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.

Calculate the midpoint \(m\) of the search interval \([i, j]\), and use it to eliminate half of the search interval.
Recursively solve the subproblem reduced by half in size, which could be \(f(i, m-1)\) or \(f(m+1, j)\).
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.

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 \(f(i, j)\):

PythonC++JavaC#GoSwiftJSTSDartRustCKotlinRubyZig

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)

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);
}

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);
}

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);
}

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)
}

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)
}

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);
}

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);
}

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);
}

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)
}

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);
}

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)
}

binary_search_recur.rb

[class]{}-[func]{dfs}

[class]{}-[func]{binary_search}

binary_search_recur.zig

[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}

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12.2 Divide and conquer search strategy¶

1. Implementing binary search based on divide-and-conquer¶

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