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🧷 문제 링크
https://school.programmers.co.kr/learn/courses/30/lessons/12911
🧭 풀이 시간
20분
👀 체감 난이도
✏️ 문제 설명
n보다 큰 자연수이면서 이진수로 변환했을 때 1의 개수가 동일한 수 중 최솟값 구하기
🔍 풀이 방법
n을 이진 문자열로 바꿔서 1의 개수 셈.
n+1부터 늘려가며 1의 개수 비교.
⏳ 회고
뭔가 더 좋은 방법이 없을지 물어보니 Integer.bitCount 라는 함수를 쓰면 그냥 1의 개수를 구할 수 있다칸다...