Skip to content

10.4   Hash Optimization Strategy

In algorithm problems, we often reduce the time complexity of algorithms by replacing linear search with hash-based search. Let's use an algorithm problem to deepen our understanding.

Question

Given an integer array nums and a target element target, search for two elements in the array whose "sum" equals target, and return their array indices. Any solution will do.

10.4.1   Linear Search: Trading Time for Space

Consider directly traversing all possible combinations. As shown in Figure 10-9, we open a two-layer loop and judge in each round whether the sum of two integers equals target. If so, return their indices.

Linear search solution for two sum

Figure 10-9   Linear search solution for two sum

The code is shown below:

two_sum.py
def two_sum_brute_force(nums: list[int], target: int) -> list[int]:
    """Method 1: Brute force enumeration"""
    # Two nested loops, time complexity is O(n^2)
    for i in range(len(nums) - 1):
        for j in range(i + 1, len(nums)):
            if nums[i] + nums[j] == target:
                return [i, j]
    return []
two_sum.cpp
/* Method 1: Brute force enumeration */
vector<int> twoSumBruteForce(vector<int> &nums, int target) {
    int size = nums.size();
    // Two nested loops, time complexity is O(n^2)
    for (int i = 0; i < size - 1; i++) {
        for (int j = i + 1; j < size; j++) {
            if (nums[i] + nums[j] == target)
                return {i, j};
        }
    }
    return {};
}
two_sum.java
/* Method 1: Brute force enumeration */
int[] twoSumBruteForce(int[] nums, int target) {
    int size = nums.length;
    // Two nested loops, time complexity is O(n^2)
    for (int i = 0; i < size - 1; i++) {
        for (int j = i + 1; j < size; j++) {
            if (nums[i] + nums[j] == target)
                return new int[] { i, j };
        }
    }
    return new int[0];
}
two_sum.cs
/* Method 1: Brute force enumeration */
int[] TwoSumBruteForce(int[] nums, int target) {
    int size = nums.Length;
    // Two nested loops, time complexity is O(n^2)
    for (int i = 0; i < size - 1; i++) {
        for (int j = i + 1; j < size; j++) {
            if (nums[i] + nums[j] == target)
                return [i, j];
        }
    }
    return [];
}
two_sum.go
/* Method 1: Brute force enumeration */
func twoSumBruteForce(nums []int, target int) []int {
    size := len(nums)
    // Two nested loops, time complexity is O(n^2)
    for i := 0; i < size-1; i++ {
        for j := i + 1; j < size; j++ {
            if nums[i]+nums[j] == target {
                return []int{i, j}
            }
        }
    }
    return nil
}
two_sum.swift
/* Method 1: Brute force enumeration */
func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
    // Two nested loops, time complexity is O(n^2)
    for i in nums.indices.dropLast() {
        for j in nums.indices.dropFirst(i + 1) {
            if nums[i] + nums[j] == target {
                return [i, j]
            }
        }
    }
    return [0]
}
two_sum.js
/* Method 1: Brute force enumeration */
function twoSumBruteForce(nums, target) {
    const n = nums.length;
    // Two nested loops, time complexity is O(n^2)
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            if (nums[i] + nums[j] === target) {
                return [i, j];
            }
        }
    }
    return [];
}
two_sum.ts
/* Method 1: Brute force enumeration */
function twoSumBruteForce(nums: number[], target: number): number[] {
    const n = nums.length;
    // Two nested loops, time complexity is O(n^2)
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            if (nums[i] + nums[j] === target) {
                return [i, j];
            }
        }
    }
    return [];
}
two_sum.dart
/* Method 1: Brute force enumeration */
List<int> twoSumBruteForce(List<int> nums, int target) {
  int size = nums.length;
  // Two nested loops, time complexity is O(n^2)
  for (var i = 0; i < size - 1; i++) {
    for (var j = i + 1; j < size; j++) {
      if (nums[i] + nums[j] == target) return [i, j];
    }
  }
  return [0];
}
two_sum.rs
/* Method 1: Brute force enumeration */
pub fn two_sum_brute_force(nums: &Vec<i32>, target: i32) -> Option<Vec<i32>> {
    let size = nums.len();
    // Two nested loops, time complexity is O(n^2)
    for i in 0..size - 1 {
        for j in i + 1..size {
            if nums[i] + nums[j] == target {
                return Some(vec![i as i32, j as i32]);
            }
        }
    }
    None
}
two_sum.c
/* Method 1: Brute force enumeration */
int *twoSumBruteForce(int *nums, int numsSize, int target, int *returnSize) {
    for (int i = 0; i < numsSize; ++i) {
        for (int j = i + 1; j < numsSize; ++j) {
            if (nums[i] + nums[j] == target) {
                int *res = malloc(sizeof(int) * 2);
                res[0] = i, res[1] = j;
                *returnSize = 2;
                return res;
            }
        }
    }
    *returnSize = 0;
    return NULL;
}
two_sum.kt
/* Method 1: Brute force enumeration */
fun twoSumBruteForce(nums: IntArray, target: Int): IntArray {
    val size = nums.size
    // Two nested loops, time complexity is O(n^2)
    for (i in 0..<size - 1) {
        for (j in i + 1..<size) {
            if (nums[i] + nums[j] == target) return intArrayOf(i, j)
        }
    }
    return IntArray(0)
}
two_sum.rb
### Method 1: Brute force enumeration ###
def two_sum_brute_force(nums, target)
  # Two nested loops, time complexity is O(n^2)
  for i in 0...(nums.length - 1)
    for j in (i + 1)...nums.length
      return [i, j] if nums[i] + nums[j] == target
    end
  end

