Minimize Manhattan Distances

Minimize Manhattan Distances - Problem

Array Math Geometry Sorting Ordered Set Hard

You are given an array points representing integer coordinates of some points on a 2D plane, where points[i] = [x_i, y_i].

The Manhattan distance between two points (x1, y1) and (x2, y2) is defined as |x1 - x2| + |y1 - y2|.

Return the minimum possible value for the maximum distance between any two points by removing exactly one point.

Input & Output

Example 1 — Basic Case

$ Input: points = [[3,10],[5,15],[10,2],[4,4]]

› Output: 12

💡 Note: Remove point [3,10]. The maximum distance among remaining points is between [5,15] and [10,2]: |5-10| + |15-2| = 5 + 13 = 18. However, removing [5,15] gives max distance 12 between [3,10] and [10,2]: |3-10| + |10-2| = 7 + 8 = 15. The optimal is actually 12.

Example 2 — Minimum Size

$ Input: points = [[1,1],[3,3]]

› Output: 0

💡 Note: With only 2 points, removing one leaves just 1 point, so maximum distance is 0.

Example 3 — Three Points

$ Input: points = [[1,1],[1,3],[3,3]]

› Output: 2

💡 Note: Remove [1,1]. Remaining points [1,3] and [3,3] have distance |1-3| + |3-3| = 2. This is optimal.

Constraints

3 ≤ points.length ≤ 10⁵
points[i].length == 2
-10⁸ ≤ points[i][0], points[i][1] ≤ 10⁸

Visualization

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Asked in

G Google 25 a Amazon 15 M Microsoft 12

The key insight is transforming Manhattan distance to Chebyshev distance using coordinate transformation (x+y, x-y). This reduces the problem to finding extreme points that are candidates for removal. Best approach is coordinate transformation with Time: O(n²), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

For each point, remove it from the array and calculate the maximum Manhattan distance among all pairs of remaining points. Track the minimum of these maximum distances.

Coordinate Transformation

⏱️ Time: O(n²) Space: O(n)

Convert points using transformation (x+y, x-y) which changes Manhattan distance to Chebyshev distance. This allows us to efficiently find extreme points and candidates for removal.

Brute Force — Algorithm Steps

For each point i, create array without point i
Calculate maximum Manhattan distance for all pairs in remaining points
Track minimum of all maximum distances

Visualization

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Step-by-Step Walkthrough

Remove Point

Try removing each point one by one

Calculate Pairs

Find maximum Manhattan distance among remaining points

Track Minimum

Keep minimum of all maximum distances

Code -

solution.c — C

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

int abs_val(int x) {
    return x < 0 ? -x : x;
}

int solution(int points[][2], int n) {
    int minMaxDist = INT_MAX;
    
    for (int skip = 0; skip < n; skip++) {
        int maxDist = 0;
        for (int i = 0; i < n; i++) {
            if (i == skip) continue;
            for (int j = i + 1; j < n; j++) {
                if (j == skip) continue;
                int dist = abs_val(points[i][0] - points[j][0]) + abs_val(points[i][1] - points[j][1]);
                if (dist > maxDist) maxDist = dist;
            }
        }
        if (maxDist < minMaxDist) minMaxDist = maxDist;
    }
    
    return minMaxDist;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int points[1000][2];
    int n = 0;
    
    char *token = strtok(line, "[],");
    while (token != NULL && n < 1000) {
        if (strlen(token) > 0 && token[0] != '\n') {
            points[n][0] = atoi(token);
            token = strtok(NULL, "[],");
            if (token != NULL) {
                points[n][1] = atoi(token);
                n++;
            }
        }
        token = strtok(NULL, "[],");
    }
    
    printf("%d\n", solution(points, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of n points, calculate O(n²) pairs of remaining points

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

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