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✅ Binary Search Approach ⚬ Create a function that takes a sorted array & target, and returns [first_index, last_index]. ⚬ Init result: [-1, -1]. Left=0, Right=len(arr)-1. ⚬ Find First Occurrence: While left <= right: mid = (left right) // 2 If arr[mid] == target: result[0] = mid; right = mid - 1 (search left). Elif arr[mid] < target: left = mid 1. Else: right = mid - 1. ⚬ Find Last Occurrence: Reset left=0, right=len(arr)-1. While left <= right: mid = (left right) // 2 If arr[mid] == target: result[1] = mid; left = mid 1 (search right). Elif arr[mid] < target: left = mid 1. Else: right = mid - 1. ⚬ Return result. ⏳Time Complexity: O(log n) – Halves the search space each step. 💾 Space Complexity: O(1) – Constant extra space.

✅ Iterative Approach ⚬ Create a function that takes the array & target, and returns [first_index, last_index]. ⚬ Init result array: [-1, -1]. ⚬ Loop through the array (i from 0 to n-1) ⚬ If arr[i] == target & result[0] == -1: Set result[0] = i (first occurrence). ⚬ If arr[i] == target & result[0] != -1: Set result[1] = i (update last occurrence). ⚬ Return the result by the loop's end. ⏳Time Complexity: O(n) – Loops over all n elements. 💾 Space Complexity: O(1) – Just a fixed 2-element array

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Binary search for peak: 1⃣Calculate the middle. 2⃣If value at the middle is bigger than middle 1 index value, the peak will be anywhere to the left, continue in that range. 3⃣else peak value will be somewhere on right. 4⃣Narrow range till found. Time: O(log n), Space: O(1).

1⃣ The iterative approach to finding a peak element in an array involves scanning through each element one by one. 2⃣If you find a larger element, you update the index. By the end, you have the index of a peak element. 👉Time complexity is O(n), and space complexity is O(1).

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