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    <p><strong>A deterministic solution (ie not random)</strong></p> <p>Sort your shows by decreasing number of episodes. </p> <p>Select the biggest and arrange a matrix with the number of columns corresponding to the number of episodes of this one, filled in the following way: </p> <pre><code>A A A A A A &lt;- First show consist of 6 episodes B B B B C C &lt;- Second and third show - 4 episodes each C C D D &lt;- Third show 2 episodes </code></pre> <p>Then collect by columns </p> <pre><code>{A,B,C}, {A,B,C}, {A,B,D}, {A,B,D}, {A,C}, {A,C} </code></pre> <p>Then Join</p> <pre><code>{A,B,C,A,B,C,A,B,D,A,B,D,A,C,A,C} </code></pre> <p>And now assign sequential numbers</p> <pre><code>{A1, B1, C1, A2, B2, C2, A3, B3, D1, A4, B4, D2, A5, C3, A6, C4} </code></pre> <p><strong>Edit</strong></p> <p>Your case</p> <pre><code>[['A'] * 2, ['L'] * 3, ['X'] * 5]) X X X X X L L L A A -&gt; {X1, L1, X2, L2, X3, L3, X4, A1, X5, A2} </code></pre> <p><strong>Edit 2</strong> </p> <p>As no Python here, perhaps a Mathematica code may be of some use: </p> <pre><code>l = {, , ,}; (* Prepare input *) l[[1]] = {a, a, a, a, a, a}; l[[2]] = {b, b, b, b}; l[[3]] = {c, c, c, c}; l[[4]] = {d, d}; le = Length@First@l; k = DeleteCases[ (*Make the matrix*) Flatten@Transpose@Partition[Flatten[l], le, le, 1, {Null}], Null]; Table[r[i] = 1, {i, k}]; (*init counters*) ReplaceAll[#, x_ :&gt; x[r[x]++]] &amp; /@ k (*assign numbers*) -&gt;{a[1], b[1], c[1], a[2], b[2], c[2], a[3], b[3], d[1], a[4], b[4], d[2], a[5], c[3], a[6], c[4]} </code></pre>
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    1. COThis algorithm definitely avoids adjacency problems. It does, however, tend to clump episodes together. I like it, overall. =) I'll edit an implementation in Python into the question. Thanks!
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    2. CO@Adeel You may get a family of different results using this same algo by choosing the number of colums greater than the number of episodes of the biggest show (but less than half the TOTAL number of episodes of all shows added up)
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