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Leetcode 1922: Count Good Numbers | Java Solution
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In this video, we solve Leetcode 1922: Count Good Numbers using an optimized Java solution that leverages the modular exponentiation technique (also known as fast power or binary exponentiation).
🔍 What you'll learn:
Understanding how to count good numbers based on digit positions.
How to handle large exponents efficiently using bit manipulation.
Why we multiply even and odd positions differently (5 and 4 choices).
Complete step-by-step breakdown of the Java code.
Optimization using base = 20 and mod = 10^9 + 7.
🧠 Concepts Covered:
Modular arithmetic
Exponentiation by squaring
Bitwise operations in Java
📘 Problem Statement Summary:
Given a number n, count how many good numbers of length n exist where:
Even indices must be even digits (0, 2, 4, 6, 8) → 5 choices
Odd indices must be prime digits (2, 3, 5, 7) → 4 choices
Return the total count modulo 10^9 + 7.
💻 Code Language: Java
📊 Difficulty Level: Medium
🔍 What you'll learn:
Understanding how to count good numbers based on digit positions.
How to handle large exponents efficiently using bit manipulation.
Why we multiply even and odd positions differently (5 and 4 choices).
Complete step-by-step breakdown of the Java code.
Optimization using base = 20 and mod = 10^9 + 7.
🧠 Concepts Covered:
Modular arithmetic
Exponentiation by squaring
Bitwise operations in Java
📘 Problem Statement Summary:
Given a number n, count how many good numbers of length n exist where:
Even indices must be even digits (0, 2, 4, 6, 8) → 5 choices
Odd indices must be prime digits (2, 3, 5, 7) → 4 choices
Return the total count modulo 10^9 + 7.
💻 Code Language: Java
📊 Difficulty Level: Medium
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