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ArrayEasyLeetCode #169

Majority Element (> n/2 Times)

Boyer-Moore Voting Algorithm finds candidate with count > n/2 in O(n) time and O(1) space.

Time O(n) Space O(1)
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Asked in interviews at

GoogleAmazon

Reference implementation (Python)

def majority_element(arr):
    candidate, count = None, 0
    for x in arr:
        if count == 0:
            candidate = x
        count += 1 if x == candidate else -1
    return candidate

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Drop a live, looping Majority Element (> n/2 Times) canvas into a blog post or course page.

<iframe src="/embed/majority-element" width="100%" height="380" style="border:1px solid #e6e8ec;border-radius:14px" loading="lazy" title="3D algorithm visualization by CodeWithMunnaX"></iframe>

Frequently asked

What is the time and space complexity of Majority Element (> n/2 Times)?

Majority Element (> n/2 Times) runs in O(n) time and O(1) auxiliary space.

Is the Majority Element (> n/2 Times) visualizer free?

The core 3D visualizer, Web Audio chimes and step intuition are free. A one-time ₹49 Pro Pass unlocks advanced algorithms, large custom arrays, Battle Arena and reel export.

Which companies ask Majority Element (> n/2 Times) in interviews?

Majority Element (> n/2 Times) shows up in interviews at Google, Amazon.

How does the 3D visualization of Majority Element (> n/2 Times) work?

Each array value becomes a 3D pillar whose height maps to its value. Comparisons glow amber, swaps flash red, and settled elements lock green while a floating pointer badge tracks the active index — synchronized line-by-line with the Python code.

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