My problem statement is this - Find the count of numbers in a sorted array that are less than a given number, and this should be done efficiently with respect to time. I wrote a program using binary search that gets the count but time complexity wise it's failing. Need help in achieving this.
import java.util.Arrays;
public class SortedSearch {
public static int countNumbers(int[] sortedArray, int lessThan) {
if(sortedArray.length ==1 || sortedArray.length == 0) {
return singleElement(sortedArray, lessThan);
}
else {
return binarySearch(sortedArray, lessThan);
}
}
public static int singleElement(int[] sortedArray, int searchVal) {
if(sortedArray.length == 0) {
return 0;
}
if(sortedArray[0] < searchVal) {
return 1;
}
return 0;
}
private static int binarySearch(int[] sortedArray, int searchVal) {
int low = 0;
int high = (sortedArray.length)-1;
int mid = (low + high)/2;
if((sortedArray.length == 0) || (sortedArray[0] > searchVal)) {
return 0;
}
if(sortedArray[high] < searchVal) {
return sortedArray.length;
}
if(sortedArray[high] == searchVal) {
return sortedArray.length-1;
}
if(sortedArray[mid] < searchVal) {
int newLow = low;
int newHigh = calculateNewHigh(sortedArray, newLow, 0, searchVal);
int[] newArray = Arrays.copyOfRange(sortedArray, newLow, newHigh+1);
return newArray.length;
}
else {
int newLow = low;
int newHigh = mid;
int[] newArray = Arrays.copyOfRange(sortedArray, newLow, newHigh+1);
return binarySearch(newArray, searchVal);
}
}
private static int calculateNewHigh(int[] sortedArray, int low, int previousHigh, int searchVal) {
int newHigh = previousHigh + (sortedArray.length-low)/2;
if(sortedArray[newHigh] < searchVal) {
newHigh = calculateNewHigh(sortedArray, newHigh, newHigh, searchVal);
}
if(sortedArray[newHigh] == searchVal) {
newHigh--;
}
if(sortedArray[newHigh] > searchVal) {
newHigh--;
}
return newHigh;
}
public static void main(String[] args) {
System.out.println(SortedSearch.countNumbers(new int[] { 1, 3, 5, 7 }, 4));
}
}