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# 12.2 Divide and conquer search strategy
We have learned that search algorithms fall into two main categories.
- **Brute-force search**: It is implemented by traversing the data structure, with a time complexity of $O(n)$.
- **Adaptive search**: It utilizes a unique data organization form or prior information, and its time complexity can reach $O(\log n)$ or even $O(1)$.
In fact, **search algorithms with a time complexity of $O(\log n)$ are usually based on the divide-and-conquer strategy**, such as binary search and trees.
- Each step of binary search divides the problem (searching for a target element in an array) into a smaller problem (searching for the target element in half of the array), continuing until the array is empty or the target element is found.
- Trees represent the divide-and-conquer idea, where in data structures like binary search trees, AVL trees, and heaps, the time complexity of various operations is $O(\log n)$.
The divide-and-conquer strategy of binary search is as follows.
- **The problem can be divided**: Binary search recursively divides the original problem (searching in an array) into subproblems (searching in half of the array), achieved by comparing the middle element with the target element.
- **Subproblems are independent**: In binary search, each round handles one subproblem, unaffected by other subproblems.
- **The solutions of subproblems do not need to be merged**: Binary search aims to find a specific element, so there is no need to merge the solutions of subproblems. When a subproblem is solved, the original problem is also solved.
Divide-and-conquer can enhance search efficiency because brute-force search can only eliminate one option per round, **whereas divide-and-conquer can eliminate half of the options**.
### 1. Implementing binary search based on divide-and-conquer
In previous chapters, binary search was implemented based on iteration. Now, we implement it based on divide-and-conquer (recursion).
!!! question
Given an ordered array `nums` of length $n$, where all elements are unique, please find the element `target`.
From a divide-and-conquer perspective, we denote the subproblem corresponding to the search interval $[i, j]$ as $f(i, j)$.
Starting from the original problem $f(0, n-1)$, perform the binary search through the following steps.
1. Calculate the midpoint $m$ of the search interval $[i, j]$, and use it to eliminate half of the search interval.
2. Recursively solve the subproblem reduced by half in size, which could be $f(i, m-1)$ or $f(m+1, j)$.
3. Repeat steps `1.` and `2.`, until `target` is found or the interval is empty and returns.
The diagram below shows the divide-and-conquer process of binary search for element $6$ in an array.
{ class="animation-figure" }
Figure 12-4 The divide-and-conquer process of binary search
In the implementation code, we declare a recursive function `dfs()` to solve the problem $f(i, j)$:
=== "Python"
```python title="binary_search_recur.py"
def dfs(nums: list[int], target: int, i: int, j: int) -> int:
"""二分查找：问题 f(i, j)"""
# 若区间为空，代表无目标元素，则返回 -1
if i > j:
return -1
# 计算中点索引 m
m = (i + j) // 2
if nums[m] < target:
# 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j)
elif nums[m] > target:
# 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1)
else:
# 找到目标元素，返回其索引
return m
def binary_search(nums: list[int], target: int) -> int:
"""二分查找"""
n = len(nums)
# 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1)
```
=== "C++"
```cpp title="binary_search_recur.cpp"
/* 二分查找：问题 f(i, j) */
int dfs(vector &nums, int target, int i, int j) {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1;
}
// 计算中点索引 m
int m = (i + j) / 2;
if (nums[m] < target) {
// 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
int binarySearch(vector &nums, int target) {
int n = nums.size();
// 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "Java"
```java title="binary_search_recur.java"
/* 二分查找：问题 f(i, j) */
int dfs(int[] nums, int target, int i, int j) {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1;
}
// 计算中点索引 m
int m = (i + j) / 2;
if (nums[m] < target) {
// 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
int binarySearch(int[] nums, int target) {
int n = nums.length;
// 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "C#"
```csharp title="binary_search_recur.cs"
/* 二分查找：问题 f(i, j) */
int DFS(int[] nums, int target, int i, int j) {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1;
}
// 计算中点索引 m
int m = (i + j) / 2;
if (nums[m] < target) {
// 递归子问题 f(m+1, j)
