Search invariants and interval conventions

Understand the idea

Choose an interval convention before writing code. For lower bound with [lo,hi), all indices below lo have values<target; all at or above hi have values≥target. The answer is a boundary, possibly n.

Work through an example

For [1,3,3,7], target 3, start lo=0,hi=4. A comparison at mid=2 sets hi=2 because mid may itself be the first valid boundary.

Time and space complexity

O(log(n+1)) comparisons/O(1) state on sorted random-access input.

Why it works

The desired first-not-smaller position lies in the closed boundary range [lo,hi].

Common mistakes

• hi=mid-1 is incompatible with this convention.
• n is a valid insertion boundary, not a valid element index.

Chapter lab · Python implementation

Binary search discards half a monotone candidate domain. The lab uses half-open [lo,hi) lower-bound search. Write the invariant and update rules together; mixing inclusive and half-open conventions creates off-by-one bugs. A logarithmic search does not make list insertion logarithmic.

This shared lab connects the chapter's topics. Follow the current lesson's worked example first, then trace the complete implementation. Additional algorithms described in topic exercises are variations you can implement independently.

def lower_bound(values,target):
    lo,hi=0,len(values)
    while lo<hi:
        mid=lo+(hi-lo)//2
        if values[mid]<target:
            lo=mid+1
        else:
            hi=mid
    return lo

def exact(values,target):
    i=lower_bound(values,target)
    return i if i<len(values) and values[i]==target else -1

def minimum_capacity(weights,days):
    if days<1 or not weights or any(w<1 for w in weights):
        raise ValueError('positive weights and days required')
    def feasible(capacity):
        used,load=1,0
        for w in weights:
            if load+w>capacity:
                used+=1; load=0
            load+=w
        return used<=days
    lo,hi=max(weights),sum(weights)
    while lo<hi:
        mid=(lo+hi)//2
        if feasible(mid): hi=mid
        else: lo=mid+1
    return lo

Run and test the lab

Copy the implementation and these assertions into a Python 3 file and run it. An assertion failure means the actual result differs from the expectation. Add the current topic's exercise and boundary tests to deepen your understanding.

assert lower_bound([1,3,3,7],3)==1
assert lower_bound([],5)==0
assert exact([1,3,3,7],4)==-1
assert exact([1,3,3,7],7)==3
assert minimum_capacity([3,2,2,4,1,4],3)==6