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Exact search

Topic 2 of 6 · Binary search

What is binary search?

Binary search narrows a search region by repeatedly checking its middle and discarding a half that cannot contain the answer. It needs sorted data, or another rule that splits possibilities into two ordered groups. Each check must justify which half is safe to discard.

Example: To find 8 in [2, 4, 8, 12, 16], check the middle value 8. If the middle were smaller than the target, continue only in the right half.

INTERACTIVE WALKTHROUGHPredict · step · explain
SHARED CHAPTER LAB · ARRAY

Search lower bound of 3.

01
13
23
37
1 / 4

Step 1 of 4 · lo=0; hi=4; mid=2

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Helpful before this chapter

If these ideas are new, begin with:

Understand the idea

Exact search can use lower bound then verify equality. Finding an insertion position alone does not prove the target exists. Check the boundary against n before indexing.

Work through an example

For target 4 in [1,3,3,7], lower bound returns 3, but value 7 differs, so exact returns -1. For target 8 it returns n and must not index.

Time and space complexity

O(log n) time/O(1) state; duplicates return the first occurrence here.

Why it works

The first possible equal element is at the lower-bound boundary.

Common mistakes

  • A returned boundary may equal n.
  • The chosen duplicate occurrence should be documented.

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
CHECK YOUR UNDERSTANDING

What must follow lower_bound for membership?

Further reading

Original documentation for the implementation details used in this chapter.

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