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| const current = a[i]; | ||
| const duplicateIndexs = []; | ||
| for (let j = i + 1; j < a.length; j++) { |
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indeed it requires 2 loop, but it has circuit breaker to exit the loop if the item already checked and has duplication, and I think I should early exit also on the second loop. no Idea how to solve this with one loop haha
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😄 since all of the solutions might have similar o notation, let's try to compare the performance when we get a chance
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