Sum of Square Numbers

Sum of Square Numbers - Problem

Given a non-negative integer c, determine whether there exist two integers a and b such that a² + b² = c.

Return true if such integers exist, otherwise return false.

Note: Both a and b can be zero or negative, but for this problem we only consider non-negative values where a ≥ 0 and b ≥ 0.

Input & Output

Example 1 — Basic True Case

$ Input: c = 5

› Output: true

💡 Note: 1² + 2² = 1 + 4 = 5, so there exist integers a=1 and b=2 such that a² + b² = c

Example 2 — Perfect Square

$ Input: c = 4

› Output: true

💡 Note: 0² + 2² = 0 + 4 = 4 (or 2² + 0² = 4), so it can be expressed as sum of two squares

Example 3 — Impossible Case

$ Input: c = 3

› Output: false

💡 Note: No combination of two squares equals 3: 0²+0²=0, 0²+1²=1, 1²+1²=2, 0²+2²=4. None equal 3.

Constraints

0 ≤ c ≤ 2³¹ - 1

Visualization

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Asked in

F Facebook 25 G Google 20 M Microsoft 15

The key insight is that we only need to check values up to √c since both squares can't exceed that. Best approach is two pointers starting from 0 and √c, adjusting based on sum comparison. Time: O(√c), Space: O(1)

Common Approaches

✓ Binary Search Optimization

⏱️ Time: O(√c log √c) Space: O(1)

For each possible value of a from 0 to √c, calculate b² = c - a² and use binary search to check if there exists an integer b such that b² equals b². This approach optimizes the perfect square checking part using binary search.

Brute Force

⏱️ Time: O(√c) Space: O(1)

For each possible value of a from 0 to √c, check if c - a² is a perfect square. We iterate through all possible values of a and for each a, we calculate b² = c - a² and check if b is an integer.

Two Pointers

⏱️ Time: O(√c) Space: O(1)

Start with left pointer at 0 and right pointer at √c. Calculate left² + right². If sum equals c, return true. If sum is less than c, move left pointer right. If sum is greater than c, move right pointer left. Continue until pointers meet.

Binary Search Optimization — Algorithm Steps

Iterate a from 0 to √c
Calculate target = c - a²
Binary search for b where b² = target
Return true if exact match found

Visualization

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Step-by-Step Walkthrough

For a=1

Need to check if c-1²=4 is perfect square

Binary Search

Search for b where b²=4 in range [0,2]

Found b=2

2²=4, so we found valid pair (1,2)

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool binarySearchSqrt(int target) {
    if (target < 0) return false;
    if (target == 0) return true;
    
    int left = 0, right = target;
    // Optimize right bound
    if (target > 46340) right = 46340; // sqrt of INT_MAX
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        long long square = (long long) mid * mid;
        if (square == target) {
            return true;
        } else if (square < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return false;
}

bool solution(int c) {
    // Instead of using sqrt(c), iterate while a*a <= c
    for (int a = 0; (long long)a * a <= c; a++) {
        int bSquared = c - a * a;
        if (binarySearchSqrt(bSquared)) {
            return true;
        }
    }
    return false;
}

int main() {
    int c;
    scanf("%d", &c);
    bool result = solution(c);
    printf("%s\n", result ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(√c log √c)

√c iterations, each with binary search taking log √c time

✓ Linear Growth

Space Complexity

O(1)

Only using variables for binary search bounds and calculations

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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