Quadrant Finder - Problem

Given two coordinates x and y, determine which quadrant the point lies in, or if it's on an axis.

Quadrant Rules:

Quadrant I: x > 0 and y > 0
Quadrant II: x < 0 and y > 0
Quadrant III: x < 0 and y < 0
Quadrant IV: x > 0 and y < 0
On Axis: x = 0 or y = 0

Return the quadrant number (1, 2, 3, or 4) or the string "On Axis" if the point lies on an axis.

Input & Output

Example 1 — Quadrant IV

$ Input: x = 3, y = -2

› Output: 4

💡 Note: Point (3, -2) has x > 0 and y < 0, which places it in Quadrant IV

Example 2 — Quadrant II

$ Input: x = -5, y = 7

› Output: 2

💡 Note: Point (-5, 7) has x < 0 and y > 0, which places it in Quadrant II

Example 3 — On Axis

$ Input: x = 0, y = 5

› Output: On Axis

💡 Note: Point (0, 5) lies on the y-axis since x = 0

Constraints

-10⁹ ≤ x ≤ 10⁹
-10⁹ ≤ y ≤ 10⁹

Visualization

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Asked in

M Microsoft 15 G Google 12 a Amazon 8

The key insight is that quadrant determination only requires checking the signs of x and y coordinates. The optimal approach uses nested conditionals: first check if either coordinate is zero (on axis), then branch on x's sign, and finally check y's sign to determine the exact quadrant. Time: O(1), Space: O(1)

Common Approaches

✓ Multiple If-Else Statements

⏱️ Time: O(1) Space: O(1)

Use separate if-else statements to check each quadrant condition one by one. First check if on axis, then check each quadrant individually.

Nested Conditional Logic

⏱️ Time: O(1) Space: O(1)

First check if the point is on an axis, then use nested conditions to efficiently determine the quadrant by checking signs systematically.

Multiple If-Else Statements — Algorithm Steps

Check if x equals 0 or y equals 0 (on axis)
Check if x > 0 and y > 0 (Quadrant I)
Check if x < 0 and y > 0 (Quadrant II)
Check remaining cases for Quadrants III and IV

Visualization

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Step-by-Step Walkthrough

Check Axis

First check if x=0 or y=0

Check Quadrant I

Then check if x>0 and y>0

Continue Until Match

Keep checking until condition matches

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

void solution(int x, int y, char* result) {
    if (x == 0 || y == 0) {
        strcpy(result, "On Axis");
    } else if (x > 0 && y > 0) {
        strcpy(result, "1");
    } else if (x < 0 && y > 0) {
        strcpy(result, "2");
    } else if (x < 0 && y < 0) {
        strcpy(result, "3");
    } else {
        strcpy(result, "4");
    }
}

int main() {
    int x, y;
    char result[20];
    scanf("%d", &x);
    scanf("%d", &y);
    solution(x, y, result);
    printf("%s\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of comparisons regardless of input

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables, no extra data structures

✓ Linear Space

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Medium Frequency

~5 min Avg. Time

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Smart Actions

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Quadrant Finder - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Multiple If-Else Statements — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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