A rational is represented as p/qb, for example 2/3. Given a sorted array of rational numbers, how to search an element using Binary Search. Use of floating point arithmetic is not allowed.
Example:
Input: arr[] = {1/5, 2/3, 3/2, 13/2}
x = 3/2
Output: Found at index 2
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To compare two rational numbers p/q and r/s, we can compare p*s with q*r.
C++
// C program for Binary Search for Rationalnal Numbers // without using floating point arithmetic #include <stdio.h> struct Rational { int p; int q; }; // Utility function to compare two Rationalnal numbers // 'a' and 'b'. It returns // 0 --> When 'a' and 'b' are same // 1 --> When 'a' is greater //-1 --> When 'b' is greate int compare(struct Rational a, struct Rational b) { // If a/b == c/d then a*d = b*c: // method to ignore division if (a.p * b.q == a.q * b.p) return 0; if (a.p * b.q > a.q * b.p) return 1; return -1; } // Returns index of x in arr[l..r] if it is present, else // returns -1. It mainly uses Binary Search. int binarySearch(struct Rational arr[], int l, int r, struct Rational x) { if (r >= l) { int mid = l + (r - l)/2; // If the element is present at the middle itself if (compare(arr[mid], x) == 0) return mid; // If element is smaller than mid, then it can // only be present in left subarray if (compare(arr[mid], x) > 0) return binarySearch(arr, l, mid-1, x); // Else the element can only be present in right // subarray return binarySearch(arr, mid+1, r, x); } return -1; } // Driver method int main() { struct Rational arr[] = {{1, 5}, {2, 3}, {3, 2}, {13, 2}}; struct Rational x = {3, 2}; int n = sizeof(arr)/sizeof(arr[0]); printf("Element found at index %d", binarySearch(arr, 0, n-1, x)); } |
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Java
// Java program for Binary Search for Rationalnal Numbers // without using floating point arithmetic class GFG { static class Rational { int p; int q; public Rational(int p, int q) { this.p = p; this.q = q; } }; // Utility function to compare two Rationalnal numbers // 'a' and 'b'. It returns // 0 -. When 'a' and 'b' are same // 1 -. When 'a' is greater //-1 -. When 'b' is greate static int compare(Rational a, Rational b) { // If a/b == c/d then a*d = b*c: // method to ignore division if (a.p * b.q == a.q * b.p) return 0; if (a.p * b.q > a.q * b.p) return 1; return -1; } // Returns index of x in arr[l..r] if it is present, else // returns -1. It mainly uses Binary Search. static int binarySearch(Rational arr[], int l, int r, Rational x) { if (r >= l) { int mid = l + (r - l)/2; // If the element is present at the middle itself if (compare(arr[mid], x) == 0) return mid; // If element is smaller than mid, then it can // only be present in left subarray if (compare(arr[mid], x) > 0) return binarySearch(arr, l, mid - 1, x); // Else the element can only be present in right // subarray return binarySearch(arr, mid + 1, r, x); } return -1; } // Driver method public static void main(String[] args) { Rational arr[] = {new Rational(1, 5), new Rational(2, 3), new Rational(3, 2), new Rational(13, 2)}; Rational x = new Rational(3, 2); int n = arr.length; System.out.printf("Element found at index %d", binarySearch(arr, 0, n - 1, x)); } } // This code is contributed by Rajput-Ji |
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C#
// C# program for Binary Search for Rationalnal Numbers // without using floating point arithmetic using System; class GFG { class Rational { public int p; public int q; public Rational(int p, int q) { this.p = p; this.q = q; } }; // Utility function to compare two Rationalnal numbers // 'a' and 'b'. It returns // 0 -. When 'a' and 'b' are same // 1 -. When 'a' is greater //-1 -. When 'b' is greate static int compare(Rational a, Rational b) { // If a/b == c/d then a*d = b*c: // method to ignore division if (a.p * b.q == a.q * b.p) return 0; if (a.p * b.q > a.q * b.p) return 1; return -1; } // Returns index of x in arr[l..r] if it is present, else // returns -1. It mainly uses Binary Search. static int binarySearch(Rational []arr, int l, int r, Rational x) { if (r >= l) { int mid = l + (r - l)/2; // If the element is present at the middle itself if (compare(arr[mid], x) == 0) return mid; // If element is smaller than mid, then it can // only be present in left subarray if (compare(arr[mid], x) > 0) return binarySearch(arr, l, mid - 1, x); // Else the element can only be present in right // subarray return binarySearch(arr, mid + 1, r, x); } return -1; } // Driver method public static void Main(String[] args) { Rational []arr = {new Rational(1, 5), new Rational(2, 3), new Rational(3, 2), new Rational(13, 2)}; Rational x = new Rational(3, 2); int n = arr.Length; Console.Write("Element found at index {0}", binarySearch(arr, 0, n - 1, x)); } } // This code is contributed by 29AjayKumar |
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Output:
Element found at index 2
Thanks to Utkarsh Trivedi for suggesting above solution.
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