✅ 🎯 📌 Simple Solution || Binary Search || Fastest Solution ✅ 🎯 📌
Code
# Code
```python3 []
class Solution:
    def minCapability(self, nums: List[int], k: int) -> int:
        def foo(cap):
            count=k
            i=0
            while i<len(nums):
                if nums[i]<=cap:
                    count-=1
                    i+=2
                else: i+=1
                if count==0: break
            return count==0
        l,r=min(nums),max(nums)
        while l<r:
            mid=(l+r)//2
            if foo(mid): r=mid
            else: l=mid+1
        return l
```
['Array', 'Binary Search', 'Python3']