633. Sum of Square Numbers


Problem Description

The problem is about determining if it's possible to find two non-negative integers a and b, such that when you square each of them and add them together, the result is equal to a given non-negative integer c. This can be expressed with the equation a^2 + b^2 = c. You are required to check if there exists at least one pair (a, b) that satisfies this equation for the given c.

Intuition

The intuition behind the solution is based on the properties of a right-angled triangle where the squares of the two shorter sides sum up to the square of the longest side (Pythagorean theorem). In this problem, the non-negative integers a and b are similar to the two shorter sides, and c is like the square of the longest side.

The solution approach is inspired by binary search, which is typically used to find an element in a sorted array. Rather than searching through all possible pairs of a and b, which would be inefficient, the algorithm starts with two potential candidates: a starting from 0 and b from the square root of c, since b can't be larger than sqrt(c) if b^2 is to stay less than or equal to c.

The while loop then examines the sum of squares of a and b. If the sum equals c, the answer is True. If the sum is less than c, a is incremented to increase the sum, exploring the possibility of larger a^2 contributions. If the sum is greater than c, b is decremented to decrease the sum, as b might be too large.

This searching process ends when a becomes greater than b because, at that point, all pairs that could potentially add up to c have already been tested. If no such pair has been found by then, the algorithm returns False, indicating no such pair exists for the given c.

Learn more about Math, Two Pointers and Binary Search patterns.

Solution Approach

The implementation uses a two-pointer technique that starts with the two pointers at the minimal and maximal possible values of a and b that could satisfy the equation a^2 + b^2 = c. Pointer a starts from 0, as it represents the smallest possible square contributing to c. Pointer b starts from int(sqrt(c)), which is the largest integer that when squared does not exceed c.

The algorithm's core logic is a loop that continues to execute as long as a is less than or equal to b. During each iteration, it computes the sum s of the squares of a and b. This sum s = a^2 + b^2 is then compared to the target sum c:

  • If s is equal to c, then we know that such a pair (a, b) exists, and the function immediately returns True.
  • If s is less than c, the sum is too small, and a is incremented by 1 to try a larger value for a^2.
  • If s is greater than c, the sum is too large, and b is decremented by 1 to try a smaller value for b^2.

The loop continues this process, incrementing a or decrementing b as necessary, stopping when a exceeds b. If the loop concludes without finding a matching pair, the function returns False, indicating there are no two perfect square numbers that add up to c.

This algorithm follows an approach similar to a binary search, as the reference approach indicates, where instead of searching in a sorted array, it "searches" through a conceptualized sorted space of squared numbers, moving one of the boundaries (either a or b) closer to a potential solution at each step.

The reason why the algorithm stops once a exceeds b is because we're dealing with non-negative integers, and squares of integers grow very quickly. Once a > b, the possible sums of squares will no longer decrease (since a^2 is now greater than b^2), which means we can be certain no subsequent iterations will result in the sum c.

The efficiency and elegance of this solution lie in its O(sqrt(c)) time complexity, as b is reduced from sqrt(c) to 0 in the worst case, and a is increased from 0 to sqrt(c) in the worst case. This is much more efficient than a brute-force approach that would be O(c), where we would have to check every pair (a, b).

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Example Walkthrough

To illustrate the solution approach, let's work through a small example with c = 5.

Initially, we start with two pointers for a and b such that:

  • a starts from 0, as it represents the smallest possible square contributing to c.
  • b starts from int(sqrt(c)), which is 2 in our example because 2^2 = 4 is the largest square less than or equal to c.

Now, let's begin the process:

  1. In the first iteration, the sum of squares s will be a^2 + b^2 = 0^2 + 2^2 = 0 + 4 = 4. Since s is less than c, we need to increase a. We increment a to 1.

  2. In the second iteration, s will be a^2 + b^2 = 1^2 + 2^2 = 1 + 4 = 5, which is exactly equal to c. At this point, we have found that the pair (a, b) = (1, 2) satisfies the equation a^2 + b^2 = c. The algorithm returns True.