return DFS(nums, target, m + 1, j);
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
return DFS(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
int BinarySearch(int[] nums, int target) {
int n = nums.Length;
// 求解问题 f(0, n-1)
return DFS(nums, target, 0, n - 1);
}
```
=== "Go"
```go title="binary_search_recur.go"
/* 二分查找：问题 f(i, j) */
func dfs(nums []int, target, i, j int) int {
// 如果区间为空，代表没有目标元素，则返回 -1
if i > j {
return -1
}
// 计算索引中点
m := i + ((j - i) >> 1)
//判断中点与目标元素大小
if nums[m] < target {
// 小于则递归右半数组
// 递归子问题 f(m+1, j)
return dfs(nums, target, m+1, j)
} else if nums[m] > target {
// 小于则递归左半数组
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m-1)
} else {
// 找到目标元素，返回其索引
return m
}
}
/* 二分查找 */
func binarySearch(nums []int, target int) int {
n := len(nums)
return dfs(nums, target, 0, n-1)
}
```
=== "Swift"
```swift title="binary_search_recur.swift"
/* 二分查找：问题 f(i, j) */
func dfs(nums: [Int], target: Int, i: Int, j: Int) -> Int {
// 若区间为空，代表无目标元素，则返回 -1
if i > j {
return -1
}
// 计算中点索引 m
let m = (i + j) / 2
if nums[m] < target {
// 递归子问题 f(m+1, j)
return dfs(nums: nums, target: target, i: m + 1, j: j)
} else if nums[m] > target {
// 递归子问题 f(i, m-1)
return dfs(nums: nums, target: target, i: i, j: m - 1)
} else {
// 找到目标元素，返回其索引
return m
}
}
/* 二分查找 */
func binarySearch(nums: [Int], target: Int) -> Int {
// 求解问题 f(0, n-1)
dfs(nums: nums, target: target, i: nums.startIndex, j: nums.endIndex - 1)
}
```
=== "JS"
```javascript title="binary_search_recur.js"
/* 二分查找：问题 f(i, j) */
function dfs(nums, target, i, j) {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1;
}
// 计算中点索引 m
const m = i + ((j - i) >> 1);
if (nums[m] < target) {
// 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
function binarySearch(nums, target) {
const n = nums.length;
// 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "TS"
```typescript title="binary_search_recur.ts"
/* 二分查找：问题 f(i, j) */
function dfs(nums: number[], target: number, i: number, j: number): number {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1;
}
// 计算中点索引 m
const m = i + ((j - i) >> 1);
if (nums[m] < target) {
// 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
function binarySearch(nums: number[], target: number): number {
const n = nums.length;
// 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "Dart"
```dart title="binary_search_recur.dart"
/* 二分查找：问题 f(i, j) */
int dfs(List nums, int target, int i, int j) {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1;
}
// 计算中点索引 m
int m = (i + j) ~/ 2;
if (nums[m] < target) {
// 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
int binarySearch(List nums, int target) {
int n = nums.length;
// 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "Rust"
```rust title="binary_search_recur.rs"
/* 二分查找：问题 f(i, j) */
fn dfs(nums: &[i32], target: i32, i: i32, j: i32) -> i32 {
// 若区间为空，代表无目标元素，则返回 -1
if i > j {
return -1;
}
let m: i32 = (i + j) / 2;
if nums[m as usize] < target {
// 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if nums[m as usize] > target {
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
fn binary_search(nums: &[i32], target: i32) -> i32 {
let n = nums.len() as i32;
// 求解问题 f(0, n-1)
dfs(nums, target, 0, n - 1)
}
```
=== "C"
```c title="binary_search_recur.c"
/* 二分查找：问题 f(i, j) */
int dfs(int nums[], int target, int i, int j) {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1;
}
// 计算中点索引 m
int m = (i + j) / 2;
if (nums[m] < target) {
// 递归子问题 f(m+1, j)
return dfs(nums, target, m + 1, j);
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
return dfs(nums, target, i, m - 1);
} else {
// 找到目标元素，返回其索引
return m;
}
}
/* 二分查找 */
int binarySearch(int nums[], int target, int numsSize) {
int n = numsSize;
// 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1);
}
```
=== "Kotlin"
```kotlin title="binary_search_recur.kt"
/* 二分查找：问题 f(i, j) */
fun dfs(
nums: IntArray,
target: Int,
i: Int,
j: Int
): Int {
// 若区间为空，代表无目标元素，则返回 -1
if (i > j) {
return -1
}
// 计算中点索引 m
val m = (i + j) / 2
return if (nums[m] < target) {
// 递归子问题 f(m+1, j)
dfs(nums, target, m + 1, j)
} else if (nums[m] > target) {
// 递归子问题 f(i, m-1)
dfs(nums, target, i, m - 1)
} else {
// 找到目标元素，返回其索引
m
}
}
/* 二分查找 */
fun binarySearch(nums: IntArray, target: Int): Int {
val n = nums.size
// 求解问题 f(0, n-1)
return dfs(nums, target, 0, n - 1)
}
```
=== "Ruby"
```ruby title="binary_search_recur.rb"
[class]{}-[func]{dfs}
[class]{}-[func]{binary_search}
```
=== "Zig"
```zig title="binary_search_recur.zig"
[class]{}-[func]{dfs}
[class]{}-[func]{binarySearch}
```
??? pythontutor "Code Visualization"