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Corrected explanations and included an example (#75)
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committed by
Oleksii Trekhleb

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2334583635
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16b6ea506a
@ -17,9 +17,10 @@ export default function integerPartition(number) {
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partitionMatrix[0][numberIndex] = 0;
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}
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// Let's fill the first row. It represents the number of way of how we can form
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// number zero out of numbers 0, 1, 2, ... Obviously there is only one way we could
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// form number 0 and it is with number 0 itself.
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// Let's fill the first column. It represents the number of ways we can form
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// number zero out of numbers 0, 0 and 1, 0 and 1 and 2, 0 and 1 and 2 and 3, ...
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// Obviously there is only one way we could form number 0
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// and it is with number 0 itself.
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for (let summandIndex = 0; summandIndex <= number; summandIndex += 1) {
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partitionMatrix[summandIndex][0] = 1;
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}
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@ -35,7 +36,18 @@ export default function integerPartition(number) {
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} else {
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// The number of combinations would equal to number of combinations of forming the same
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// number but WITHOUT current summand number plus number of combinations of forming the
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// previous number but WITH current summand.
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// <current number - current summand> number but WITH current summand.
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// Example: number of ways to form number 4 using summands 1, 2 and 3 is the sum of
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// {number of ways to form 4 with sums that begin with 1 +
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// number of ways to form 4 with sums that begin with 2 and include 1} +
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// {number of ways to form 4 with sums that begin with 3 and include 2 and 1}
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// Taking these sums to proceed in descending order of intergers, this gives us:
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// With 1: 1+1+1+1 -> 1 way
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// With 2: 2+2, 2+1+1 -> 2 ways
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// With 3: 3 + (4-3) <= convince yourself that number of ways to form 4 starting
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// with 3 is == number of ways to form 4-3 where 4-3 == <current number-current summand>
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// Helper: if there are n ways to get (4-3) then 4 can be represented as 3 + first way,
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// 3 + second way, and so on until the 3 + nth way. So answer for 4 is: 1 + 2 + 1 = 4 ways
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const combosWithoutSummand = partitionMatrix[summandIndex - 1][numberIndex];
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const combosWithSummand = partitionMatrix[summandIndex][numberIndex - summandIndex];
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