Algorithm - 이분탐색

1. while (left <= right)

2. 조건 따져가며 left=target+1 or right=target-1 으로 업데이트

3. return left

def binary(num):
    left = 0
    right = len(LIS)-1
    mid = (left + right) // 2
    while(left <= right):
        if LIS[mid] == num:
            return mid
        elif LIS[mid] > num:
            right = mid - 1
        elif LIS[mid] < num:
            left = mid + 1
    return left

1. while (left + 1 < right)

2. 조건 따져가며 left=target or right=target 으로 업데이트

def solution(n, times):
    answer = 10e14
    times.sort()
    left,right=-1,times[-1]*n
    
    while(left+1<right):
        
        total=0
        target=(left+right)//2
        
        for time in times:
            total+=(target//time)         
        if total>=n:
            answer=min(answer,target)
            right=target
        else:   
            left=target
    return answer

def can_move(stones,target,k):
    stack=0
    for stone in stones:
        if stone<target: 
            stack+=1
            if stack==k: return False
        else:
            stack=0
    return True


def solution(stones, k):
    answer = 0
    max,min=0,200000
    for stone in stones:
        if stone>max: max=stone
        if stone<min: min=stone
    while(min<=max):
        target=(min+max)//2
        if can_move(stones,target,k):
            if target>answer: answer=target
            min=target+1
        else:
            max=target-1
    return answer