In this example, we found that c can indeed be expressed as the sum of two squared integers. The algorithm works efficiently and quickly identifies the correct pair without needing to check every single possibility.

If we had a larger c where no squares add up to c, the algorithm would continue incrementing a and decrementing b until a surpasses b. If no valid pair is found by the time a exceeds b, the algorithm will return False, indicating no such pair exists for the given c.

Solution Implementation

1from math import sqrt
2
3class Solution:
4    def judgeSquareSum(self, target: int) -> bool:
5        # Initialize two pointers. `left` starts at 0, and `right` starts at the square root of target,
6        # truncated to an integer, which is the largest possible value for a or b if a^2 + b^2 == target.
7        left, right = 0, int(sqrt(target))
8      
9        # Loop until the two pointers meet
10        while left <= right:
11            # Calculate the sum of squares of the two pointers
12            current_sum = left ** 2 + right ** 2
13          
14            # If the sum equals the target, we found a solution
15            if current_sum == target:
16                return True
17            # If the sum is less than the target,
18            # increment the left pointer to try a larger square for `a^2`
19            elif current_sum < target:
20                left += 1
21            # If the sum is greater than the target,
22            # decrement the right pointer to try a smaller square for `b^2`
23            else:
24                right -= 1
25      
26        # If no pair of squares was found that sums up to the target, return False
27        return False
28
29# Example usage:
30# solution = Solution()
31# print(solution.judgeSquareSum(5))  # Output: True
32# print(solution.judgeSquareSum(3))  # Output: False
33
1class Solution {
2    public boolean judgeSquareSum(int c) {
3        // Initialize two pointers. 'smallest' starts at 0 and 'largest' starts at the square root of 'c'
4        long smallest = 0;
5        long largest = (long) Math.sqrt(c);
6
7        // Use a two-pointer approach to find if there exist two numbers 'a' and 'b'
8        // such that a^2 + b^2 equals to 'c'
9        while (smallest <= largest) {
10            // Calculate the sum of squares of 'smallest' and 'largest'
11            long sumOfSquares = smallest * smallest + largest * largest;
12
13            // Check if the current sum of squares equals to 'c'
14            if (sumOfSquares == c) {
15                // If yes, then we found that 'c' can be expressed as a sum of squares.
16                return true;
17            }
18
19            // If the sum of squares is less than 'c', we increment 'smallest' to get a larger sum
20            if (sumOfSquares < c) {
21                ++smallest;
22            } else {
23                // If the sum of squares is greater than 'c', we decrement 'largest' to get a smaller sum
24                --largest;
25            }
26        }
27      
28        // If we exit the loop, then there are no such 'a' and 'b' that satisfy a^2 + b^2 = 'c'
29        return false;
30    }
31}
32
1class Solution {
2public:
3    // This method checks if the input 'c' can be expressed as the sum of squares of two integers.
4    bool judgeSquareSum(int c) {
5        // Start one pointer 'a' from the smallest possible square, 0.
6        long a = 0;
7        // Start another pointer 'b' from the largest possible square that is less than or equal to 'c'.
8        long b = static_cast<long>(sqrt(c));
9      
10        // Continue the search as long as 'a' is less than or equal to 'b'.
11        while (a <= b) {
12            // Calculate the sum of the squares of 'a' and 'b'.
13            long sumOfSquares = a * a + b * b;
14          
15            // If the sum equals to 'c', we found the two numbers whose squares sum up to 'c'.
16            if (sumOfSquares == c) return true;
17          
18            // If the sum is less than 'c', then we need to increase 'a' to get a larger sum.
19            if (sumOfSquares < c)
20                ++a;
21            // If the sum is greater than 'c', we need to decrease 'b' to get a smaller sum.
22            else
23                --b;
24        }
25      
26        // If no pair of integers has been found whose squares sum up to 'c', return false.
27        return false;
28    }
29};
30
1// This function checks if a given number c can be written as the sum
2// of the squares of two integers (c = a^2 + b^2).
3function judgeSquareSum(c: number): boolean {
4    // Starting with a at the smallest square (0) and b at the largest
5    // square not greater than the square root of c.
6    let a = 0;
7    let b = Math.floor(Math.sqrt(c));
8
9    // Loop until a and b meet or cross each other.
10    while (a <= b) {
11        // Calculating the sum of squares of both a and b.
12        let sumOfSquares = a ** 2 + b ** 2;
13
14        // If the sum of squares is equal to c, we found a valid pair.
15        if (sumOfSquares == c) {
16            return true;
17        }
18
19        // If the sum of squares is less than c, we need to try a larger value of a.
20        if (sumOfSquares < c) {
21            ++a; // Increment a to increase the sum.
22        } else {
23            --b; // Decrement b to decrease the sum.
24        }
25    }
26
27    // If we exit the loop, no valid pair was found that fulfills the condition.
28    return false;
29}
30

