Nice, creative solution. But since we know that range is 1..n what about
array_sum = [1, 2, 3, 4, 1].sum consecutive_sum = n∗(n+1)/2
duplicate = array_sum - consecutive_sum
Oh shoot! I can’t believe I didn’t think of that. You’re totally right. Nice one! 👍🏻
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Nice, creative solution. But since we know that range is 1..n what about
array_sum = [1, 2, 3, 4, 1].sum
consecutive_sum = n∗(n+1)/2
duplicate = array_sum - consecutive_sum
Oh shoot! I can’t believe I didn’t think of that. You’re totally right. Nice one! 👍🏻