  []
end

This method has a time complexity of \(O(n^2)\) and a space complexity of \(O(1)\), which is very time-consuming with large data volumes.

10.4.2   Hash-Based Search: Trading Space for Time

Consider using a hash table where key-value pairs are array elements and element indices respectively. Loop through the array, performing the steps shown in Figure 10-10 in each round:

  1. Check if the number target - nums[i] is in the hash table. If so, directly return the indices of these two elements.
  2. Add the key-value pair nums[i] and index i to the hash table.

Hash table solution for two sum

two_sum_hashtable_step2

two_sum_hashtable_step3

Figure 10-10   Hash table solution for two sum

The implementation code is shown below, requiring only a single loop:

two_sum.py
def two_sum_hash_table(nums: list[int], target: int) -> list[int]:
    """Method 2: Auxiliary hash table"""
    # Auxiliary hash table, space complexity is O(n)
    dic = {}
    # Single loop, time complexity is O(n)
    for i in range(len(nums)):
        if target - nums[i] in dic:
            return [dic[target - nums[i]], i]
        dic[nums[i]] = i
    return []
two_sum.cpp
/* Method 2: Auxiliary hash table */
vector<int> twoSumHashTable(vector<int> &nums, int target) {
    int size = nums.size();
    // Auxiliary hash table, space complexity is O(n)
    unordered_map<int, int> dic;
    // Single loop, time complexity is O(n)
    for (int i = 0; i < size; i++) {
        if (dic.find(target - nums[i]) != dic.end()) {
            return {dic[target - nums[i]], i};
        }
        dic.emplace(nums[i], i);
    }
    return {};
}
two_sum.java
/* Method 2: Auxiliary hash table */
int[] twoSumHashTable(int[] nums, int target) {
    int size = nums.length;
    // Auxiliary hash table, space complexity is O(n)
    Map<Integer, Integer> dic = new HashMap<>();
    // Single loop, time complexity is O(n)
    for (int i = 0; i < size; i++) {
        if (dic.containsKey(target - nums[i])) {
            return new int[] { dic.get(target - nums[i]), i };
        }
        dic.put(nums[i], i);
    }
    return new int[0];
}
two_sum.cs
/* Method 2: Auxiliary hash table */
int[] TwoSumHashTable(int[] nums, int target) {
    int size = nums.Length;
    // Auxiliary hash table, space complexity is O(n)
    Dictionary<int, int> dic = [];
    // Single loop, time complexity is O(n)
    for (int i = 0; i < size; i++) {
        if (dic.ContainsKey(target - nums[i])) {
            return [dic[target - nums[i]], i];
        }
        dic.Add(nums[i], i);
    }
    return [];
}
two_sum.go
/* Method 2: Auxiliary hash table */
func twoSumHashTable(nums []int, target int) []int {
    // Auxiliary hash table, space complexity is O(n)
    hashTable := map[int]int{}
    // Single loop, time complexity is O(n)
    for idx, val := range nums {
        if preIdx, ok := hashTable[target-val]; ok {
            return []int{preIdx, idx}
        }
        hashTable[val] = idx
    }
    return nil
}
two_sum.swift
/* Method 2: Auxiliary hash table */
func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
    // Auxiliary hash table, space complexity is O(n)
    var dic: [Int: Int] = [:]
    // Single loop, time complexity is O(n)
    for i in nums.indices {
        if let j = dic[target - nums[i]] {
            return [j, i]
        }
        dic[nums[i]] = i
    }
    return [0]
}
two_sum.js
/* Method 2: Auxiliary hash table */
function twoSumHashTable(nums, target) {
    // Auxiliary hash table, space complexity is O(n)
    let m = {};
    // Single loop, time complexity is O(n)
    for (let i = 0; i < nums.length; i++) {
        if (m[target - nums[i]] !== undefined) {
            return [m[target - nums[i]], i];
        } else {
            m[nums[i]] = i;
        }
    }
    return [];
}
two_sum.ts
/* Method 2: Auxiliary hash table */
function twoSumHashTable(nums: number[], target: number): number[] {
    // Auxiliary hash table, space complexity is O(n)
    let m: Map<number, number> = new Map();
    // Single loop, time complexity is O(n)
    for (let i = 0; i < nums.length; i++) {
        let index = m.get(target - nums[i]);
        if (index !== undefined) {
            return [index, i];
        } else {
            m.set(nums[i], i);
        }
    }
    return [];
}
two_sum.dart
/* Method 2: Auxiliary hash table */
List<int> twoSumHashTable(List<int> nums, int target) {
  int size = nums.length;
  // Auxiliary hash table, space complexity is O(n)
  Map<int, int> dic = HashMap();
  // Single loop, time complexity is O(n)
  for (var i = 0; i < size; i++) {
    if (dic.containsKey(target - nums[i])) {
      return [dic[target - nums[i]]!, i];
    }
    dic.putIfAbsent(nums[i], () => i);
  }
  return [0];
}
two_sum.rs
/* Method 2: Auxiliary hash table */
pub fn two_sum_hash_table(nums: &Vec<i32>, target: i32) -> Option<Vec<i32>> {
    // Auxiliary hash table, space complexity is O(n)
    let mut dic = HashMap::new();
    // Single loop, time complexity is O(n)
    for (i, num) in nums.iter().enumerate() {
        match dic.get(&(target - num)) {
            Some(v) => return Some(vec![*v as i32, i as i32]),
            None => dic.insert(num, i as i32),
        };
    }
    None
}
two_sum.c
/* Hash table */
typedef struct {
    int key;
    int val;
    UT_hash_handle hh; // Implemented using uthash.h
} HashTable;

