Pr 5: isAllUniqueElements #4
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| array.forEach((currentElement, currentIndex) => { | ||
| array.slice(currentIndex + 1).forEach((elementToCompare) => { | ||
| if (elementToCompare === currentElement) { | ||
| isAllUnique = false; | ||
| } | ||
| }); | ||
| }); |
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Is there a more time/space efficient way to do this? without making a new arrays/nested loops?
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This solution improves space complexity.
| array.forEach((currentElement, currentIndex) => { | |
| array.slice(currentIndex + 1).forEach((elementToCompare) => { | |
| if (elementToCompare === currentElement) { | |
| isAllUnique = false; | |
| } | |
| }); | |
| }); | |
| array.forEach((currentElement, currentIndex) => { | |
| if (array[currentIndex + 1, -1].includes(currentElement)) { | |
| isAllUnique = false; | |
| } | |
| }); | |
| }); |
| let isAllUnique = true; | ||
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| array.forEach((currentElement, currentIndex) => { | ||
| array.slice(currentIndex + 1).forEach((elementToCompare) => { |
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What's the time and space complexity? Can we improve either of them?
| array.slice(currentIndex + 1).forEach((elementToCompare) => { | |
| if (array[currentIndex + 1, -1].includes(currentElement)) { | |
| isAllUnique = false; | |
| } |
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| array.forEach((currentElement, currentIndex) => { | ||
| array.slice(currentIndex + 1).forEach((elementToCompare) => { | ||
| if (elementToCompare === currentElement) { |
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The '===' operator could cause some issues if the elements are not the exact match (even in type) and this might create some issues, Using '==' operator would help in avoiding these issues.
| let isAllUnique = true; | ||
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| array.forEach((currentElement, currentIndex) => { | ||
| array.slice(currentIndex + 1).forEach((elementToCompare) => { |
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I am not sure of the purpose of both slice and forEach on this line of code.
Please clarify.
| array.forEach((currentElement, currentIndex) => { | ||
| array.slice(currentIndex + 1).forEach((elementToCompare) => { |
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Can we avoid having a nested loop if possible? Maybe we can first populate a hash with the elements and then use that to lookup each element.
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Is it possible to return early if we find a duplicate element?
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