Finding the nearest, lower number in a sorted NSArray - ios

Question

I have a sorted array of times like so
[0.0, 1.2, 4.3, 5.9, 7.2, 8.0]
While an audio file plays, I want to be able to take the current time and find what the nearest, lower number is in the array.
My approach would be to traverse the array, possible in reverse order as it feels like it should be faster. Is there a better way?
The playback SHOULD be linear, but might be fast-forwarded/rewound, so I would like to come up with a solution that takes that into account, but I'm not really sure how else to approach the problem.

Answer 1

The method you are looking for is -[NSArray indexOfObject:inSortedRange:options:usingComparator:]. It performs a binary search. With the options:NSBinarySearchingInsertionIndex option, if the value isn't found exactly, it returns the index where the object would be inserted, which is the index of the least larger element, or the count of items in the array.
NSTimeInterval currentTime = ...;
NSUInteger index = [times indexOfObject:#(currentTime)
inSortedRange:NSMakeRange(0, times.count)
options:NSBinarySearchingInsertionIndex
usingComparator:^(id object0, id object1) {
NSTimeInterval time0 = [object0 doubleValue];
NSTimeInterval time1 = [object1 doubleValue];
if (time0 < time1) return NSOrderedAscending;
else if (time0 > time1) return NSOrderedDescending;
else return NSOrderedSame;
}];
// If currentTime was not found exactly, then index is the next larger element
// or array count..
if (index == times.count || [times[index] doubleValue] > currentTime) {
--index;
}

Answer 2

The fastest* way to find something in a sorted array is binary search: if there are n items, check the element at index n/2. If that element is greater than what you're looking for, check the element at index n/4; otherwise, if it's less than what you're looking for, check the element at index 3n/4. Continue subdividing in this fashion until you've found what you want, i.e. the position where the current time should be. Then you can pick the preceding element, as that's the closest element that's less than the current time.
However, once you've done that once, you can keep track of where you are in the list. As the user plays through the file, keep checking to see if the time has passed the next element and so on. In other words, remember where you were, and use that when you check again. If the user rewinds, check the preceding elements.
*Arguably, this isn't strictly true -- there are surely faster ways if you can make a good guess as to the probable location of the element in question. But if you don't know anything other than that the element appears somewhere in the array, it's usually the right approach.

Answer 3

I'm not sure if it's the best approach, but I think it'll get the job done (assuming your array is always ascending order).
- (NSNumber *) incrementalClosestLowestNumberForNumber:(NSNumber *)aNumber inArray:(NSArray *)array {
for (int i = 0; i < array.count; i++) {
if ([array[i] floatValue] == [aNumber floatValue]) {
return aNumber;
}
else if ([array[i] floatValue] > [aNumber floatValue]) {
int index = (i > 0) ? i - 1 : 0;
return array[index];
}
}
return #0;
}
Then call it like this:
NSArray * numbArray = #[#0.0, #1.2, #4.3, #5.9, #7.2, #8.0];
NSNumber * closestNumber = [self closestLowestNumberForNumber:#2.4 inArray:numbArray];
NSLog(#"closest number: %#", closestNumber);
I'm not sure if someone else knows a special way that is much faster.
Based on some of the other answers / comments, I came up with this, perhaps one of them can point out if a whole is somewhere.
- (NSNumber *) quartalClosestLowestNumberForNumber:(NSNumber *)compareNumber inArray:(NSArray *)array {
int low = 0;
int high = array.count - 1;
NSNumber * lastNumber;
int currentIndex = 0;
for (currentIndex = low + (high - low) / 2; low <= high; currentIndex = low + (high - low) / 2) {
NSNumber * numb = array[currentIndex];
if (numb.floatValue < compareNumber.floatValue) {
low = currentIndex + 1;
}
else if (numb.floatValue > compareNumber.floatValue) {
high = currentIndex - 1;
}
else if (numb.floatValue == compareNumber.floatValue) {
return numb;
}
lastNumber = numb;
}
if (lastNumber.floatValue > compareNumber.floatValue && currentIndex != 0) {
lastNumber = array[currentIndex - 1];
}
return lastNumber;
}
I'm really bored right now, so I'm trying to test the fastest method. Here's how I did it.
NSMutableArray * numbersArray = [NSMutableArray new];
for (int i = 0; i < 100000; i++) {
float floater = i / 100.0;
[numbersArray addObject: #(floater)];
}
// courtesy #RobMayoff
NSDate * binaryDate = [NSDate date];
NSNumber * closestNumberBinary = [self binaryClosestLowestNumberForNumber:#4.4 inArray:numbersArray];
NSLog(#"Found closest number binary: %# in: %f seconds", closestNumberBinary, -[binaryDate timeIntervalSinceNow]);
// The Quartal Version
NSDate * quartalDate = [NSDate date];
NSNumber * closestNumberQuartal = [self quartalClosestLowestNumberForNumber:#4.4 inArray:numbersArray];
NSLog(#"Found closest number quartal: %# in: %f seconds", closestNumberQuartal, -[quartalDate timeIntervalSinceNow]);
// The incremental version
NSDate * incrementalDate = [NSDate date];
NSNumber * closestNumberIncremental = [self incrementalClosestLowestNumberForNumber:#4.4 inArray:numbersArray];
NSLog(#"Found closest number incremental: %# in: %f seconds", closestNumberIncremental, -[incrementalDate timeIntervalSinceNow]);
And here's the output:
Found closest number binary: 4.4 in: 0.000030 seconds
Found closest number quartal: 4.4 in: 0.000015 seconds
Found closest number incremental: 4.4 in: 0.000092 seconds
And another test case:
Found closest number binary: 751.48 in: 0.000030 seconds
Found closest number quartal: 751.48 in: 0.000016 seconds
Found closest number incremental: 751.48 in: 0.013042 seconds