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Array-3 > seriesUp
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Given n>=0, create an array with the pattern {1,    1, 2,    1, 2, 3,   ... 1, 2, 3 .. n} (spaces added to show the grouping). Note that the length of the array will be 1 + 2 + 3 ... + n, which is known to sum to exactly n*(n + 1)/2.

seriesUp(3) → [1, 1, 2, 1, 2, 3]
seriesUp(4) → [1, 1, 2, 1, 2, 3, 1, 2, 3, 4]
seriesUp(2) → [1, 1, 2]

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public int[] seriesUp(int n) { }


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Difficulty: 321.0

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