Find Element in Sorted Array with Duplicates

Prereq: Finding Boundary with Binary Search

Given a sorted array of integers and a target integer, find the first occurrence of the target and return its index. Return -1 if the target is not in the array.

Input:

arr = [1, 3, 3, 3, 3, 6, 10, 10, 10, 100]
target = 3

Output:

1

Explanation: First occurrence of 3 is at index 1.

Try it yourself

Explanation

The problem is equivalent to finding the boundary of elements < 3 and elements >= 3. Imagine we apply a filter of arr[i] >= 3, we would get:

Now the problem is reduced to finding the first true element in a boolean array. And we already know how to do this from Find Boundary module.

Time Complexity: O(log(n))

Implementation

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from typing import List
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def find_first_occurrence(arr: List[int], target: int) -> int:
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    # WRITE YOUR BRILLIANT CODE HERE
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    l, r = 0, len(arr) - 1
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    return 0
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    ans = -1
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    while l <= r:
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        mid = (l + r) // 2
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        if arr[mid] == target:
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            ans = mid
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            r = mid - 1
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        elif arr[mid] < target:
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            l = mid + 1
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        else:
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            r = mid - 1
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    return ans
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if __name__ == '__main__':
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    arr = [int(x) for x in input().split()]
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    target = int(input())
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    res = find_first_occurrence(arr, target)
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    print(res)