LeetCode - Algorithms - 633. Sum of Square Numbers

Problem

633. Sum of Square Numbers

Java

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class Solution {
    public boolean judgeSquareSum(int c) {
        if (c == 0 || c == 1)
            return true;
        int root = mySqrt(c);
        for (int i = root; i >= 0; i--) {
            int re = c - i * i;
            if (isPerfectSquare(re)) {
                return true;
            }
        }
        return false;
    }

    private int mySqrt(int x) {
        if (x == 0 || x == 1)
            return x;

        long start = 1, end = x, ans = 0;
        while (start <= end) {
            long mid = (start + end) / 2;

            if (mid * mid == x)
                return new Long(mid).intValue();

            if (mid * mid < x) {
                start = mid + 1;
                ans = mid;
            } else
                end = mid - 1;
        }
        return new Long(ans).intValue();
    }

    private boolean isPerfectSquare(int num) {
        long lo = 1, hi = num;
        while (lo <= hi) {
            long mid = (lo + hi) / 2;
            if (mid * mid == num)
                return true;
            if (mid * mid > num) {
                hi = mid - 1;
            } else
                lo = mid + 1;
        }
        return false;
    }
}

Submission Detail

  • 124 / 124 test cases passed.
  • Runtime: 242 ms, faster than 5.54% of Java online submissions for Sum of Square Numbers.
  • Memory Usage: 37.6 MB, less than 14.48% of Java online submissions for Sum of Square Numbers.

2

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class Solution {
    public boolean judgeSquareSum(int c) {
        if (c == 0 || c == 1)
            return true;
        for (int i = 0; i <= c/2; i++) {
            int re = c - i * i;
            if (re < 0)
                break;
            if (isPerfectSquare(re)) {
                return true;
            }
        }
        return false;
    }

    private boolean isPerfectSquare(int num) {
        long lo = 1, hi = num;
        while (lo <= hi) {
            long mid = (lo + hi) / 2;
            if (mid * mid == num)
                return true;
            if (mid * mid > num) {
                hi = mid - 1;
            } else
                lo = mid + 1;
        }
        return false;
    }
}

Submission Detail

  • 124 / 124 test cases passed.
  • Runtime: 206 ms, faster than 7.18% of Java online submissions for Sum of Square Numbers.
  • Memory Usage: 35.5 MB, less than 93.83% of Java online submissions for Sum of Square Numbers.

number theory

An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no term \( p^k \), where prime \( p \equiv 3{\pmod {4}} \) and k is odd.

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class Solution {
    public boolean judgeSquareSum(int c) {
        if (c == 0 || c == 1)
            return true;

        while (c % 2 == 0) {
            c = c >> 1;
        }

        for (int p = 3; p * p <= c; p += 2) {
            int k = 0;
            if (c % p == 0) {
                while (c % p == 0) {
                    k++;
                    c /= p;
                }
                if (p % 4 == 3 && (k & 1) == 1)
                    return false;
            }
        }
        return c % 4 != 3;
    }
}

Submission Detail

  • 124 / 124 test cases passed.
  • Runtime: 0 ms, faster than 100.00% of Java online submissions for Sum of Square Numbers.
  • Memory Usage: 35.7 MB, less than 74.91% of Java online submissions for Sum of Square Numbers.

References