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Leetcode #2936: Number of Equal Numbers Blocks

In this guide, we solve Leetcode #2936 Number of Equal Numbers Blocks in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given a 0-indexed array of integers, nums. The following property holds for nums: All occurrences of a value are adjacent.

Quick Facts

  • Difficulty: Medium
  • Premium: Yes
  • Tags: Array, Binary Search, Interactive

Intuition

The problem structure suggests a monotonic decision, which makes binary search a natural fit.

By halving the search space each step, we reach the answer efficiently.

Approach

Search either directly on a sorted array or on the answer space using a check function.

Each check is fast, and the logarithmic search keeps the overall runtime low.

Steps:

  • Define the search bounds.
  • Check the mid point condition.
  • Narrow the bounds until convergence.

Example

Input: nums = [3,3,3,3,3] Output: 1 Explanation: There is only one block here which is the whole array (because all numbers are equal) and that is: [3,3,3,3,3]. So the answer would be 1.

Python Solution

# Definition for BigArray. # class BigArray: # def at(self, index: long) -> int: # pass # def size(self) -> long: # pass class Solution(object): def countBlocks(self, nums: Optional["BigArray"]) -> int: i, n = 0, nums.size() ans = 0 while i < n: ans += 1 x = nums.at(i) if i + 1 < n and nums.at(i + 1) != x: i += 1 else: i += bisect_left(range(i, n), True, key=lambda j: nums.at(j) != x) return ans

Complexity

The time complexity is O(m×log⁡n)O(m \times \log n)O(m×logn), where mmm is the number of different elements in the array numnumnum, and nnn is the length of the array numnumnum. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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