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Question:Given a sorted array A of n elements, possibly with duplicates, find the index of the first occurrence of a number in O(log n) time.

Input:4, 4, 8, 8, 8, 15, 16, 23, 23, 42. Find 8

Output:Index = 2 (0 based indexing)

This problem is very much similar to the binary search problem. We can apply the same methodology with just an added condition. For a number to be the first occurrence in the array, the number to the left of it must be smaller. We just add this condition to our loop.

if((mid == low && arr[mid] == data) || (arr[mid] == data && arr[mid-1] < data))

Here is the sample code for the same:-

#include<stdio.h> int binarySearchFirstOccurrence(int arr[], int low, int high, int data) { int mid; // A simple implementation of Binary Search if(high >= low) { mid = low + (high - low)/2; // To avoid overflow if((mid == low && arr[mid] == data) || (arr[mid] == data && arr[mid-1] < data)) return mid; // We need to give preference to left part of the array // since we are concerned with the first occurrence else if(arr[mid] >= data) return binarySearchFirstOccurrence(arr, low, mid-1, data); else // We need to search in the right half return binarySearchFirstOccurrence(arr, mid+1, high, data); } } int getFirstOccurrence(int arr, int len, int data) { int low = 0; int high = len - 1; int index = binarySearchFirstOccurrence(arr, low, high, data); return index; } // Driver program to test the code int main(void) { int arr[] = {4, 4, 8, 8, 8, 15, 15, 16, 23, 42, 42}; int result = getFirstOccurrence(arr, 11, 8); printf("The first occurrence is at = %d", result); return 0; }

*Time Complexity:-* O(log n)

*Space Complexity:-* O(1)