I was wondering if it is possible to find the closest element in a sorted List
for a element that is not in the list.
For example if we had the values [1,3,6,7] and we are looking for the element closest to 4, it should return 3, because 3 is the biggest number in the array, that is smaller than 4.
I hope it makes sense, because English is not my native language.
Because the collection is sorted, you can do a modified binary search in O( log n )
:
public static int search(int value, int[] a) {
if(value < a[0]) {
return a[0];
}
if(value > a[a.length-1]) {
return a[a.length-1];
}
int lo = 0;
int hi = a.length - 1;
while (lo <= hi) {
int mid = (hi + lo) / 2;
if (value < a[mid]) {
hi = mid - 1;
} else if (value > a[mid]) {
lo = mid + 1;
} else {
return a[mid];
}
}
// lo == hi + 1
return (a[lo] - value) < (value - a[hi]) ? a[lo] : a[hi];
}
Since most of the code above is binary search, you can leverage the binarySearch(...)
provided in the std library and examine the value of the insertion point
:
public static int usingBinarySearch(int value, int[] a) {
if (value <= a[0]) { return a[0]; }
if (value >= a[a.length - 1]) { return a[a.length - 1]; }
int result = Arrays.binarySearch(a, value);
if (result >= 0) { return a[result]; }
int insertionPoint = -result - 1;
return (a[insertionPoint] - value) < (value - a[insertionPoint - 1]) ?
a[insertionPoint] : a[insertionPoint - 1];
}