LeetCode #1779 — EASY

Find Nearest Point That Has the Same X or Y Coordinate

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

The Problem

Problem Statement

You are given two integers, x and y, which represent your current location on a Cartesian grid: (x, y). You are also given an array points where each points[i] = [a_i, b_i] represents that a point exists at (a_i, b_i). A point is valid if it shares the same x-coordinate or the same y-coordinate as your location.

Return the index (0-indexed) of the valid point with the smallest Manhattan distance from your current location. If there are multiple, return the valid point with the smallest index. If there are no valid points, return -1.

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

Example 1:

Input: x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]]
Output: 2
Explanation: Of all the points, only [3,1], [2,4] and [4,4] are valid. Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. [2,4] has the smallest index, so return 2.

Example 2:

Input: x = 3, y = 4, points = [[3,4]]
Output: 0
Explanation: The answer is allowed to be on the same location as your current location.

Example 3:

Input: x = 3, y = 4, points = [[2,3]]
Output: -1
Explanation: There are no valid points.

Constraints:

1 <= points.length <= 10⁴
points[i].length == 2
1 <= x, y, a_i, b_i <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given two integers, x and y, which represent your current location on a Cartesian grid: (x, y). You are also given an array points where each points[i] = [ai, bi] represents that a point exists at (ai, bi). A point is valid if it shares the same x-coordinate or the same y-coordinate as your location. Return the index (0-indexed) of the valid point with the smallest Manhattan distance from your current location. If there are multiple, return the valid point with the smallest index. If there are no valid points, return -1. The Manhattan distance between two points (x1, y1) and (x2, y2) is abs(x1 - x2) + abs(y1 - y2).

Baseline thinking

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

Pattern signal: Array

Example 1

3
4
[[1,2],[3,1],[2,4],[2,3],[4,4]]

Example 2

3
4
[[3,4]]

Example 3

3
4
[[2,3]]

Core Insight

What unlocks the optimal approach

Iterate through each point, and keep track of the current point with the smallest Manhattan distance from your current location.

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 #1779: Find Nearest Point That Has the Same X or Y Coordinate
class Solution {
    public int nearestValidPoint(int x, int y, int[][] points) {
        int ans = -1, mi = 1000000;
        for (int i = 0; i < points.length; ++i) {
            int a = points[i][0], b = points[i][1];
            if (a == x || b == y) {
                int d = Math.abs(a - x) + Math.abs(b - y);
                if (d < mi) {
                    mi = d;
                    ans = i;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1779: Find Nearest Point That Has the Same X or Y Coordinate
func nearestValidPoint(x int, y int, points [][]int) int {
	ans, mi := -1, 1000000
	for i, p := range points {
		a, b := p[0], p[1]
		if a == x || b == y {
			d := abs(a-x) + abs(b-y)
			if d < mi {
				ans, mi = i, d
			}
		}
	}
	return ans
}

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

# Accepted solution for LeetCode #1779: Find Nearest Point That Has the Same X or Y Coordinate
class Solution:
    def nearestValidPoint(self, x: int, y: int, points: List[List[int]]) -> int:
        ans, mi = -1, inf
        for i, (a, b) in enumerate(points):
            if a == x or b == y:
                d = abs(a - x) + abs(b - y)
                if mi > d:
                    ans, mi = i, d
        return ans

// Accepted solution for LeetCode #1779: Find Nearest Point That Has the Same X or Y Coordinate
impl Solution {
    pub fn nearest_valid_point(x: i32, y: i32, points: Vec<Vec<i32>>) -> i32 {
        let n = points.len();
        let mut min_dif = i32::MAX;
        let mut res = -1;
        for i in 0..n {
            let (p_x, p_y) = (points[i][0], points[i][1]);
            if p_x != x && p_y != y {
                continue;
            }
            let dif = (p_x - x).abs() + (p_y - y).abs();
            if dif < min_dif {
                min_dif = dif;
                res = i as i32;
            }
        }
        res
    }
}

// Accepted solution for LeetCode #1779: Find Nearest Point That Has the Same X or Y Coordinate
function nearestValidPoint(x: number, y: number, points: number[][]): number {
    let res = -1;
    let midDif = Infinity;
    points.forEach(([px, py], i) => {
        if (px != x && py != y) {
            return;
        }
        const dif = Math.abs(px - x) + Math.abs(py - y);
        if (dif < midDif) {
            midDif = dif;
            res = i;
        }
    });
    return res;
}

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.