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You don't need to check for duplicates for all 3 elements
Ariel Davies
May 19, 2024
You only need to ensure two of the three numbers are not repeating. Because if 2 of the 3 are unique then the triplet is inherently unique. The second while loop for adjusting the right pointer is not necessary and makes the solution a more confusing.
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Sidd Mittala year ago
good point - but the second while loop also increases efficiency. If there are duplicate right pointer elements, you may as well skip them since only 1 valid left pointer element will be able to sum all numbers to 0. (see there test example, it illustrates it quite well...
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