Time and Space Complexity

The time complexity of the given code is O(sqrt(c)) because the while loop runs with a starting from 0 and b starting from int(sqrt(c)), moving closer to each other with each iteration. The loop terminates when a exceeds b, and since b decrements with a magnitude that is at most the square root of c (as that is its starting value), we can bound the number of iterations by 2*sqrt(c) (since both a and b can move sqrt(c) times at most).

The space complexity of the code is O(1) because the space used does not scale with the value of c. Only a constant amount of additional memory is used for the variables a, b, and s.

Learn more about how to find time and space complexity quickly using problem constraints.


Fast Track Your Learning with Our Quick Skills Quiz:

What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?

1def fun(arr: List[int]) -> List[int]:
2    import heapq
3    heapq.heapify(arr)
4    res = []
5    for i in range(3):
6        res.append(heapq.heappop(arr))
7    return res
8
1public static int[] fun(int[] arr) {
2    int[] res = new int[3];
3    PriorityQueue<Integer> heap = new PriorityQueue<>();
4    for (int i = 0; i < arr.length; i++) {
5        heap.add(arr[i]);
6    }
7    for (int i = 0; i < 3; i++) {
8        res[i] = heap.poll();
9    }
10    return res;
11}
12
1class HeapItem {
2    constructor(item, priority = item) {
3        this.item = item;
4        this.priority = priority;
5    }
6}
7
8class MinHeap {
9    constructor() {
10        this.heap = [];
11    }
12
13    push(node) {
14        // insert the new node at the end of the heap array
15        this.heap.push(node);
16        // find the correct position for the new node
17        this.bubble_up();
18    }
19
20    bubble_up() {
21        let index = this.heap.length - 1;
22
23        while (index > 0) {
24            const element = this.heap[index];
25            const parentIndex = Math.floor((index - 1) / 2);
26            const parent = this.heap[parentIndex];
27
28            if (parent.priority <= element.priority) break;
29            // if the parent is bigger than the child then swap the parent and child
30            this.heap[index] = parent;
31            this.heap[parentIndex] = element;
32            index = parentIndex;
33        }
34    }
35
36    pop() {
37        const min = this.heap[0];
38        this.heap[0] = this.heap[this.size() - 1];
39        this.heap.pop();
40        this.bubble_down();
41        return min;
42    }
43
44    bubble_down() {
45        let index = 0;
46        let min = index;
47        const n = this.heap.length;
48
49        while (index < n) {
50            const left = 2 * index + 1;
51            const right = left + 1;
52
53            if (left < n && this.heap[left].priority < this.heap[min].priority) {
54                min = left;
55            }
56            if (right < n && this.heap[right].priority < this.heap[min].priority) {
57                min = right;
58            }
59            if (min === index) break;
60            [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61            index = min;
62        }
63    }
64
65    peek() {
66        return this.heap[0];
67    }
68
69    size() {
70        return this.heap.length;
71    }
72}
73
74function fun(arr) {
75    const heap = new MinHeap();
76    for (const x of arr) {
77        heap.push(new HeapItem(x));
78    }
79    const res = [];
80    for (let i = 0; i < 3; i++) {
81        res.push(heap.pop().item);
82    }
83    return res;
84}
85

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