Apple Interview Question for Staff Engineers


Country: United States




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0
of 0 vote

Elegant 4 line solution in Python (for small strings of course):

def sortStrings(strs, col):
  data = [tuple(word.split()) for word in strs]
  data = [(first_name, last_name, int(age)) for first_name, last_name, age in data]
  result = sorted(data, key=lambda x: x[col-1])
  return result

Test Code:

strings = ['john doe 33', 'smith black 9', 'diana yale 12']
print(sortStrings(strings, 1)) # Order by first name
print(sortStrings(strings, 2)) # Order by last name
print(sortStrings(strings, 3)) # Order by age
'''
Output:
[('diana', 'yale', 12), ('john', 'doe', 33), ('smith', 'black', 9)]
[('smith', 'black', 9), ('john', 'doe', 33), ('diana', 'yale', 12)]
[('smith', 'black', 9), ('diana', 'yale', 12), ('john', 'doe', 33)]
'''

However, in a distributed setting, when we have large amounts of strings to process, we can divide up the column that we are trying to sort by and process per chunks. We would create multiple chunks and use a sorting algorithm such as quicksort or mergesort to combine the data into a sorted order.

- prudent_programmer March 19, 2018 | Flag Reply
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0
of 0 vote

If you want the output in the same string format as the input:

def sort_giant_string(giant, col, col_type=(str, str, int)):
  data = [row.split(" ") for row in giant.split("\n")]
  data.sort(key=lambda x: col_type[col - 1](x[col - 1]))
  return "\n".join([" ".join([str(cell) for cell in row]) for row in data])

- noammohr June 24, 2018 | Flag Reply
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0
of 0 vote

-(NSArray *) sortedArrayWithColumn:(NSUInteger) column {
    return [self sortedArrayUsingComparator:^NSComparisonResult(NSString *obj1, NSString *obj2){
        NSArray *components1 = [obj1 componentsSeparatedByString:@","];
        NSArray *components2 = [obj2 componentsSeparatedByString:@","];
        
        if (column >= components1.count && column >= components2.count) {
            return NSOrderedSame;
        }
        
        if (column >= components1.count) {
            return NSOrderedDescending;
        }
        
        if (column >= components2.count) {
            return NSOrderedAscending;
        }
        
        NSString *component1 = components1[column];
        NSString *component2 = components2[column];
        
        return [component1 caseInsensitiveCompare:component2];
    }];
}

- Rajiv Singh August 02, 2018 | Flag Reply


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