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Asked: 2026-05-12T23:07:27+00:00

I am trying to output all the possible unique integer combinations from 1 to

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I am trying to output all the possible unique integer combinations from 1 to max for a set number of integers. So for 3 integers and a max of 4 I would get:

123
124
134
234

I am doing this with nested for loops but I want to allow the user to input the number of integers at runtime. right now I have 

if(numInts >6);
for(int x = 1; x < max; x++);
if(numInts >5);
for(int y = 1; y < max; y++);
...

Is there a way to clean this up so I don’t have to write out each possible integer for loop.

PS: I know the code above will not output the requested output. This is for a programing competition so I am not asking for code solutions just the idea that would make this possible.

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1 Answer

  1. Editorial Team

    Looking at your comments on the original post, you want an iterative solution. A recursive solution will be just as fast as an iterative solution when you have a language that supports tail call optimization. But if you’re working with Java/C#, again, this isn’t available, so here’s another way to look at the problem.

    You are generating combinations. A combination is just a subset with a certain number of elements. Subsets with small-ish sets can be expressed with bitmasks.

    If I have the set [1, 2, 3, 4] and I want to describe the subset [1, 3, 4], I can do so by going through each element and asking “True or False: is this element in the subset?” So, for [1, 3, 4], the result is [True, False, True, True]. If I am working with a set less than 32 (or 64) bytes, I can encode this as an integer: 1011b = 11. This is very compact in memory and computers tend to have very fast bitmath operators.

    So what is a combination then, in terms of these binary numbers? If I want all subsets with N members, I can translate that as “I want all numbers with N bits set.”

    Take [1, 2, 3, 4] as we have been. We want all subsets with 3 elements. How many 4-bit numbers are there with exactly 3 bits set? Answer: 1110b, 1101b, 1011b, and 0111b. If we turn these integers into subsets, we get your solutions: [1, 2, 3], [1, 2, 4], [1, 3, 4], and [2, 3, 4].

    You can start thinking in terms of the bits only. You start off with the lowest number with N bits set. That corresponds to a solution. You then start flipping bits one-for-one. In a systematic way such that each iteration always results in the next highest number.

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