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Consider the following Python function implementing the optimal wiggle subsequence algorithm. What is the value of count after the loop finishes when the input is [1, 5, 4]?

easy🧾 Code Trace Q12 of Q15

Greedy Algorithms - Wiggle Subsequence

Consider the following Python function implementing the optimal wiggle subsequence algorithm. What is the value of count after the loop finishes when the input is [1, 5, 4]?

def wiggleMaxLength(nums):
    if not nums:
        return 0
    count = 1
    last_diff = 0
    for i in range(1, len(nums)):
        diff = nums[i] - nums[i - 1]
        if (diff > 0 and last_diff <= 0) or (diff < 0 and last_diff >= 0):
            count += 1
            last_diff = diff
    return count

A2

B3

C1

D0

Step-by-Step Solution

Solution:

Step 1: Trace loop iterations
Input: [1, 5, 4] - i=1: diff=5-1=4 >0 and last_diff=0 ≤0 -> count=2, last_diff=4 - i=2: diff=4-5=-1 <0 and last_diff=4 ≥0 -> count=3, last_diff=-1
Step 2: Final count value
After loop, count=3, which is the length of the longest wiggle subsequence.
Final Answer:
Option B -> Option B
Quick Check:
Count increments twice for valid wiggles [OK]

Quick Trick: Count increments on sign change of diff from last_diff

Common Mistakes:

MISTAKES

Off-by-one counting
Ignoring initial count=1
Not updating last_diff correctly

Trap Explanation:

PITFALL

Some think count stays 1 or increments incorrectly due to sign checks.

Interviewer Note:

CONTEXT

Tests ability to mentally execute code and track variable updates precisely.

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