 ; i<100 ; i++) {
    // do something

{ // if you are pedantic about block scope
int i = 0;
while(i < 100) {
    // do something
} // variable i will never leave this block

The for loop is meant to process things over and over, so a nested for loop can be used to iterate all possible permutations. For example,

for (int i=0 ; i<m ; i++)
    for (int j=0; j<n ; j++)

is guaranteed to run m x n times, creating all possible pairs (combinations) of values of i and j . This is the easiest way to create such pairs. However, careful observations can be made for the sake of optimisation whether all such pairs are indeed required to be calculated.

The nesting of for loops we had on the screen was 13 levels deep, with variable loop limits. That takes it to the order of n^13 , which is pretty gigantic in theory. So how do we optimise this? The approach lies in understanding the problem, rather than speculating approaches based on abstract mathematical applications.

Understanding the problem background

The colleague who wrote the original piece of code was researching into the spread of diseases. To that end, she wanted to test the various combinations of environmental factors and calculate a score for how infectious it is. The problem the for-loops meant to solve were to combine all factors in every possible way - hence the nesting. The key factor here is combination , and not permutation . The latter is where the order does not matter, therefore, it is fewer in number, and more efficient to calculate.

Each environmental factor comes from a group which is distinct from the other groups, which are things like season and geographical location. If we have n distinct groups each with m distinct items, then the total possible ways of combining them becomes n x n x n x ... (m times) or n ^ m. This is how we get a nesting of 13 for-loops.


Memoization is a technique where previous computations are stored to speed up processing for repetitive calculations. A good way to describe it would be to write this function which stores the results of previously computed calculation in a cache . When a pre-computed result is asked for, it returns the value from the cache in constant time - O(1) . Pretty efficient. The only downside is the lookup cost and the storage for the cache. Even this can be further optimized using hashing .

int[] cache = new int[100];

// calculate something for 0 <= n < 100
int calculate_something(int n) {
    if (cache[n] != 0) {
        return cache[n];
    int result = perform_calculations();
    cache[n] = result;
    return result;

For our use case, we can use memoization to save pre-computed combinations. This can be applied as saving the results of the inner loops so that they are not iterated over and over again.

Java String-ops

In Java, Strings are immutable , which means that once assigned, a string is essentially a constant, and cannot be changed. Any operation that acts on the string creates a new String object. In the given scenario, each environmental condition was expressed as a String, and all loops were combining them through concatenation . Which meant that there were new String objects being created over and over 13 levels down.

Perhaps a better design would be to represent the environmental conditions as enumerations or some other constant factor which speeds up their combination operation. But given the scenario, let us assume that they have to be Strings. In this case, the objective is to prevent the repetitive creation of new Strings. The goal of each nested loop to add an item from a group to the final String. Instead, we make use of List to hold the results for us, as they are much better in terms of performance and efficiency. If desired, they can be turned into a String at the end.

We require one loop to go over all of the environmental groups. This will be the outer loop. We then require another loop to go over the items in each group and add them to a list of items. We store the results in a list of lists. Each list is a combination of items.

The approach is pretty simple -

  1. For every list, add the current item to it
  2. Collect all lists and replace the result with this new list
// create a list of lists
List<List<String>> results = new ArrayList<ArrayList<String>>();

// put the items in the first group to populate the list
for (i=0 ; i<group[0].length ; i++) {
    // create a new list for each item
    List<String> newlist = new ArrayList<String>();
    // add the item to the list; it will be the sole member
    // add this new list to the results grouplist

// iterate over the rest of groups
for (i=1 ; i<no_groups ; i++) {
    // create an empty grouplist (list of lists)
    // to preserve the original during processing
    List<List<String>> newresults = new ArrayList<ArrayList<String>>();
    // iterate over the items in group
    for (j=0 ; j<group[i].length ; j++) {
        // iterate over the items in results
        for (k=0 ; k<results.length ; k++) {
            // create a copy of the list, and add the current item to it
            List<String> newlist = new ArrayList<String>(results[k]);
            // add this list to the new group of results
    // replace the results with the new ones
    results = newresults;

The way this works is, after copying over items from the first group, the contents of results will be -


Then, we iterate m times, once for each of the remaining groups (outer loop); and then for each item in the group, we create a new list by adding the current item to every list in the results. Which gives -


The cache here is the stored result of every combination of the previous iterations, and we only add the current items to each of them without repeating the calculations. Additionally, there are no new String operations, therefore, no new Strings are being created. However, we create a lot of lists . Turns out that this is quite efficient because lists are better in terms of memory management than Strings as they can be expanded with any free space on the heap whereas a String requires continuous allocations.

Though the number of combinations remain the same - m x n, the number of loops required to process them has reduced them from 13 to 3. The optimization trick here is to avoid repeating the same computation over and over again by using memoization . This reduced the runtime of this part from several minutes to a few seconds .

More optimizations

The original program also removed combinations which were impossible - such as a certain region having a certain season (which never happens). The way these were done was by taking the impossible condition in a list, generating all possible combinations of them and concatenating them as Strings. Then it checked whether these occurred as substring in the combinations of results. Since we deal with lists here, expressing these as lists of items that are impossible together makes sense. So we store the impossible conditions together as a list, and check whether the item is a sub-list of every list. If it is, then we remove that list as a violated condition. This operation is not as efficient as calculating substrings because lists deal in object references whereas Strings are allocated as continuous arrays and have byte checking. But the offset achieved from not creating Strings in the original loop problem is enough to make this a trivial delay in comparison.


Through the optimisations, the total runtime went down a bit, which I’m sure can be reduced further down. What I learnt through this little exercise was the application of what concepts I learned through college, and then later while doing little projects. I tailored the solution based on the problem, which is expected of an engineer - to solve real-world problems through practical solutions. It is through various such problems that collective experience is gained and one levels up in knowledge.