CF 102697008 - N-Dimensional Distance

CF 102697008 - N-Dimensional Distance

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Solve time: 49s
Verified: yes

Solution

Problem Understanding

The input describes two points living in a D dimensional space. Instead of having only an x coordinate and a y coordinate, each point has one coordinate for every dimension from 1 to D. The goal is to compute the straight line distance between these two points.

For each dimension, we compare the two coordinates, find their difference, square it, and add it to the total. After all dimensions have contributed, taking the square root of this sum gives the final distance.

The dimension can reach 500, which means the algorithm must be essentially linear. Any approach that tries to generate combinations of dimensions or simulate the space would be unnecessary and would quickly become impossible. A single pass over the coordinates is enough, so the running time grows only with the input size.

The coordinate values can be negative, so treating the difference as an unsigned value or forgetting the square operation can produce incorrect results. For example, consider a two dimensional case where the points are (1, -1) and (-1, 1). The correct distance is:

sqrt((-1 - 1)^2 + (1 - (-1))^2)
= sqrt(4 + 4)
= 2.828427...

A careless implementation that adds raw differences would get 0, because the positive and negative changes cancel each other.

Another edge case is when both points are identical. For input:

2
5
5
5
5

the answer must be:

0.0

An implementation that assumes there is always some movement between points may accidentally produce a nonzero value or fail because it does not handle a zero square root result.

Approaches

The direct brute force idea is to calculate the distance formula exactly as written. For every dimension, we compute the difference between the two coordinates, square it, and add it to a running sum. Since every coordinate must affect the answer, this is already the minimum amount of work needed.

A slower interpretation might repeatedly recalculate the formula or build unnecessary intermediate structures. If the dimension is 500, even a few extra passes are not needed. The optimal solution performs exactly one calculation per dimension.

The observation that makes the problem simple is that the distance formula is independent across dimensions. Each coordinate pair contributes one term to the final sum, so there is no interaction between dimensions. We can accumulate the squared differences as we read the input and only compute the square root once at the end.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(D)	O(D)	Accepted, if implemented directly
Optimal	O(D)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of dimensions. The value tells us how many coordinate pairs will contribute to the distance.
2. Read each coordinate from the first point and the matching coordinate from the second point. For that dimension, calculate the difference, square it, and add it to the accumulated squared distance.

The square removes the sign, so a coordinate that moves in the negative direction contributes the same amount as one moving in the positive direction. 3. After all dimensions have been processed, take the square root of the accumulated value. This is the Euclidean distance between the two points.

Why it works:

The algorithm maintains the exact value of the expression inside the square root. Every dimension contributes exactly one term (q[i] - p[i])^2, and the algorithm adds every such term once. Since the final distance formula is the square root of this complete sum, the produced answer is exactly the required distance.

Python Solution

import sys
import math

input = sys.stdin.readline

def solve():
    d = int(input())
    
    first = [int(input()) for _ in range(d)]
    
    total = 0
    for i in range(d):
        second = int(input())
        diff = second - first[i]
        total += diff * diff
    
    print(math.sqrt(total))

if __name__ == "__main__":
    solve()

The first list stores the coordinates of the first point because the input gives all of those coordinates before the second point begins. The algorithm then reads the second point one coordinate at a time and immediately combines it with the matching coordinate from the stored point.

The variable total stores the squared distance, not the final distance. Keeping the value as an integer avoids unnecessary floating point operations while the summation is happening. The square root is applied only once after the complete sum is known.

Python integers do not overflow, so the maximum possible squared sum is safe. The only floating point operation is the final square root, which matches the required output format.

Worked Examples

Consider the input:

2
1
1
1
2

The first point is (1, 1) and the second point is (1, 2).

Dimension	First coordinate	Second coordinate	Difference squared	Current sum
1	1	1	0	0
2	1	2	1	1

The final distance is sqrt(1) = 1.0. This trace shows that dimensions contribute independently.

For a three dimensional example:

3
7
3
6
4
17
2

The points are (7, 3, 6) and (4, 17, 2).

Dimension	First coordinate	Second coordinate	Difference squared	Current sum
1	7	4	9	9
2	3	17	196	205
3	6	2	16	221

The answer is sqrt(221) = 14.866068747318506. This confirms that negative or positive movement does not matter after squaring.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(D)	Every dimension is processed exactly once
Space	O(D)	The first point's coordinates are stored

The dimension limit is small enough that storing one point is safe. The algorithm does not create any structure related to the number of possible positions in the space, so it easily fits within the limits.

Test Cases

import sys
import io
import math

def solution(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)

    d = int(sys.stdin.readline())
    a = [int(sys.stdin.readline()) for _ in range(d)]
    s = 0
    for i in range(d):
        b = int(sys.stdin.readline())
        s += (b - a[i]) ** 2

    ans = str(math.sqrt(s))

    sys.stdin = old_stdin
    return ans

assert solution("""2
1
1
1
2
""") == "1.0", "sample 1"

assert solution("""3
7
3
6
4
17
2
""") == str(math.sqrt(221)), "sample 2"

assert solution("""2
5
5
5
5
""") == "0.0", "identical points"

assert solution("""2
1
-1
-1
1
""") == str(math.sqrt(8)), "negative coordinates"

assert solution("""5
0
0
0
0
0
1
1
1
1
1
""") == str(math.sqrt(5)), "many dimensions"

Test input	Expected output	What it validates
Two points differing in one coordinate	1.0	Basic distance calculation
Three dimensional points	sqrt(221)	Generalization beyond 2D
Identical points	0.0	Zero distance handling
Mixed positive and negative values	sqrt(8)	Correct use of squaring
Five dimensions	sqrt(5)	Processing arbitrary dimensions

Edge Cases

When coordinates have opposite signs, the subtraction must happen before squaring. For input:

2
1
-1
-1
1

the algorithm calculates (-1 - 1)^2 + (1 - (-1))^2, which becomes 4 + 4. The final answer is 2.8284271247461903. A method that ignores signs before subtraction would lose this information.

When both points are the same, every difference is zero. For:

2
5
5
5
5

the accumulated squared distance remains zero through every iteration. The final square root is 0.0, so no special handling is required.

When the dimension is larger than two, the same formula still applies without modification. For a five dimensional input where every coordinate changes by one, the accumulated value is 1 + 1 + 1 + 1 + 1 = 5, and the output is sqrt(5). The loop naturally handles this because it never assumes a fixed number of coordinates.