LeetCode #3694 — MEDIUM

Distinct Points Reachable After Substring Removal

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

You are given a string s consisting of characters 'U', 'D', 'L', and 'R', representing moves on an infinite 2D Cartesian grid.

'U': Move from (x, y) to (x, y + 1).
'D': Move from (x, y) to (x, y - 1).
'L': Move from (x, y) to (x - 1, y).
'R': Move from (x, y) to (x + 1, y).

You are also given a positive integer k.

You must choose and remove exactly one contiguous substring of length k from s. Then, start from coordinate (0, 0) and perform the remaining moves in order.

Return an integer denoting the number of distinct final coordinates reachable.

Example 1:

Input: s = "LUL", k = 1

Output: 2

Explanation:

After removing a substring of length 1, s can be "UL", "LL" or "LU". Following these moves, the final coordinates will be (-1, 1), (-2, 0) and (-1, 1) respectively. There are two distinct points (-1, 1) and (-2, 0) so the answer is 2.

Example 2:

Input: s = "UDLR", k = 4

Output: 1

Explanation:

After removing a substring of length 4, s can only be the empty string. The final coordinates will be (0, 0). There is only one distinct point (0, 0) so the answer is 1.

Example 3:

Input: s = "UU", k = 1

Output: 1

Explanation:

After removing a substring of length 1, s becomes "U", which always ends at (0, 1), so there is only one distinct final coordinate.

Constraints:

1 <= s.length <= 10⁵
s consists of only 'U', 'D', 'L', and 'R'.
1 <= k <= s.length

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of characters 'U', 'D', 'L', and 'R', representing moves on an infinite 2D Cartesian grid. 'U': Move from (x, y) to (x, y + 1). 'D': Move from (x, y) to (x, y - 1). 'L': Move from (x, y) to (x - 1, y). 'R': Move from (x, y) to (x + 1, y). You are also given a positive integer k. You must choose and remove exactly one contiguous substring of length k from s. Then, start from coordinate (0, 0) and perform the remaining moves in order. Return an integer denoting the number of distinct final coordinates reachable.

Baseline thinking

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

Pattern signal: Hash Map · Sliding Window

Example 1

"LUL"
1

Example 2

"UDLR"
4

Example 3

"UU"
1

Step 02

Core Insight

What unlocks the optimal approach

Use prefix sums for the <code>x</code>-coordinate and <code>y</code>-coordinate.
Using the prefix sum arrays, compute the <code>x</code>-coordinate and <code>y</code>-coordinate after removing a subarray of size <code>k</code>.

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 #3694: Distinct Points Reachable After Substring Removal
class Solution {
    public int distinctPoints(String s, int k) {
        int n = s.length();
        int[] f = new int[n + 1];
        int[] g = new int[n + 1];
        int x = 0, y = 0;
        for (int i = 1; i <= n; ++i) {
            char c = s.charAt(i - 1);
            if (c == 'U') {
                ++y;
            } else if (c == 'D') {
                --y;
            } else if (c == 'L') {
                --x;
            } else {
                ++x;
            }
            f[i] = x;
            g[i] = y;
        }
        Set<Long> st = new HashSet<>();
        for (int i = k; i <= n; ++i) {
            int a = f[n] - (f[i] - f[i - k]);
            int b = g[n] - (g[i] - g[i - k]);
            st.add(1L * a * n + b);
        }
        return st.size();
    }
}

// Accepted solution for LeetCode #3694: Distinct Points Reachable After Substring Removal
func distinctPoints(s string, k int) int {
	n := len(s)
	f := make([]int, n+1)
	g := make([]int, n+1)
	x, y := 0, 0
	for i := 1; i <= n; i++ {
		c := s[i-1]
		if c == 'U' {
			y++
		} else if c == 'D' {
			y--
		} else if c == 'L' {
			x--
		} else {
			x++
		}
		f[i] = x
		g[i] = y
	}
	st := make(map[int64]struct{})
	for i := k; i <= n; i++ {
		a := f[n] - (f[i] - f[i-k])
		b := g[n] - (g[i] - g[i-k])
		key := int64(a)*int64(n) + int64(b)
		st[key] = struct{}{}
	}
	return len(st)
}

# Accepted solution for LeetCode #3694: Distinct Points Reachable After Substring Removal
class Solution:
    def distinctPoints(self, s: str, k: int) -> int:
        n = len(s)
        f = [0] * (n + 1)
        g = [0] * (n + 1)
        x = y = 0
        for i, c in enumerate(s, 1):
            if c == "U":
                y += 1
            elif c == "D":
                y -= 1
            elif c == "L":
                x -= 1
            else:
                x += 1
            f[i] = x
            g[i] = y
        st = set()
        for i in range(k, n + 1):
            a = f[n] - (f[i] - f[i - k])
            b = g[n] - (g[i] - g[i - k])
            st.add((a, b))
        return len(st)

// Accepted solution for LeetCode #3694: Distinct Points Reachable After Substring Removal
use std::collections::HashSet;

impl Solution {
    pub fn distinct_points(s: String, k: i32) -> i32 {
        let n = s.len();
        let mut f = vec![0; n + 1];
        let mut g = vec![0; n + 1];
        let mut x = 0;
        let mut y = 0;
        let bytes = s.as_bytes();
        for i in 1..=n {
            match bytes[i - 1] as char {
                'U' => y += 1,
                'D' => y -= 1,
                'L' => x -= 1,
                _ => x += 1,
            }
            f[i] = x;
            g[i] = y;
        }
        let mut st = HashSet::new();
        let k = k as usize;
        for i in k..=n {
            let a = f[n] - (f[i] - f[i - k]);
            let b = g[n] - (g[i] - g[i - k]);
            st.insert((a as i64) * (n as i64) + (b as i64));
        }
        st.len() as i32
    }
}

// Accepted solution for LeetCode #3694: Distinct Points Reachable After Substring Removal
function distinctPoints(s: string, k: number): number {
    const n = s.length;
    const f = new Array(n + 1).fill(0);
    const g = new Array(n + 1).fill(0);
    let x = 0,
        y = 0;
    for (let i = 1; i <= n; ++i) {
        const c = s[i - 1];
        if (c === 'U') ++y;
        else if (c === 'D') --y;
        else if (c === 'L') --x;
        else ++x;
        f[i] = x;
        g[i] = y;
    }
    const st = new Set<number>();
    for (let i = k; i <= n; ++i) {
        const a = f[n] - (f[i] - f[i - k]);
        const b = g[n] - (g[i] - g[i - k]);
        st.add(a * n + b);
    }
    return st.size;
}

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(n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.