LeetCode #1266 — EASY

Minimum Time Visiting All Points

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

On a 2D plane, there are n points with integer coordinates points[i] = [x_i, y_i]. Return the minimum time in seconds to visit all the points in the order given by points.

You can move according to these rules:

In 1 second, you can either:
- move vertically by one unit,
- move horizontally by one unit, or
- move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second).
You have to visit the points in the same order as they appear in the array.
You are allowed to pass through points that appear later in the order, but these do not count as visits.

Example 1:

Input: points = [[1,1],[3,4],[-1,0]]
Output: 7
Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0]   
Time from [1,1] to [3,4] = 3 seconds 
Time from [3,4] to [-1,0] = 4 seconds
Total time = 7 seconds

Example 2:

Input: points = [[3,2],[-2,2]]
Output: 5

Constraints:

points.length == n
1 <= n <= 100
points[i].length == 2
-1000 <= points[i][0], points[i][1] <= 1000

Step 01

Brute Force Baseline

Problem summary: On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return the minimum time in seconds to visit all the points in the order given by points. You can move according to these rules: In 1 second, you can either: move vertically by one unit, move horizontally by one unit, or move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second). You have to visit the points in the same order as they appear in the array. You are allowed to pass through points that appear later in the order, but these do not count as visits.

Baseline thinking

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

Pattern signal: Array · Math

Example 1

[[1,1],[3,4],[-1,0]]

Example 2

[[3,2],[-2,2]]

Step 02

Core Insight

What unlocks the optimal approach

To walk from point A to point B there will be an optimal strategy to walk ?
Advance in diagonal as possible then after that go in straight line.
Repeat the process until visiting all the points.

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

Upper-end input sizes

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 #1266: Minimum Time Visiting All Points
class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int ans = 0;
        for (int i = 1; i < points.length; ++i) {
            int dx = Math.abs(points[i][0] - points[i - 1][0]);
            int dy = Math.abs(points[i][1] - points[i - 1][1]);
            ans += Math.max(dx, dy);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1266: Minimum Time Visiting All Points
func minTimeToVisitAllPoints(points [][]int) (ans int) {
	for i, p := range points[1:] {
		dx := abs(p[0] - points[i][0])
		dy := abs(p[1] - points[i][1])
		ans += max(dx, dy)
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1266: Minimum Time Visiting All Points
class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        return sum(
            max(abs(p1[0] - p2[0]), abs(p1[1] - p2[1])) for p1, p2 in pairwise(points)
        )

// Accepted solution for LeetCode #1266: Minimum Time Visiting All Points
impl Solution {
    pub fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 {
        let mut ans = 0;
        for i in 1..points.len() {
            let dx = (points[i][0] - points[i - 1][0]).abs();
            let dy = (points[i][1] - points[i - 1][1]).abs();
            ans += dx.max(dy);
        }
        ans
    }
}

// Accepted solution for LeetCode #1266: Minimum Time Visiting All Points
function minTimeToVisitAllPoints(points: number[][]): number {
    let ans = 0;
    for (let i = 1; i < points.length; i++) {
        const dx = Math.abs(points[i][0] - points[i - 1][0]);
        const dy = Math.abs(points[i][1] - points[i - 1][1]);
        ans += Math.max(dx, dy);
    }
    return ans;
}

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(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.