I have a use case where if a number lies between 0-10 it should return 0 and if it lies between 11-20 it should return 1 etc
0 => 0-3, (0 and 3 are inclusive) 1 => 4-15, (4 and 15 are inclusive) 2 => 16-40, (16 and 40 are inclusive) 3 => 41-88, (41 and 88 are inclusive) 5 => 89-300 (89 and 300 are inclusive)
I was thinking how could I implement and was thinking java maps, but it does not allow range searching
I am interested in something like this , I have a function
int foo() { }
if foo returns 5 , since it lies between 0 to 10 I would use 0, if foo return 25 it would use 2.
Any ideas
Edit : Actually the ranges are not as simple as 0-10, 11-20. I want to be able to do range searches. Sorry about the confusion. Based on the queries I have added the correct example, the numbers are continous
I can think of a number of possible solutions for the more general problem where the ranges are not uniform and there are 'holes'. The simplest are:
floorEntry(key)
to do the lookups. This should be less time efficient (O(log(N) lookups) but more space efficient.Here's a solution using NavigableMaps that allows for 'holes' in the mapping.
private static class Range { public int upper, value; ... } NavigableMap<Integer, Range> map = new TreeMap<Integer, Range>(); map.put(0, new Range(3, 0)); // 0..3 => 0 map.put(5, new Range(10, 1)); // 5..10 => 1 map.put(100, new Range(200, 2)); // 100..200 => 2 // To do a lookup for some value in 'key' Map.Entry<Integer,Range> entry = map.floorEntry(key); if (entry == null) { // too small } else if (key <= entry.getValue().upper) { return entry.getValue().value; } else { // too large or in a hole }
On the other hand, if there are no 'holes' the solution is simpler:
NavigableMap<Integer, Integer> map = new TreeMap<Integer, Integer>(); map.put(0, 0); // 0..4 => 0 map.put(5, 1); // 5..10 => 1 map.put(11, 2); // 11..200 => 2 // To do a lookup for some value in 'key' if (key < 0 || key > 200) { // out of range } else { return map.floorEntry(key).getValue(); }
Pseudo-code:
new int[] {0, 3, 5, 15, 100, 300}
.Arrays.binarySearch()
.10
in the above array would be 3
, placing it between 5
and 15
, so it belongs in the second range.If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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