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    <p><strong>Brute force:</strong> if time complexity doesn't matter it would be a solution for 0 &lt; M &lt;= N invariant. nextRandom(N) is a function which returns random integer in [0..N):</p> <pre><code>init() { for (int idx = 0; idx &lt; N; idx++) { a[idx] = -1; } for (int idx = 0; idx &lt; M; idx++) { getNext(); } } int getNext() { for (int idx = 1; idx &lt; M; idx++) { a[idx -1] = a[idx]; } while (true) { r = nextRandom(N); idx = 0; while (idx &lt; M &amp;&amp; a[idx] != r) idx++; if (idx == M) { a[idx - 1] = r; return r; } } } </code></pre> <p><strong>O(M) solution:</strong> It is recursive solution for simplicity. It supposes to run nextRandom() which returns a random number in [0..1):</p> <pre><code>rnd(0, 0, N, M); // to get next M distinct random numbers int rnd(int idx, int n1, int n2, int m) { if (n1 &gt;= n2 || m &lt;= 0) return idx; int r = nextRandom(n2 - n1) + n1; int m1 = (int) ((m-1.0)*(r-n1)/(n2-n1) + nextRandom()); // gives [0..m-1] int m2 = m - m1 - 1; idx = rnd(idx, n1, r-1, m1); print r; return rnd(idx+1, r+1, n2, m2); } </code></pre> <p>the idea is to select a random r in between [0..N) on first step which splits the range on two sub-ranges by N1 and N2 elements in each (N1+N2==N-1). We need to repeat the same step for [0..r) which has N1 elements and [r+1..N) (N2 elements) choosing M1 and M2 (M1+M2==M-1) so as M1/M2 == N1/N2. M1 and M2 must be integers, but the proportion can give real results, we need to round values with probabilities (1.2 will give 1 with p=0.8 and 2 with p=0.2 etc.). </p>
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    1. POGenerating M distinct random numbers (one at a time) from a given range 0..N-1 in less than O(M) memory
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    1. COa[] is of size M so it's not less than O(M)
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    2. COI added second solution which doesn't require a[] at all if you print r instead, which gives result for O(M). Recursive is for the code readability.
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    3. COIt's still not less than O(M) in memory.
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