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Title: Vinjunho/feature/boj 2075 N번째큰수 by Rekalux · Pull Request #298 · molgorithm/molgorithm2 · GitHub

Open Graph Title: Vinjunho/feature/boj 2075 N번째큰수 by Rekalux · Pull Request #298 · molgorithm/molgorithm2

X Title: Vinjunho/feature/boj 2075 N번째큰수 by Rekalux · Pull Request #298 · molgorithm/molgorithm2

Description: 백준 2075 N번째 큰 수 문제 링크 1. 설계 로직 n^2개의 수 중에서 n번째의 수를 구하는 문제이다. n^2개의 데이터를 모두 관리할 필요가 없다 상위 n개만 관리하여 n번째 수를 뽑아내면 되겠다. 이분탐색을 통해 랭킹을 관리하면 더 효율적. (n^2)log(n) , 1<=n<=1500 2. 코드 import sys, bisect in...

Open Graph Description: 백준 2075 N번째 큰 수 문제 링크 1. 설계 로직 n^2개의 수 중에서 n번째의 수를 구하는 문제이다. n^2개의 데이터를 모두 관리할 필요가 없다 상위 n개만 관리하여 n번째 수를 뽑아내면 되겠다. 이분탐색을 통해 랭킹을 관리하면 더 효율적. (n^2)log(n) , 1<=n<=1500 2. 코드 import sys, bise...

X Description: 백준 2075 N번째 큰 수 문제 링크 1. 설계 로직 n^2개의 수 중에서 n번째의 수를 구하는 문제이다. n^2개의 데이터를 모두 관리할 필요가 없다 상위 n개만 관리하여 n번째 수를 뽑아내면 되겠다. 이분탐색을 통해 랭킹을 관리하면 더 효율적. (n^2)log(n) , 1&lt;=n&lt;=1500 2. 코드 import s...

Opengraph URL: https://github.com/molgorithm/molgorithm2/pull/298

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og:image:alt백준 2075 N번째 큰 수 문제 링크 1. 설계 로직 n^2개의 수 중에서 n번째의 수를 구하는 문제이다. n^2개의 데이터를 모두 관리할 필요가 없다 상위 n개만 관리하여 n번째 수를 뽑아내면 되겠다. 이분탐색을 통해 랭킹을 관리하면 더 효율적. (n^2)log(n) , 1<=n<=1500 2. 코드 import sys, bise...
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