How to Find the Closest Sum of Three in an Array using Two Point

  • 时间：2020-09-25 11:32:47
  • 分类：网络文摘
  • 阅读：147 次

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

The straightforward solution has to be the bruteforce algorithm, that exhausts every three numbers using O(N^3) loop – which is obviously too slow.

Two Pointer Algorithm in O(nlogN)

We first need to sort the entire array which takes O(nlogN). Once we have determined the first two numbers in O(N^2), we can search the rest in (logN) as the array is sorted. The following algorithm takes O(n*n*logN).

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class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            for (int j = i + 1; j < nums.size() - 1; ++ j) {
                int k = nums.size() - 1;
                while (j < k) {
                    int cur = nums[i] + nums[j] + nums[k];
                    if (cur == target) {
                        return target;
                    } else if (cur < target) {
                        j ++;
                    } else {
                        k --;
                    }
                    if (abs(cur - target) < abs(sum - target)) {
                        sum = cur;
                    }
                }
            }
        }
        return sum;
    }
};
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            for (int j = i + 1; j < nums.size() - 1; ++ j) {
                int k = nums.size() - 1;
                while (j < k) {
                    int cur = nums[i] + nums[j] + nums[k];
                    if (cur == target) {
                        return target;
                    } else if (cur < target) {
                        j ++;
                    } else {
                        k --;
                    }
                    if (abs(cur - target) < abs(sum - target)) {
                        sum = cur;
                    }
                }
            }
        }
        return sum;
    }
};

However, we don’t need to bruteforce the second number. Once the first number is settled, we can using two pointer algorithm to determine the remainding two numbers. If at anytime, we find a sum that is equal to the target, we immediately return the sum otherwise, we need to iteratively store the minimal sum difference.

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class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            int j = i + 1;
            int k = nums.size() - 1;
            while (j < k) {
                int cur = nums[i] + nums[j] + nums[k];
                if (cur == target) {
                    return target;
                } else if (cur < target) {
                    j ++;
                } else {
                    k --;
                }
                if (abs(cur - target) < abs(sum - target)) {
                    sum = cur;
                }
            }
        }
        return sum;
    }
};
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            int j = i + 1;
            int k = nums.size() - 1;
            while (j < k) {
                int cur = nums[i] + nums[j] + nums[k];
                if (cur == target) {
                    return target;
                } else if (cur < target) {
                    j ++;
                } else {
                    k --;
                }
                if (abs(cur - target) < abs(sum - target)) {
                    sum = cur;
                }
            }
        }
        return sum;
    }
};

The optimal algorithm using two pointer algorithm is O(nlogN).

–EOF (The Ultimate Computing & Technology Blog) —

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