That seems like an very complex way to frame the problem. Given the two arrays of length n and m, each of the result arrays in your answer are going to be leng

That seems like an very complex way to frame the problem. Given the two arrays of length n and m, each of the result arrays in your answer are going to be length n+m. Think of each element in those arrays as a binary decision between choosing the next element of each array - there will be C(n+m,n) possible outcomes - and this is just a bit string where a 0 means take the next element from the first array and a 1 from the second array. Once you see this, it is just a matter of generating all length n+m bit strings with exactly n 0-bits. This is the next permutation problem which can be done in constant space and linear time and very efficiently too. C++ even has an STL function for it: http://www.cplusplus.com/reference/algorithm/next_permutation/ http://www.cplusplus.com/reference/algorithm/next_permutatio... Your starting array is '0011' in his example, then keep generating the next permutation in a loop until exhausted so '0101', '0110', etc... edit: here's a cute bit hack for it too: https://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation https://graphics.stanford.edu/~seander/bithacks.html#NextBit...