/* Hash table lookup */
HashTable *find(HashTable *h, int key) {
    HashTable *tmp;
    HASH_FIND_INT(h, &key, tmp);
    return tmp;
}

/* Hash table element insertion */
void insert(HashTable **h, int key, int val) {
    HashTable *t = find(*h, key);
    if (t == NULL) {
        HashTable *tmp = malloc(sizeof(HashTable));
        tmp->key = key, tmp->val = val;
        HASH_ADD_INT(*h, key, tmp);
    } else {
        t->val = val;
    }
}

/* Method 2: Auxiliary hash table */
int *twoSumHashTable(int *nums, int numsSize, int target, int *returnSize) {
    HashTable *hashtable = NULL;
    for (int i = 0; i < numsSize; i++) {
        HashTable *t = find(hashtable, target - nums[i]);
        if (t != NULL) {
            int *res = malloc(sizeof(int) * 2);
            res[0] = t->val, res[1] = i;
            *returnSize = 2;
            return res;
        }
        insert(&hashtable, nums[i], i);
    }
    *returnSize = 0;
    return NULL;
}
two_sum.kt
/* Method 2: Auxiliary hash table */
fun twoSumHashTable(nums: IntArray, target: Int): IntArray {
    val size = nums.size
    // Auxiliary hash table, space complexity is O(n)
    val dic = HashMap<Int, Int>()
    // Single loop, time complexity is O(n)
    for (i in 0..<size) {
        if (dic.containsKey(target - nums[i])) {
            return intArrayOf(dic[target - nums[i]]!!, i)
        }
        dic[nums[i]] = i
    }
    return IntArray(0)
}
two_sum.rb
### Method 2: Auxiliary hash table ###
def two_sum_hash_table(nums, target)
  # Auxiliary hash table, space complexity is O(n)
  dic = {}
  # Single loop, time complexity is O(n)
  for i in 0...nums.length
    return [dic[target - nums[i]], i] if dic.has_key?(target - nums[i])

    dic[nums[i]] = i
  end

  []
end

This method reduces the time complexity from \(O(n^2)\) to \(O(n)\) through hash-based search, greatly improving runtime efficiency.

Since an additional hash table needs to be maintained, the space complexity is \(O(n)\). Nevertheless, this method achieves a more balanced overall time-space efficiency, making it the optimal solution for this problem.

Feel free to drop your insights, questions or suggestions