How to Determine Sum of Two Square Numbers?

  • 时间：2020-10-05 13:15:44
  • 分类：网络文摘
  • 阅读：97 次

Given a non-negative integer c, your task is to decide whether there’re two integers a and b such that a^2 + b^2 = c.

Example 1:
Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5

Example 2:
Input: 3
Output: False

This is a typical/classic math problem, and the computer is very good at solving it by bruteforce algorithms.

Bruteforce Algorithm to Check Sum of Two Square Numbers

The following C++ implements the bruteforce with no optimisation, which runs at O(C) complexity.

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class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a <= c; ++ a) {
            for (unsigned int b = 0; b * b <= c; ++ b) {
                if (a * a + b * b == c)
                    return true;
            }
        }
        return false;
    }    
};
class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a <= c; ++ a) {
            for (unsigned int b = 0; b * b <= c; ++ b) {
                if (a * a + b * b == c)
                    return true;
            }
        }
        return false;
    }    
};

Optimised Bruteforce Algorithm

We know that n-th square number is the sum of the n-th odd number. For example, 9 = 1 + 3 + 5, and 16 = 1 + 3 + 5 + 7 … Therefore, we can reduce the runtime for the inner loop.

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class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a <= c; ++ a) {
            int b =  c - a * a;
            if (isSquare(b)) {
                return true;
            }
        }
        return false;
    }    
 
private:
    inline bool isSquare(unsigned int b) {
        unsigned int i = 1, sum = 0;
        while (sum < b) {
            sum += i;
            i += 2;
        }
        return sum == b;
    }
};
class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a <= c; ++ a) {
            int b =  c - a * a;
            if (isSquare(b)) {
                return true;
            }
        }
        return false;
    }    

private:
    inline bool isSquare(unsigned int b) {
        unsigned int i = 1, sum = 0;
        while (sum < b) {
            sum += i;
            i += 2;
        }
        return sum == b;
    }
};

This optimisation is great, however, the approach still runs at O(C) complexity.

Using SQRT

In fact, we can compute the b value once a is determined, we just need to check if b is an integer, which can be done using the sqrt function, that runs at O(logC). 

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class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a <= c; ++ a) {
            int b = (int)sqrt(c - a * a);
            if ((unsigned long)b * b + (unsigned long)a * a == c) {
                return true;
            }
        }
        return false;
    }    
};
class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a <= c; ++ a) {
            int b = (int)sqrt(c - a * a);
            if ((unsigned long)b * b + (unsigned long)a * a == c) {
                return true;
            }
        }
        return false;
    }    
};

The overall complexity is O(sqrt(c)log(c)).

Using Binary Search to Determine if An integer is sum of two square integers

Last but not least, we can use the binary search to determine if an integer is a square – that runs at O(logN) complexity.

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class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a < = c; ++ a) {
            unsigned int b = c - a * a;
            if (binarySearch(0, b, b)) {
                return true;
            }
        }
        return false;
    }    
    
    bool binarySearch(unsigned int left, unsigned int right, unsigned int n) {
        if (left > right) return false;
        int64_t mid = left + (right - left) / 2;
        if (mid * mid == n) return true;
        if (mid * mid > n) {
            return binarySearch(left, mid - 1, n);
        }
        return binarySearch(mid + 1, right, n);
    }
};
class Solution {
public:
    bool judgeSquareSum(int c) {
        for (unsigned int a = 0; a * a < = c; ++ a) {
            unsigned int b = c - a * a;
            if (binarySearch(0, b, b)) {
                return true;
            }
        }
        return false;
    }    
    
    bool binarySearch(unsigned int left, unsigned int right, unsigned int n) {
        if (left > right) return false;
        int64_t mid = left + (right - left) / 2;
        if (mid * mid == n) return true;
        if (mid * mid > n) {
            return binarySearch(left, mid - 1, n);
        }
        return binarySearch(mid + 1, right, n);
    }
};

This approach also runs at O(log(C)sqrt(C)) complexity.

--EOF (The Ultimate Computing & Technology Blog) --

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