Binary search to find the range in which the number lies

Question

I have an array

Values array: 12 20 32 40 52
              ^  ^  ^  ^  ^
              0  1  2  3  4

on which I have to perform binary sear

Answer 1

I know this is an old thread, but since I had to solve a similar problem I thought I would share it. Given a set of non-overlapping ranges of integers, I need to test if a given value lies in any of those ranges. The following (in Java), uses a modified binary search to test if a value lies within the sorted (lowest to highest) set of integer ranges.

/**
 * Very basic Range representation for long values
 *
 */
public class Range {

private long low;
private long high;

public Range(long low, long high) {
    this.low = low;
    this.high = high;
}

public boolean isInRange(long val) {
    return val >= low && val <= high;
}

public long getLow() {
    return low;
}

public void setLow(long low) {
    this.low = low;
}

public long getHigh() {
    return high;
}

public void setHigh(long high) {
    this.high = high;
}

@Override
public String toString() {
    return "Range [low=" + low + ", high=" + high + "]";
}
}



import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

//Java implementation of iterative Binary Search over Ranges
class BinaryRangeSearch { 
// Returns index of x if it is present in the list of Range, 
// else return -1 
int binarySearch(List ranges, int x) 
{ 

    Range[] arr = new Range[ranges.size()];
    arr = ranges.toArray(arr);
    int low = 0, high = arr.length - 1; 
    int iters = 0;
    while (low <= high) { 
        int mid = low + (high - low) / 2; // find mid point

        // Check if x is present a
        if (arr[mid].getLow() == x) {
            System.out.println(iters + " iterations");
            return mid;                 
        }

        // If x greater, ignore left half 
        if (x > arr[mid].getHigh()) {
            low = mid + 1; 
        }
        else if (x >= arr[mid].getLow()) {
            System.out.println(iters + " iterations");
            return mid;
        }

        // If x is smaller, ignore right half of remaining Ranges
        else
            high = mid - 1; 
        iters++;
    } 

    return -1; // not in any of the given Ranges
} 

// Driver method to test above 
public static void main(String args[]) 
{ 
    BinaryRangeSearch ob = new BinaryRangeSearch(); 

    // make a test list of long Range
    int multiplier = 1;

    List ranges = new ArrayList<>();
    int high = 0;
    for(int i = 0; i <7; i++) {

        int low = i + high;
        high = (i+10) * multiplier;
        Range r = new Range(low, high);
        multiplier *= 10;
        ranges.add(r);
    }

    System.out.println(Arrays.toString(ranges.toArray()));

    int result = ob.binarySearch(ranges, 11); 
    if (result == -1) 
        System.out.println("Element not present"); 
    else
        System.out.println("Element found at "
                        + "index " + result); 
} 
}