Find Last Occurrence of an Element in a Sorted Array Using Binary Search

Problem Statement❯

Given a sorted array of integers and a target number key, your task is to find the last occurrence (last index) of the key in the array.

If the key appears multiple times, return the index of its last appearance.
If the key does not exist in the array, return -1.
If the array is empty, return -1.

This problem is commonly solved using a modified version of binary search for better efficiency.

Examples❯

Input Array	Key	Last Occurrence Index	Description
[5, 10, 10, 10, 20, 20]	10	3	Key 10 appears at indices 1, 2, and 3. Last is at index 3.
[1, 2, 3, 4, 5]	3	2	Key 3 occurs only once at index 2.
[1, 1, 1, 1, 1]	1	4	All elements are the key. Last is at index 4.
[1, 2, 4, 5, 6]	3	-1	Key 3 is not present in the array.
[1, 2, 3, 4, 5]	5	4	Key is the last element in the array.
[10, 20, 30, 40, 50]	10	0	Key is the first element, and occurs only once.
[]	7	-1	Empty array. No elements to search.

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Solution❯

Understanding the Problem

We are given a sorted array and a target number called key. Our goal is to find the last occurrence (also called the rightmost index) of this key in the array.

Since the array is sorted, we can use a powerful technique called binary search to solve this efficiently in O(log n) time. However, we need to slightly modify the traditional binary search so that instead of stopping at the first match, we continue looking toward the right side of the array. This way, we ensure that we don’t miss any later occurrences of the same number.

Step-by-Step Solution with Example

Example

Input: arr = [2, 4, 10, 10, 10, 18, 20], key = 10  
Expected Output: 4 (since the last occurrence of 10 is at index 4)

Step 1: Initialize Variables

We begin by setting two pointers: start = 0 and end = length of array - 1. We also set a result variable res = -1 to store the index of the last occurrence (if found).

Step 2: Apply Binary Search

We now enter a loop while start <= end. In each iteration, we calculate the middle index: mid = Math.floor((start + end) / 2).

Then, we compare arr[mid] with the key and take action:

Case 1: arr[mid] == key
We have found an occurrence! But we still want to find if there’s another occurrence further to the right. So, we set res = mid and update start = mid + 1.
Case 2: arr[mid] < key
The key must be on the right, so we move the start pointer to mid + 1.
Case 3: arr[mid] > key
The key must be on the left, so we move the end pointer to mid - 1.

Step 3: End of Search

The loop ends when start > end. At this point, the value of res will either hold the last occurrence index or remain -1 if the key wasn't found.

How the Code Handles Edge Cases

If the key occurs once: We find and return that index.
If the key occurs multiple times: The algorithm continues to search toward the right, updating the result every time it finds a match, and ultimately returns the last index.
If the key does not exist: No match will be found, and the result stays at -1.
If the array is empty: The loop never runs, and -1 is returned immediately.

By continuing the search even after finding a match, we make sure we don’t stop too early. This makes our approach robust, especially when the target number appears multiple times. And because we are cutting the search space in half each time, the time complexity is only O(log n), which is very efficient for large arrays.

Algorithm Steps❯

Initialize start = 0, end = n - 1, and res = -1 to store the result.
While start ≤ end, calculate mid = start + (end - start) / 2.
If arr[mid] == key: update res = mid and move start = mid + 1 to search towards the right.
If arr[mid] < key: move start = mid + 1.
If arr[mid] > key: move end = mid - 1.
After the loop, res will contain the last occurrence index or -1 if not found.

Code

C++

Python

Java

Rust

Kotlin

Swift

#include <stdio.h>

int lastOccurrence(int arr[], int n, int key) {
  int start = 0, end = n - 1, res = -1;
  while (start <= end) {
    int mid = start + (end - start) / 2;
    if (arr[mid] == key) {
      res = mid;
      start = mid + 1;
    } else if (key < arr[mid]) {
      end = mid - 1;
    } else {
      start = mid + 1;
    }
  }
  return res;
}

int main() {
  int arr[] = {5, 10, 10, 10, 20, 20};
  int n = sizeof(arr) / sizeof(arr[0]);
  int key = 10;
  printf("Last Occurrence Index: %d\n", lastOccurrence(arr, n, key));
  return 0;
}

Time Complexity❯

Case	Time Complexity	Explanation
Best Case	O(1)	When the key is found at the first mid computation and no further checking is needed.
Average Case	O(log n)	Each iteration halves the search space, leading to logarithmic performance.
Worst Case	O(log n)	In the worst case, the search checks each level of the binary search tree until no more right occurrences are found.

Space Complexity❯

O(1)

Explanation: Only a fixed number of variables are used, with no extra memory required.

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