Skip to main content

Binary Search on Answer Pattern – Must Solve LeetCode Problems

Binary Search on Answer Pattern (Complete Practice Guide)

Binary Search on Answer is a powerful problem-solving pattern frequently used in coding interviews and competitive programming. Instead of searching inside an array, we perform binary search on the range of possible answers.

  • Minimum or maximum value is asked
  • Answer lies within a numeric range
  • A feasibility function exists
  • Feasibility is monotonic

📌 LeetCode Problems Using Binary Search on Answer

Problem Difficulty Core Idea
1283 – Find the Smallest Divisor Given a Threshold Easy Binary search on divisor value
1011 – Capacity To Ship Packages Within D Days Medium Binary search on ship capacity
875 – Koko Eating Bananas Medium Binary search on eating speed
1482 – Minimum Number of Days to Make m Bouquets Medium Binary search on number of days
1760 – Minimum Limit of Balls in a Bag Medium Binary search on max balls per bag
410 – Split Array Largest Sum Hard Binary search on maximum subarray sum
1552 – Magnetic Force Between Two Balls Medium Binary search on minimum distance
1891 – Cutting Ribbons Medium Binary search on ribbon length

🎯 Final Takeaway

Mastering Binary Search on Answer turns many complex-looking problems into predictable templates. Once you identify the range and feasibility check, the solution becomes straightforward.

Comments

Popular posts from this blog

LeetCode 88 Explained: Four Approaches, Mistakes, Fixes & the Final Optimal Python Solution

Evolving My Solution to “Merge Sorted Array” A practical, beginner-friendly walkthrough showing four versions of my code (from a naive approach to the optimal in-place two-pointer solution). Includes explanations, complexity and ready-to-paste code. Problem Summary You are given two sorted arrays: nums1 with size m + n (first m are valid) nums2 with size n Goal: Merge nums2 into nums1 in sorted order in-place . Version 1 — Beginner Approach (Extra List) I merged into a new list then copied back. Works, but not in-place and uses extra memory. class Solution: def merge(self, nums1, m, nums2, n): result = [] p1 = 0 p2 = 0 for _ in range(m+n): if p1 >= m: result.extend(nums2[p2:n]) break elif p2 >= n: result.extend(nums1[p1:m]) break elif nu...

Introducing CodeMad: Your Ultimate Universal IDE with Custom Shortcuts

Introducing CodeMad: Your Ultimate Multi-Language IDE with Custom Shortcuts Welcome to the world of CodeMad, your all-in-one Integrated Development Environment (IDE) that simplifies coding and boosts productivity. Developed in Python, CodeMad is designed to make your coding experience smoother and more efficient across a variety of programming languages, including C, C++, Java, Python, and HTML. Whether you're a beginner or an experienced programmer, CodeMad is your go-to tool. In this blog, we'll dive deep into the workings of CodeMad, highlighting its unique features and easy installation process. The Power of Shortcuts CodeMad's intuitive interface is built around a set of powerful keyboard shortcuts that make coding a breeze. Here are some of the key shortcuts you'll find in CodeMad: Copy (Ctrl+C) : Duplicate text with ease. Paste (Ctrl+V) : Quickly insert copied content into your code. Undo (Ctrl+Z) and Redo (Ctrl+Y) : Correct mistakes and s...

Product of Array Except Self in Python | Prefix & Suffix Explained (LeetCode 238)

Problem Overview The Product of Array Except Self is a classic problem that tests your understanding of array traversal and optimization. The task is simple to state but tricky to implement efficiently. Given an integer array nums , you need to return an array such that each element at index i is equal to the product of all the elements in nums except nums[i] . The challenge is that: Division is not allowed The solution must run in O(n) time Initial Thoughts At first glance, it feels natural to compute the total product of the array and divide it by the current element. However, this approach fails because division is forbidden and handling zeroes becomes messy. This pushed me to think differently — instead of excluding the current element, why not multiply everything around it? That’s where the prefix and suffix product pattern comes in. Key Insight: Prefix & Suffix Products For every index i : Prefix product → product of all elements to t...