hello-algo/en/docs/chapter_divide_and_conquer/binary_search_recur.md

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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<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(vector<int> &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<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(List<int> 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"

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