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There is a little bit less elegant, but alternative solution:
It's not needed to call the 2nd letterBits() for strB, it's enough to check if at least 1 letter of strB exists in bitsA .
letterBits()
strB
bitsA
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Two Strings
You are viewing a single comment's thread. Return to all comments →
There is a little bit less elegant, but alternative solution:
It's not needed to call the 2nd
letterBits()
forstrB
, it's enough to check if at least 1 letter ofstrB
exists inbitsA
.