LeetCode #3464 — HARD

Maximize the Distance Between Points on a Square

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer side, representing the edge length of a square with corners at (0, 0), (0, side), (side, 0), and (side, side) on a Cartesian plane.

You are also given a positive integer k and a 2D integer array points, where points[i] = [x_i, y_i] represents the coordinate of a point lying on the boundary of the square.

You need to select k elements among points such that the minimum Manhattan distance between any two points is maximized.

Return the maximum possible minimum Manhattan distance between the selected k points.

The Manhattan Distance between two cells (x_i, y_i) and (x_j, y_j) is |x_i - x_j| + |y_i - y_j|.

Example 1:

Input: side = 2, points = [[0,2],[2,0],[2,2],[0,0]], k = 4

Output: 2

Explanation:

Select all four points.

Example 2:

Input: side = 2, points = [[0,0],[1,2],[2,0],[2,2],[2,1]], k = 4

Output: 1

Explanation:

Select the points (0, 0), (2, 0), (2, 2), and (2, 1).

Example 3:

Input: side = 2, points = [[0,0],[0,1],[0,2],[1,2],[2,0],[2,2],[2,1]], k = 5

Output: 1

Explanation:

Select the points (0, 0), (0, 1), (0, 2), (1, 2), and (2, 2).

Constraints:

1 <= side <= 10⁹
4 <= points.length <= min(4 * side, 15 * 10³)
points[i] == [xi, yi]
The input is generated such that:
- points[i] lies on the boundary of the square.
- All points[i] are unique.
4 <= k <= min(25, points.length)

Step 01

Brute Force Baseline

Problem summary: You are given an integer side, representing the edge length of a square with corners at (0, 0), (0, side), (side, 0), and (side, side) on a Cartesian plane. You are also given a positive integer k and a 2D integer array points, where points[i] = [xi, yi] represents the coordinate of a point lying on the boundary of the square. You need to select k elements among points such that the minimum Manhattan distance between any two points is maximized. Return the maximum possible minimum Manhattan distance between the selected k points. The Manhattan Distance between two cells (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Binary Search

Example 1

2
[[0,2],[2,0],[2,2],[0,0]]
4

Example 2

2
[[0,0],[1,2],[2,0],[2,2],[2,1]]
4

Example 3

2
[[0,0],[0,1],[0,2],[1,2],[2,0],[2,2],[2,1]]
5

Core Insight

What unlocks the optimal approach

Can we use binary search for this problem?
Think of the coordinates on a straight line in clockwise order.
Binary search on the minimum Manhattan distance <code>x</code>.
During the binary search, for each coordinate, find the immediate next coordinate with distance >= <code>x</code>.
Greedily select up to <code>k</code> coordinates.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3464: Maximize the Distance Between Points on a Square
class Solution {
  public int maxDistance(int side, int[][] points, int k) {
    List<int[]> ordered = getOrderedPoints(side, points);
    int l = 0;
    int r = side;

    while (l < r) {
      final int m = (l + r + 1) / 2;
      if (isValidDistance(ordered, k, m))
        l = m;
      else
        r = m - 1;
    }

    return l;
  }

  private record Sequence(int startX, int startY, int endX, int endY, int length) {}

  // Returns true if we can select `k` points such that the minimum Manhattan
  // distance between any two consecutive chosen points is at least `m`.
  private boolean isValidDistance(List<int[]> ordered, int k, int d) {
    Deque<Sequence> dq = new ArrayDeque<>(List.of(new Sequence(
        ordered.get(0)[0], ordered.get(0)[1], ordered.get(0)[0], ordered.get(0)[1], 1)));
    int maxLength = 1;

    for (int i = 1; i < ordered.size(); ++i) {
      final int x = ordered.get(i)[0];
      final int y = ordered.get(i)[1];
      int startX = x;
      int startY = y;
      int length = 1;
      while (!dq.isEmpty() &&
             (Math.abs(x - dq.peekFirst().endX()) + Math.abs(y - dq.peekFirst().endY()) >= d)) {
        if (Math.abs(x - dq.peekFirst().startX()) + Math.abs(y - dq.peekFirst().startY()) >= d &&
            dq.peekFirst().length() + 1 >= length) {
          startX = dq.peekFirst().startX();
          startY = dq.peekFirst().startY();
          length = dq.peekFirst().length() + 1;
          maxLength = Math.max(maxLength, length);
        }
        dq.pollFirst();
      }
      dq.addLast(new Sequence(startX, startY, x, y, length));
    }

