Which algorithmic approach guarantees an optimal solution with time complexity better than O(n*m), where n is the number of intervals and m is the number of points?
easy🔍 Pattern Recognition Q11 of Q15
Intervals - Count of Intervals Containing Each Point
You are given a list of intervals and a list of points. For each point, you need to find how many intervals contain it. Which algorithmic approach guarantees an optimal solution with time complexity better than O(n*m), where n is the number of intervals and m is the number of points?
AUse a line sweep algorithm by sorting interval start and end events along with points, then sweeping through them.
BUse a brute force nested loop checking each point against every interval.
CUse dynamic programming to precompute counts for all points in the range.
DSort intervals and points, then for each point use binary search to count intervals starting before and ending after the point.
Step-by-Step Solution
Step 1: Understand brute force and binary search limitations
Brute force is O(n*m), too slow for large inputs. Binary search approach requires careful handling of interval ends and starts but is less optimal than line sweep.
Step 2: Recognize line sweep advantages
Line sweep sorts all events (starts, points, ends) and processes them in one pass, maintaining active intervals count efficiently in O((n+m) log (n+m)) time.
Final Answer:
Option A -> Option A
Quick Check:
Line sweep processes all events sorted by coordinate with correct event ordering [OK]
Quick Trick:Line sweep processes starts, points, ends in sorted order [OK]
Common Mistakes:
MISTAKES
Assuming binary search alone is optimal
Thinking DP applies here
Using brute force for large inputs
Trap Explanation:
PITFALL
Binary search approach looks efficient but handling ends and starts precisely is tricky and less optimal than line sweep.
Interviewer Note:
CONTEXT
Tests if candidate can identify the best pattern for interval counting problems.
Master "Count of Intervals Containing Each Point" in Intervals
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