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I've solved this using an array of the size of alphabet to count the occurrences of each letter. If the count of a letter is one, then it does not have a partner.
I also had to deal with the situation where there is no underscores in the input.
Happy Ladybugs
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I've solved this using an array of the size of alphabet to count the occurrences of each letter. If the count of a letter is one, then it does not have a partner. I also had to deal with the situation where there is no underscores in the input.