    return maxLength >= k;
  }

  // Returns the ordered points on the perimeter of a square of side length
  // `side`, starting from left, top, right, and bottom boundaries.
  private List<int[]> getOrderedPoints(int side, int[][] points) {
    List<int[]> left = new ArrayList<>();
    List<int[]> top = new ArrayList<>();
    List<int[]> right = new ArrayList<>();
    List<int[]> bottom = new ArrayList<>();

    for (int[] point : points) {
      final int x = point[0];
      final int y = point[1];
      if (x == 0 && y > 0)
        left.add(point);
      else if (x > 0 && y == side)
        top.add(point);
      else if (x == side && y < side)
        right.add(point);
      else
        bottom.add(point);
    }

    left.sort(Comparator.comparingInt(a -> a[1]));
    top.sort(Comparator.comparingInt(a -> a[0]));
    right.sort(Comparator.comparingInt(a -> - a[1]));
    bottom.sort(Comparator.comparingInt(a -> - a[0]));
    List<int[]> ordered = new ArrayList<>();
    ordered.addAll(left);
    ordered.addAll(top);
    ordered.addAll(right);
    ordered.addAll(bottom);
    return ordered;
  }
}

// Accepted solution for LeetCode #3464: Maximize the Distance Between Points on a Square
package main

import (
	"math"
	"math/bits"
	"slices"
	"sort"
)

// https://space.bilibili.com/206214
func maxDistance(side int, points [][]int, k int) int {
	n := len(points)
	a := make([]int, n)
	for i, p := range points {
		x, y := p[0], p[1]
		if x == 0 {
			a[i] = y
		} else if y == side {
			a[i] = side + x
		} else if x == side {
			a[i] = side*3 - y
		} else {
			a[i] = side*4 - x
		}
	}
	slices.Sort(a)

	f := make([]int, n+1)
	end := make([]int, n) // 改成保存 a[i]，但 C++/Java 这样写就得用 64 位整数了

	ans := sort.Search(side*4/k, func(low int) bool {
		low++
		j := n
		for i := n - 1; i >= 0; i-- {
			x := a[i]
			for a[j-1] >= x+low {
				j--
			}
			f[i] = f[j] + 1
			if f[i] == 1 {
				end[i] = x // i 自己就是最后一个点
			} else {
				end[i] = end[j]
			}
			if f[i] == k && end[i]-x <= side*4-low {
				return false
			}
		}
		return true
	})
	return ans
}

func maxDistance4(side int, points [][]int, k int) int {
	n := len(points)
	a := make([]int, n, n+1)
	for i, p := range points {
		x, y := p[0], p[1]
		if x == 0 {
			a[i] = y
		} else if y == side {
			a[i] = side + x
		} else if x == side {
			a[i] = side*3 - y
		} else {
			a[i] = side*4 - x
		}
	}
	slices.Sort(a)
	a = append(a, math.MaxInt) // 哨兵

	g := make([][]int, n+1)
	ans := sort.Search(side*4/k, func(low int) bool {
		low++
		clear(g)
		j := n
		for i := n - 1; i >= 0; i-- {
			for a[j-1] >= a[i]+low {
				j--
			}
			g[j] = append(g[j], i) // 建树
		}

		st := []int{}
		var dfs func(int) bool
		dfs = func(x int) bool {
			st = append(st, a[x])
			m := len(st)
			// 注意栈中多了一个 a[n]，所以是 m > k 不是 >=
			if m > k && st[m-k]-a[x] <= side*4-low {
				return true
			}
			for _, y := range g[x] {
				if dfs(y) {
					return true
				}
			}
			st = st[:m-1] // 恢复现场
			return false
		}
		return !dfs(n)
	})
	return ans
}

