leetcode 633. Sum of Square Numbers

Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a2 + b2 = c.

Example 1:

Input: 5Output: TrueExplanation: 1 * 1 + 2 * 2 = 5

Example 2:

Input: 3Output: False

这道题要注意两个测试用例：
2 , return true 。
2147482647，return false。

public boolean judgeSquareSum(int c) {     HashSet<Integer> set = new HashSet<Integer>();             for (int i = 0; i <= Math.sqrt(c); i++) {         set.add(i * i);         if (set.contains(c - i * i)) {             return true;         }     }     return false;}

有大神用了类似于二分查找的思想。

public class Solution {    public boolean judgeSquareSum(int c) {        if (c < 0) {            return false;        }        int left = 0, right = (int)Math.sqrt(c);        while (left <= right) {            int cur = left * left + right * right;            if (cur < c) {                left++;            } else if (cur > c) {                right--;            } else {                return true;            }        }        return false;    }}

这道题有solutions：https://leetcode.com/problems/sum-of-square-numbers/solution/

Approach #3 Using sqrt function [Accepted]

Algorithm

Check if \sqrt{c - a^2}√​c−a​2​​​​​ turns out to be an integer.

Java

public class Solution {    public boolean judgeSquareSum(int c) {        for (long a = 0; a * a <= c; a++) {            double b = Math.sqrt(c - a * a);            if (b == (int) b)                return true;        }        return false;    }}

Complexity Analysis

• Time complexity : O\big(\sqrt{c}log(c)\big)O(√​c​​​log(c)). We iterate over \sqrt{c}√​c​​​ values for choosing $a$. For every $a$ chosen, finding square root of c - a^2c−a​2​​ takes $O(log(c))$ time in the worst case.

• Space complexity : $O(1)$. Constant extra space is used.

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Approach #4 Using Binary Search [Accepted]

Algorithm

Another method to check if c - a^2c−a​2​​ is a perfect square, is by making use of Binary Search. 

Java

public class Solution {    public boolean judgeSquareSum(int c) {        for (long a = 0; a * a <= c; a++) {            int b = c - (int)(a * a);            if (binary_search(0, b, b))                return true;        }        return false;    }    public boolean binary_search(long s, long e, int n) {        if (s > e)            return false;        long mid = s + (e - s) / 2;        if (mid * mid == n)            return true;        if (mid * mid > n)            return binary_search(s, mid - 1, n);        return binary_search(mid + 1, e, n);    }}

Complexity Analysis

• Time complexity : O\big(\sqrt{c}log(c)\big)O(√​c​​​log(c)). Binary search taking $O(log(c))$ in the worst case is done for \sqrt{c}√​c​​​ values of $a$.

• Space complexity : $O(log(c))$. Binary Search will take $O(log(c))$ space.

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Approach #5 Fermat Theorem [Accepted]:

Algorithm

这个方法基于费马的平方和定理:

任何正整数 $n$ 是两个平方数之和的充分必要条件是: $n$ 的任意满足 $(4k+3)$  的素数因子都满足出现了偶数次。

Interested reader can refer to this documentation.

Java

public class Solution {    public boolean judgeSquareSum(int c) {        for (int i = 2; i * i <= c; i++) {            int count = 0;            if (c % i == 0) {                while (c % i == 0) {                    count++;                    c /= i;                }                if (i % 4 == 3 && count % 2 != 0)                    return false;            }        }        return c % 4 != 3;    }}

Complexity Analysis

• Time complexity : O\big(\sqrt{c}log(c)\big)O(√​c​​​log(c)). We find the factors of $c$ and their count using repeated division. We check for the factors in the range [0, \sqrt{c}][0,√​c​​​]. The maximum number of times a factor can occur(repeated division can be done) is $log(n)$(considering 2 as the only factor, c=2^xc=2​x​​. Thus, $x=log(c)$).

• Space complexity : $O(1)$. Constant space is used.