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Evaluate the given code snippet. Ignoring the order, which line of code will you use in the place of "# LINE A" to generate the output shown here ?keys = ['key1', 'key2', 'key3', 'key4', 'key5'] vals = ['val1', 'val2', 'val3', 'val4', 'val5']     # LINE Aprint(mydict)# Output: {'key1': 'val1', 'key2': 'val2', 'key3': 'val3', 'key4': 'val4', 'key5': 'val5'}Select one:mydict = dict()for x in keys:       mydict[x] = vals.pop()mydict = dict(keys.join(vals))mydict = dict(keys, vals)mydict = dict()mydict.fromkeys(keys, vals)mydict = dict()for x in range(len(keys)):       mydict[keys[x]] = vals[x]

Question

Evaluate the given code snippet. Ignoring the order, which line of code will you use in the place of "# LINE A" to generate the output shown here ?keys = ['key1', 'key2', 'key3', 'key4', 'key5'] vals = ['val1', 'val2', 'val3', 'val4', 'val5']     # LINE Aprint(mydict)# Output: {'key1': 'val1', 'key2': 'val2', 'key3': 'val3', 'key4': 'val4', 'key5': 'val5'}Select one:mydict = dict()for x in keys:       mydict[x] = vals.pop()mydict = dict(keys.join(vals))mydict = dict(keys, vals)mydict = dict()mydict.fromkeys(keys, vals)mydict = dict()for x in range(len(keys)):       mydict[keys[x]] = vals[x]

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Solution

The correct line of code to replace "# LINE A" and generate the desired output is:

mydict = dict() for x in range(len(keys)): mydict[keys[x]] = vals[x]

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