func maxDistance3(side int, points [][]int, k int) int {
	n := len(points)
	a := make([]int, n)
	for i, p := range points {
		x, y := p[0], p[1]
		if x == 0 {
			a[i] = y
		} else if y == side {
			a[i] = side + x
		} else if x == side {
			a[i] = side*3 - y
		} else {
			a[i] = side*4 - x
		}
	}
	slices.Sort(a)

	k-- // 往后跳 k-1 步，这里先减一，方便计算
	highBit := bits.Len(uint(k)) - 1
	nxt := make([][5]int, n+1) // 5 可以改为 highBit+1（用 array 而不是 slice，提高访问效率）
	for j := range nxt[n] {
		nxt[n][j] = n // 哨兵
	}

	ans := sort.Search(side*4/k, func(low int) bool {
		low++
		// 预处理倍增数组 nxt
		j := n
		for i := n - 1; i >= 0; i-- {
			for a[j-1] >= a[i]+low {
				j--
			}
			nxt[i][0] = j
			for k := 1; k <= highBit; k++ {
				nxt[i][k] = nxt[nxt[i][k-1]][k-1]
			}
		}

		// 枚举起点
		for i, start := range a {
			// 往后跳 k-1 步（注意上面把 k 减一了）
			cur := i
			for j := highBit; j >= 0; j-- {
				if k>>j&1 > 0 {
					cur = nxt[cur][j]
				}
			}
			if cur == n { // 出界
				break
			}
			if a[cur]-start <= side*4-low {
				return false
			}
		}
		return true
	})
	return ans
}

func maxDistance22(side int, points [][]int, k int) int {
	a := make([]int, len(points))
	for i, p := range points {
		x, y := p[0], p[1]
		if x == 0 {
			a[i] = y
		} else if y == side {
			a[i] = side + x
		} else if x == side {
			a[i] = side*3 - y
		} else {
			a[i] = side*4 - x
		}
	}
	slices.Sort(a)

	ans := sort.Search(side*4/k, func(low int) bool {
		low++
		idx := make([]int, k)
		cur := a[0]
		for j, i := 1, 0; j < k; j++ {
			i += sort.Search(len(a)-i, func(j int) bool { return a[i+j] >= cur+low })
			if i == len(a) {
				return true
			}
			idx[j] = i
			cur = a[i]
		}
		if cur-a[0] <= side*4-low {
			return false
		}

		// 第一个指针移动到第二个指针的位置，就不用继续枚举了
		end0 := idx[1]
		for idx[0]++; idx[0] < end0; idx[0]++ {
			for j := 1; j < k; j++ {
				for a[idx[j]] < a[idx[j-1]]+low {
					idx[j]++
					if idx[j] == len(a) {
						return true
					}
				}
			}
			if a[idx[k-1]]-a[idx[0]] <= side*4-low {
				return false
			}
		}
		return true
	})
	return ans
}

func maxDistance21(side int, points [][]int, k int) int {
	a := make([]int, len(points))
	for i, p := range points {
		x, y := p[0], p[1]
		if x == 0 {
			a[i] = y
		} else if y == side {
			a[i] = side + x
		} else if x == side {
			a[i] = side*3 - y
		} else {
			a[i] = side*4 - x
		}
	}
	slices.Sort(a)

	ans := sort.Search(side*4/k, func(low int) bool {
		low++
		idx := make([]int, k)
		for {
			for j := 1; j < k; j++ {
				for a[idx[j]] < a[idx[j-1]]+low {
					idx[j]++
					if idx[j] == len(a) {
						return true
					}
				}
			}
			if a[idx[k-1]]-a[idx[0]] <= side*4-low {
				return false
			}
			idx[0]++
		}
	})
	return ans
}

# Accepted solution for LeetCode #3464: Maximize the Distance Between Points on a Square
from dataclasses import dataclass


