65 lines
2.0 KiB
Java
65 lines
2.0 KiB
Java
// Algorithm BinarySearch Iterative method
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/*binarySearch(arr, x, low, high)
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repeat till low = high
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mid = (low + high)/2
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if (x == arr[mid])
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return mid
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else if (x > arr[mid]) // x is on the right side
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low = mid + 1
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else // x is on the left side
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high = mid - 1
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*/
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// Time Complexity : O(log(n))
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public class BinarySearch {
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static int binarySearch(int arr[], int key)
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{
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int start = 0;
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int end = arr.length - 1;
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while (start <= end) {
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// We use (start + (end - start)/2) rather than using (start + end)/2 to avoid
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// arithmetic overflow.
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// Arithmetic overflow is the situation when the value of a variable increases
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// beyond the maximum value of the memory location, and wraps around.
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int mid = start + (end - start) / 2; // optimised way
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if (arr[mid] == key)// key element is found at the middle of the array
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return mid;
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else if (arr[mid] < key) {// so the key lies in the right hand side of array
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start = mid + 1;
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}
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else {// so the key lies in the left subarray
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end = mid - 1;
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}
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}
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// we reach here when the key element is not present
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return -1;
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}
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public static void main(String[] args)
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{
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int arr[] = { 1, 3, 4, 5, 6 };
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/*
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* List<ArrayList<Integer>> arr = new ArrayList<>();
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* arr.add(new ArrayList<Integer>(Arrays.asList( 1, 3, 4, 5, 6 )));
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*/
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int key = 4; // element to search
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int index = binarySearch(arr, key);
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if (index == -1) {
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System.out.println("key element not found");
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}
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else {
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System.out.println("key element found at index :" + index);
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}
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}
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}
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