Problem Statement |
| | The cardinality of a binary number is given as the total number of "ones" it contains. For example, the cardinality of binary 10100 (decimal 20) is 2, because there are 2 "ones". The cardinality of binary 11110 (decimal 30) is 4, because there are 4 "ones".
Given a int[] of decimal numbers arrange them in ascending order of binary cardinality and return this arranged int[]. If two numbers have the same binary cardinality, then the smaller number must come first in the arranged int[]. |
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Definition |
| | | Class: | BinaryCardinality | | Method: | arrange | | Parameters: | int[] | | Returns: | int[] | | Method signature: | int[] arrange(int[] numbers) | | (be sure your method is public) |
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Constraints |
| - | numbers will have between 1 and 50 elements inclusive. |
| - | each element of numbers will be between 0 and 1000000 inclusive. |
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Examples |
| 0) | |
| | | Returns: { 4 } | | There is only one element in the array, so it must be returned. |
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| 1) | |
| | | Returns: { 2, 3, 7, 15, 31 } | | We start by converting the above set of decimals into binary numbers. We get the following array: {11111, 1111, 111, 11, 10}. Now we can calculate the cardinality of each number: {5, 4, 3, 2, 1}. So 2 has the lowest binary cardinality, followed by 3, 7, 15 and finally 31. |
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| 2) | |
| | | Returns: { 1, 2, 4, 8, 3, 5, 6, 9, 10, 7 } | | The cardinality array is {2, 2, 1, 3, 2, 2, 1, 2, 1, 1}. Note that although 10 and 3 have the same cardinality of 2, 3 must come earlier because it is smaller in value. |
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| 3) | |
| | {811385,340578,980086,545001,774872,855585,13848,863414,419523,190151,784903,127461} |
| Returns:
{ 13848, 340578, 545001, 855585, 419523, 811385, 127461, 190151, 774872, 863414, 784903, 980086 } | |
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