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Question

Formatted question description: https://leetcode.ca/all/1.html


Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target,
where index1 must be less than index2. Please note that your returned answers (both index1 and
index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9 Output: index1=1, index2=2

Algorithm

In general, increasing time complexity comes at the cost of using more space, creating a trade-off. However, in this particular problem, I aim to achieve linear time complexity by only traversing the array once and using a HashMap to establish a mapping between each number and its coordinate position.

Since the search efficiency of a HashMap is constant, while traversing the array, we can subtract the current number from the target to find the other number needed. We can then look up this other number directly in the HashMap. Note that the number found may not be the immediate preceding number. For example, if the target is 4 and we have traversed a 2, the other number needed may not be the previous number, which is also 2.

The implementation steps for this approach are as follows: first traverse the array to create the HashMap mapping, and then traverse it again to search for the target numbers and record their indices.

Code

  • import java.util.HashMap;
    
    class Two_Sum {
    
        public static void main(String[] args) {
    
            Two_Sum out = new Two_Sum();
            Solution s = out.new Solution();
    
            int[] a = {2, 7, 11, 15};
            int target = 9;
    
            int[] result = s.twoSum(a, target);
    
            for (int each : result) {
                System.out.println(each);
            }
        }
    
        // time: O(N)
        // space: O(N)
        public class Solution {
    
            public int[] twoSum(int[] nums, int target) {
    
                // set up hashmap: remaining to original. thread-unsafe but for here is ok
                HashMap<Integer, Integer> hm = new HashMap<>();
    
                for (int i = 0; i < nums.length; i++) {
    
                    // if key in hm, result found.
                    // so that only one pass
                    if (hm.containsKey(nums[i])) {
                        int otherIndex = hm.get(nums[i]);
    
                        return new int[]{Math.min(i, otherIndex) + 1, Math.max(i, otherIndex) + 1};
                    }
                    // put new record into hashmap
                    else {
                        hm.put(target - nums[i], i);
                    }
                }
    
                return null;
            }
    
        }
    }
    
    ############
    
    class Solution {
        public int[] twoSum(int[] nums, int target) {
            Map<Integer, Integer> m = new HashMap<>();
            for (int i = 0;; ++i) {
                int x = nums[i];
                int y = target - x;
                if (m.containsKey(y)) {
                    return new int[] {m.get(y), i};
                }
                m.put(x, i);
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/two-sum/
    // Time: O(NlogN)
    // Space: O(N)
    class Solution {
    public:
        vector<int> twoSum(vector<int>& A, int target) {
            vector<vector<int>> v;
            int N = A.size(), L = 0, R = N - 1;
            for (int i = 0; i < N; ++i) v.push_back({ A[i], i });
            sort(begin(v), end(v));
            while (L < R) {
                int sum = v[L][0] + v[R][0];
                if (sum == target) return { v[L][1], v[R][1] };
                if (sum < target) ++L;
                else --R;
            }
            return {};
        }
    };
    
  • class Solution:
        def twoSum(self, nums: List[int], target: int) -> List[int]:
            m = {}
            for i, v in enumerate(nums):
                x = target - v
                if x in m:
                    return [m[x], i]
                m[v] = i
    
    ############
    
    class Solution(object):
      def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        d = {}
        for i, num in enumerate(nums):
          if target - num in d:
            return [d[target - num], i]
          d[num] = i
        # no special case handling because it's assumed that it has only one solution
    
    
  • func twoSum(nums []int, target int) []int {
    	m := map[int]int{}
    	for i := 0; ; i++ {
    		x := nums[i]
    		y := target - x
    		if j, ok := m[y]; ok {
    			return []int{j, i}
    		}
    		m[x] = i
    	}
    }
    
  • /**
     * @param {number[]} nums
     * @param {number} target
     * @return {number[]}
     */
    var twoSum = function (nums, target) {
        const m = new Map();
        for (let i = 0; ; ++i) {
            const x = nums[i];
            const y = target - x;
            if (m.has(y)) {
                return [m.get(y), i];
            }
            m.set(x, i);
        }
    };
    
    
  • class Solution {
    
        /**
         * @param Integer[] $nums
         * @param Integer $target
         * @return Integer[]
         */
        function twoSum($nums, $target) {
            $len = count($nums);
            for ($i = 0; $i < $len; $i++) {
                for ($j = 1 + $i; $j < $len; $j++) {
                    if ($nums[$i] + $nums[$j] == $target) {
                        return [$i, $j];
                    }
                }
            }
        }
    }
    
  • # @param {Integer[]} nums
    # @param {Integer} target
    # @return {Integer[]}
    def two_sum(nums, target)
      nums.each_with_index do |x, idx|
        if nums.include? target - x
          return [idx, nums.index(target - x)] if nums.index(target - x) != idx
        end
        next
      end
    end
    
    
  • import scala.collection.mutable
    
    object Solution {
      def twoSum(nums: Array[Int], target: Int): Array[Int] = {
        var map = new mutable.HashMap[Int, Int]()
        for (i <- 0 to nums.length) {
          if (map.contains(target - nums(i))) {
            return Array(map(target - nums(i)), i)
          } else {
            map += (nums(i) -> i)
          }
        }
        Array(0, 0)
      }
    }
    
    
  • public class Solution {
        public int[] TwoSum(int[] nums, int target) {
            var m = new Dictionary<int, int>();
            for (int i = 0, j; ; ++i) {
                int x = nums[i];
                int y = target - x;
                if (m.TryGetValue(y, out j)) {
                    return new [] {j, i};
                }
                if (!m.ContainsKey(x)) {
                    m.Add(x, i);
                }
            }
        }
    }
    
  • #[
        Author: @joe733
    ]#
    
    import std/enumerate
    
    proc twoSum(nums: seq[int], target: int): seq[int] =
        var
            bal: int
            tdx: int
        for idx, val in enumerate(nums):
            bal = target - val
            if bal in nums:
                tdx = nums.find(bal)
                if idx != tdx:
                    return @[idx, tdx]
    
    
  • use std::collections::HashMap;
    
    pub fn soluation(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let mut map = HashMap::new();
        for (i, item) in nums.iter().enumerate() {
            if map.contains_key(item) {
                return vec![i as i32, map[item]];
            } else {
                let x = target - nums[i];
                map.insert(x, i as i32);
            }
        }
        unreachable!()
    }
    
    
  • class Solution {
        func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
            var m = [Int: Int]()
            var i = 0
            while true {
                let x = nums[i]
                let y = target - nums[i]
                if let j = m[target - nums[i]] {
                    return [j, i]
                }
                m[nums[i]] = i
                i += 1
            }
        }
    }
    

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