Skip to content

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.

Figure 12-4 shows the divide-and-conquer process of binary search for element \(6\) in an array.

The divide-and-conquer process of binary search

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 \(f(i, j)\):

binary_search_recur.py
def dfs(nums: list[int], target: int, i: int, j: int) -> int:
    """Binary search: problem f(i, j)"""
    # If the interval is empty, indicating no target element, return -1
    if i > j:
        return -1
    # Calculate midpoint index m
    m = (i + j) // 2
    if nums[m] < target:
        # Recursive subproblem f(m+1, j)
        return dfs(nums, target, m + 1, j)
    elif nums[m] > target:
        # Recursive subproblem f(i, m-1)
        return dfs(nums, target, i, m - 1)
    else:
        # Found the target element, thus return its index
        return m

def binary_search(nums: list[int], target: int) -> int:
    """Binary search"""
    n = len(nums)
    # Solve problem f(0, n-1)
    return dfs(nums, target, 0, n - 1)
binary_search_recur.cpp
/* Binary search: problem f(i, j) */
int dfs(vector<int> &nums, int target, int i, int j) {
    // If the interval is empty, indicating no target element, return -1
    if (i > j) {
        return -1;
    }
    // Calculate midpoint index m
    int m = i + (j - i) / 2;
    if (nums[m] < target) {
        // Recursive subproblem f(m+1, j)
        return dfs(nums, target, m + 1, j);
    } else if (nums[m] > target) {
        // Recursive subproblem f(i, m-1)
        return dfs(nums, target, i, m - 1);
    } else {
        // Found the target element, thus return its index
        return m;
    }
}

/* Binary search */
int binarySearch(vector<int> &nums, int target) {
    int n = nums.size();
    // Solve problem f(0, n-1)
    return dfs(nums, target, 0, n - 1);
}
binary_search_recur.java
/* Binary search: problem f(i, j) */
int dfs(int[] nums, int target, int i, int j) {
    // If the interval is empty, indicating no target element, return -1
    if (i > j) {
        return -1;
    }
    // Calculate midpoint index m
    int m = i + (j - i) / 2;
    if (nums[m] < target) {
        // Recursive subproblem f(m+1, j)
        return dfs(nums, target, m + 1, j);
    } else if (nums[m] > target) {
        // Recursive subproblem f(i, m-1)
        return dfs(nums, target, i, m - 1);
    } else {
        // Found the target element, thus return its index
        return m;
    }
}

/* Binary search */
int binarySearch(int[] nums, int target) {
    int n = nums.length;
    // Solve problem f(0, n-1)
    return dfs(nums, target, 0, n - 1);
}
binary_search_recur.cs
[class]{binary_search_recur}-[func]{DFS}

[class]{binary_search_recur}-[func]{BinarySearch}
binary_search_recur.go
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
binary_search_recur.swift
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
binary_search_recur.js
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
binary_search_recur.ts
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
binary_search_recur.dart
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
binary_search_recur.rs
[class]{}-[func]{dfs}

[class]{}-[func]{binary_search}
binary_search_recur.c
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
binary_search_recur.kt
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
binary_search_recur.rb
[class]{}-[func]{dfs}

[class]{}-[func]{binary_search}
binary_search_recur.zig
[class]{}-[func]{dfs}

[class]{}-[func]{binarySearch}
Feel free to drop your insights, questions or suggestions