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Collections are a way for you to organize kata so that you can create your own training routines. Every collection you create is public and automatically sharable with other warriors. After you have added a few kata to a collection you and others can train on the kata contained within the collection.
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This comment is hidden because it contains spoiler information about the solution
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting
It looks like many solutions here are O(n^2) ones, but here complexity is just more obvious.
Do you count yours solution as o(n)? For me it seems to be O(n * log(n)), cause "counts[ch] = (counts[ch] || 0) + 1" should have at least O(log(n)) complexity when the size of 'counts' grows.
exuse me for necroposting