FIRST AND LAST OCCURRENCE

 

First and Last Occurrence of an Element in a Sorted Array

Introduction

In this problem, we are given a sorted array that may contain duplicate elements. The goal is to find the first and last occurrence of a given element.

Since the array is sorted, we can use Binary Search to solve this efficiently.

Problem Statement

Given a sorted array arr[] and a target x, return:

  • Index of the first occurrence of x

  • Index of the last occurrence of x

If x is not present, return [-1, -1].

Example

Input:

arr = [1, 3, 5, 5, 5, 5, 67, 123, 125]
x = 5

Output:

[2, 5]

Solution Code

class Solution:
    def find(self, arr, x):
        n = len(arr)

       
        low, high = 0, n - 1
        first = -1

        while low <= high:
            mid = (low + high) // 2

            if arr[mid] == x:
                first = mid
                high = mid - 1
            elif arr[mid] < x:
                low = mid + 1
            else:
                high = mid - 1

        low, high = 0, n - 1
        last = -1

        while low <= high:
            mid = (low + high) // 2

            if arr[mid] == x:
                last = mid
                low = mid + 1
            elif arr[mid] < x:
                low = mid + 1
            else:
                high = mid - 1

        return [first, last]

Approach Used: Modified Binary Search

Idea

We use Binary Search twice:

  • First to find the first occurrence

  • Second to find the last occurrence

How It Works

Finding First Occurrence

  • When arr[mid] == x:

    • Store index

    • Move left (high = mid - 1) to check for earlier occurrence

Finding Last Occurrence

  • When arr[mid] == x:

    • Store index

    • Move right (low = mid + 1) to check for later occurrence

Why This Approach Was Chosen

  • Efficient for sorted arrays

  • Avoids scanning entire array

  • Finds both positions in logarithmic time

Alternative Approach

  • Linear Search

    for i in range(n)

     Time Complexity: O(n)

Why not used?
Too slow for large arrays (up to 10⁶ elements)

Time and Space Complexity

  • Time Complexity: O(log n)

  • Space Complexity: O(1)

Important Observation

  • We don’t stop at first match

  • We continue searching to find boundary positions

  • That’s the key difference from normal binary search

Edge Cases

  • Element not present = return [-1, -1]

  • Only one occurrence = both indices same

  • All elements same

Conclusion

Using modified binary search allows us to efficiently find the first and last occurrences of an element in a sorted array. By slightly adjusting the search direction after finding the target, we can locate the exact boundaries in logarithmic time.

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