# Lower Bound

# Lower Bound

Lower bound finds the **first index** at which a value $x$ can be inserted in a sorted array while preserving order. It returns the smallest index $i$ such that

$$
A[i] \ge x
$$

If all elements are smaller than $x$, it returns $n$.

This operation forms the basis of many ordered data structure queries, including range queries and duplicate handling.

## Problem

Given a sorted array $A$ of length $n$ and a value $x$, find the smallest index $i$ such that

$$
A[i] \ge x
$$

Return $n$ if no such index exists.

## Key Idea

The predicate

$$
A[i] \ge x
$$

is **monotone**:

* false for small indices
* true for large indices

Binary search can locate the first true position.

## Algorithm

Maintain a half-open interval $[l, r)$.

```id="l0w2bd"
lower_bound(A, x):
    l = 0
    r = length(A)

    while l < r:
        m = l + (r - l) // 2

        if A[m] < x:
            l = m + 1
        else:
            r = m

    return l
```

## Example

Let

$$
A = [1, 3, 3, 5, 7]
$$

and

$$
x = 3
$$

Steps:

| step | l | r | m | A[m] | action     |
| ---- | - | - | - | ---- | ---------- |
| 1    | 0 | 5 | 2 | 3    | move left  |
| 2    | 0 | 2 | 1 | 3    | move left  |
| 3    | 0 | 1 | 0 | 1    | move right |

Return index $1$.

## Interpretation

* All indices $< i$ satisfy $A[j] < x$
* All indices $\ge i$ satisfy $A[j] \ge x$

This partitions the array.

## Correctness

Invariant:

* The answer always lies in $[l, r)$

Each step reduces the interval while preserving:

* left side contains only false values
* right side contains only possible true values

When $l = r$, the first true index has been isolated.

## Complexity

| case      | time        |
| --------- | ----------- |
| all cases | $O(\log n)$ |

Space:

$$
O(1)
$$

## Use Cases

* Insert position in sorted array
* Count elements less than $x$
* Range queries: combine with upper bound
* Deduplication boundaries

## Implementation

```id="q8b2w1"
def lower_bound(a, x):
    l, r = 0, len(a)
    while l < r:
        m = l + (r - l) // 2
        if a[m] < x:
            l = m + 1
        else:
            r = m
    return l
```

```id="d7m3k9"
func LowerBound(a []int, x int) int {
    l, r := 0, len(a)
    for l < r {
        m := l + (r-l)/2
        if a[m] < x {
            l = m + 1
        } else {
            r = m
        }
    }
    return l
}
```