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.and2., untiltargetis 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 \(f(i, j)\):
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: 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) / 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);
}