Minimum Operations to Exceed Threshold Value I - Problem

You're tasked with cleaning up an array by removing elements that don't meet a minimum quality standard! 🧹

Given a 0-indexed integer array nums and an integer k (the threshold), you can perform operations where each operation removes one occurrence of the smallest element from the array.

Goal: Find the minimum number of operations needed so that all remaining elements in the array are greater than or equal to k.

Example: If nums = [2, 11, 10, 1, 3] and k = 3, you need to remove elements 1 and 2 (both < 3), requiring 2 operations.

Input & Output

example_1.py — Basic Case
$ Input: nums = [2, 11, 10, 1, 3], k = 3
› Output: 2
💡 Note: Elements 1 and 2 are less than 3, so we need 2 operations to remove them. After removal, remaining elements [11, 10, 3] are all ≥ 3.
example_2.py — All Elements Valid
$ Input: nums = [1, 2, 3, 4, 5], k = 1
› Output: 0
💡 Note: All elements are already ≥ 1 (the threshold), so no operations are needed.
example_3.py — Remove All Elements
$ Input: nums = [1, 2, 3], k = 10
› Output: 3
💡 Note: All elements (1, 2, 3) are less than 10, so we need to remove all of them, requiring 3 operations.

Visualization

Tap to expand
🏭 Factory Quality ControlProduction Line: Quality Scores2REJECT11PASS10PASS1REJECT3PASS📋 Quality Control ReportMinimum Quality Standard: k = 3Products Below Standard: 2Actions Required: 2 removals🎯 Optimization InsightInstead of removing products one by one,we can simply count how many failthe quality check in one inspection!✨ Final ResultTime Complexity: O(n) - Single pass through production lineSpace Complexity: O(1) - Only need a counter for rejects💡 The key insight: Count failures instead of simulating removals!
Understanding the Visualization
1
Products Arrive
Products with quality scores [2, 11, 10, 1, 3] arrive at quality control
2
Set Threshold
Management sets minimum quality threshold k = 3
3
Inspect Products
Go through each product and check if quality < 3
4
Count Rejects
Products with scores 2 and 1 fail inspection - that's 2 products to discard
Key Takeaway
🎯 Key Insight: Rather than simulating the removal process, we can directly count elements below the threshold - that's exactly how many operations we need!

Time & Space Complexity

Time Complexity
⏱️
O(n)

Single pass through the array to count elements less than k

n
2n
✓ Linear Growth
Space Complexity
O(1)

Only using a single counter variable, no extra data structures

n
2n
✓ Linear Space

Related Problems

Constraints

  • 1 ≤ nums.length ≤ 50
  • 1 ≤ nums[i] ≤ 109
  • 1 ≤ k ≤ 109
  • All array elements are positive integers
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