@dataclass(frozen=True)
class Sequence:
  startX: int
  startY: int
  endX: int
  endY: int
  length: int

  def __iter__(self):
    yield self.startX
    yield self.startY
    yield self.endX
    yield self.endY
    yield self.length


class Solution:
  def maxDistance(self, side: int, points: list[list[int]], k: int) -> int:
    ordered = self._getOrderedPoints(side, points)

    def isValidDistance(m: int) -> bool:
      """
      Returns True if we can select `k` points such that the minimum Manhattan
      distance between any two consecutive chosen points is at least `m`.
      """
      dq = collections.deque([Sequence(*ordered[0], *ordered[0], 1)])
      maxLength = 1

      for i in range(1, len(ordered)):
        x, y = ordered[i]
        startX, startY = ordered[i]
        length = 1
        while dq and abs(x - dq[0].endX) + abs(y - dq[0].endY) >= m:
          if (abs(x - dq[0].startX) + abs(y - dq[0].startY) >= m
                  and dq[0].length + 1 >= length):
            startX = dq[0].startX
            startY = dq[0].startY
            length = dq[0].length + 1
            maxLength = max(maxLength, length)
          dq.popleft()
        dq.append(Sequence(startX, startY, x, y, length))

      return maxLength >= k

    l = 0
    r = side

    while l < r:
      m = (l + r + 1) // 2
      if isValidDistance(m):
        l = m
      else:
        r = m - 1

    return l

  def _getOrderedPoints(self, side: int, points: list[list[int]]) -> list[list[int]]:
    """
    Returns the ordered points on the perimeter of a square of side length
    `side`, starting from left, top, right, and bottom boundaries.
    """
    left = sorted([(x, y) for x, y in points if x == 0 and y > 0])
    top = sorted([(x, y) for x, y in points if x > 0 and y == side])
    right = sorted([(x, y) for x, y in points if x == side and y < side],
                   reverse=True)
    bottom = sorted([(x, y) for x, y in points if y == 0], reverse=True)
    return left + top + right + bottom

// Accepted solution for LeetCode #3464: Maximize the Distance Between Points on a Square
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3464: Maximize the Distance Between Points on a Square
// class Solution {
//   public int maxDistance(int side, int[][] points, int k) {
//     List<int[]> ordered = getOrderedPoints(side, points);
//     int l = 0;
//     int r = side;
// 
//     while (l < r) {
//       final int m = (l + r + 1) / 2;
//       if (isValidDistance(ordered, k, m))
//         l = m;
//       else
//         r = m - 1;
//     }
// 
//     return l;
//   }
// 
//   private record Sequence(int startX, int startY, int endX, int endY, int length) {}
// 
//   // Returns true if we can select `k` points such that the minimum Manhattan
//   // distance between any two consecutive chosen points is at least `m`.
//   private boolean isValidDistance(List<int[]> ordered, int k, int d) {
//     Deque<Sequence> dq = new ArrayDeque<>(List.of(new Sequence(
//         ordered.get(0)[0], ordered.get(0)[1], ordered.get(0)[0], ordered.get(0)[1], 1)));
//     int maxLength = 1;
// 
//     for (int i = 1; i < ordered.size(); ++i) {
//       final int x = ordered.get(i)[0];
//       final int y = ordered.get(i)[1];
//       int startX = x;
//       int startY = y;
//       int length = 1;
//       while (!dq.isEmpty() &&
//              (Math.abs(x - dq.peekFirst().endX()) + Math.abs(y - dq.peekFirst().endY()) >= d)) {
//         if (Math.abs(x - dq.peekFirst().startX()) + Math.abs(y - dq.peekFirst().startY()) >= d &&
//             dq.peekFirst().length() + 1 >= length) {
//           startX = dq.peekFirst().startX();
//           startY = dq.peekFirst().startY();
//           length = dq.peekFirst().length() + 1;
//           maxLength = Math.max(maxLength, length);
//         }
//         dq.pollFirst();
//       }
//       dq.addLast(new Sequence(startX, startY, x, y, length));
//     }
// 
//     return maxLength >= k;
//   }
// 
//   // Returns the ordered points on the perimeter of a square of side length
//   // `side`, starting from left, top, right, and bottom boundaries.
//   private List<int[]> getOrderedPoints(int side, int[][] points) {
//     List<int[]> left = new ArrayList<>();
//     List<int[]> top = new ArrayList<>();
//     List<int[]> right = new ArrayList<>();
//     List<int[]> bottom = new ArrayList<>();
// 
//     for (int[] point : points) {
//       final int x = point[0];
//       final int y = point[1];
//       if (x == 0 && y > 0)
//         left.add(point);
//       else if (x > 0 && y == side)
//         top.add(point);
//       else if (x == side && y < side)
//         right.add(point);
//       else
//         bottom.add(point);
//     }
// 
//     left.sort(Comparator.comparingInt(a -> a[1]));
//     top.sort(Comparator.comparingInt(a -> a[0]));
//     right.sort(Comparator.comparingInt(a -> - a[1]));
//     bottom.sort(Comparator.comparingInt(a -> - a[0]));
//     List<int[]> ordered = new ArrayList<>();
//     ordered.addAll(left);
//     ordered.addAll(top);
//     ordered.addAll(right);
//     ordered.addAll(bottom);
//     return ordered;
//   }
// }

// Accepted solution for LeetCode #3464: Maximize the Distance Between Points on a Square
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3464: Maximize the Distance Between Points on a Square
// class Solution {
//   public int maxDistance(int side, int[][] points, int k) {
//     List<int[]> ordered = getOrderedPoints(side, points);
//     int l = 0;
//     int r = side;
// 
//     while (l < r) {
//       final int m = (l + r + 1) / 2;
//       if (isValidDistance(ordered, k, m))
//         l = m;
//       else
//         r = m - 1;
//     }
// 
//     return l;
//   }
// 
//   private record Sequence(int startX, int startY, int endX, int endY, int length) {}
// 
//   // Returns true if we can select `k` points such that the minimum Manhattan
//   // distance between any two consecutive chosen points is at least `m`.
//   private boolean isValidDistance(List<int[]> ordered, int k, int d) {
//     Deque<Sequence> dq = new ArrayDeque<>(List.of(new Sequence(
//         ordered.get(0)[0], ordered.get(0)[1], ordered.get(0)[0], ordered.get(0)[1], 1)));
//     int maxLength = 1;
// 
//     for (int i = 1; i < ordered.size(); ++i) {
//       final int x = ordered.get(i)[0];
//       final int y = ordered.get(i)[1];
//       int startX = x;
//       int startY = y;
//       int length = 1;
//       while (!dq.isEmpty() &&
//              (Math.abs(x - dq.peekFirst().endX()) + Math.abs(y - dq.peekFirst().endY()) >= d)) {
//         if (Math.abs(x - dq.peekFirst().startX()) + Math.abs(y - dq.peekFirst().startY()) >= d &&
//             dq.peekFirst().length() + 1 >= length) {
//           startX = dq.peekFirst().startX();
//           startY = dq.peekFirst().startY();
//           length = dq.peekFirst().length() + 1;
//           maxLength = Math.max(maxLength, length);
//         }
//         dq.pollFirst();
//       }
//       dq.addLast(new Sequence(startX, startY, x, y, length));
//     }
// 
//     return maxLength >= k;
//   }
// 
//   // Returns the ordered points on the perimeter of a square of side length
//   // `side`, starting from left, top, right, and bottom boundaries.
//   private List<int[]> getOrderedPoints(int side, int[][] points) {
//     List<int[]> left = new ArrayList<>();
//     List<int[]> top = new ArrayList<>();
//     List<int[]> right = new ArrayList<>();
//     List<int[]> bottom = new ArrayList<>();
// 
//     for (int[] point : points) {
//       final int x = point[0];
//       final int y = point[1];
//       if (x == 0 && y > 0)
//         left.add(point);
//       else if (x > 0 && y == side)
//         top.add(point);
//       else if (x == side && y < side)
//         right.add(point);
//       else
//         bottom.add(point);
//     }
// 
//     left.sort(Comparator.comparingInt(a -> a[1]));
//     top.sort(Comparator.comparingInt(a -> a[0]));
//     right.sort(Comparator.comparingInt(a -> - a[1]));
//     bottom.sort(Comparator.comparingInt(a -> - a[0]));
//     List<int[]> ordered = new ArrayList<>();
//     ordered.addAll(left);
//     ordered.addAll(top);
//     ordered.addAll(right);
//     ordered.addAll(bottom);
//     return ordered;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Maximize the Distance Between Points on a Square

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Overflow in intermediate arithmetic

Boundary update without `+1` / `-1`