LeetCode #3443 — MEDIUM

Maximum Manhattan Distance After K Changes

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 the characters 'N', 'S', 'E', and 'W', where s[i] indicates movements in an infinite grid:

'N' : Move north by 1 unit.
'S' : Move south by 1 unit.
'E' : Move east by 1 unit.
'W' : Move west by 1 unit.

Initially, you are at the origin (0, 0). You can change at most k characters to any of the four directions.

Find the maximum Manhattan distance from the origin that can be achieved at any time while performing the movements in order.

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: s = "NWSE", k = 1

Output: 3

Explanation:

Change s[2] from 'S' to 'N'. The string s becomes "NWNE".

Movement	Position (x, y)	Manhattan Distance	Maximum
s[0] == 'N'	(0, 1)	0 + 1 = 1	1
s[1] == 'W'	(-1, 1)	1 + 1 = 2	2
s[2] == 'N'	(-1, 2)	1 + 2 = 3	3
s[3] == 'E'	(0, 2)	0 + 2 = 2	3

The maximum Manhattan distance from the origin that can be achieved is 3. Hence, 3 is the output.

Example 2:

Input: s = "NSWWEW", k = 3

Output: 6

Explanation:

Change s[1] from 'S' to 'N', and s[4] from 'E' to 'W'. The string s becomes "NNWWWW".

The maximum Manhattan distance from the origin that can be achieved is 6. Hence, 6 is the output.

Constraints:

1 <= s.length <= 10⁵
0 <= k <= s.length
s consists of only 'N', 'S', 'E', and 'W'.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of the characters 'N', 'S', 'E', and 'W', where s[i] indicates movements in an infinite grid: 'N' : Move north by 1 unit. 'S' : Move south by 1 unit. 'E' : Move east by 1 unit. 'W' : Move west by 1 unit. Initially, you are at the origin (0, 0). You can change at most k characters to any of the four directions. Find the maximum Manhattan distance from the origin that can be achieved at any time while performing the movements in order. 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: Hash Map · Math

Example 1

"NWSE"
1

Example 2

"NSWWEW"
3

Core Insight

What unlocks the optimal approach

We can brute force all the possible directions (NE, NW, SE, SW).
Change up to <code>k</code> characters to maximize the distance in the chosen direction.

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 #3443: Maximum Manhattan Distance After K Changes
class Solution {
    private char[] s;
    private int k;

    public int maxDistance(String s, int k) {
        this.s = s.toCharArray();
        this.k = k;
        int a = calc('S', 'E');
        int b = calc('S', 'W');
        int c = calc('N', 'E');
        int d = calc('N', 'W');
        return Math.max(Math.max(a, b), Math.max(c, d));
    }

    private int calc(char a, char b) {
        int ans = 0, mx = 0, cnt = 0;
        for (char c : s) {
            if (c == a || c == b) {
                ++mx;
            } else if (cnt < k) {
                ++mx;
                ++cnt;
            } else {
                --mx;
            }
            ans = Math.max(ans, mx);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3443: Maximum Manhattan Distance After K Changes
func maxDistance(s string, k int) int {
	calc := func(a rune, b rune) int {
		var ans, mx, cnt int
		for _, c := range s {
			if c == a || c == b {
				mx++
			} else if cnt < k {
				mx++
				cnt++
			} else {
				mx--
			}
			ans = max(ans, mx)
		}
		return ans
	}
	a := calc('S', 'E')
	b := calc('S', 'W')
	c := calc('N', 'E')
	d := calc('N', 'W')
	return max(a, b, c, d)
}

# Accepted solution for LeetCode #3443: Maximum Manhattan Distance After K Changes
class Solution:
    def maxDistance(self, s: str, k: int) -> int:
        def calc(a: str, b: str) -> int:
            ans = mx = cnt = 0
            for c in s:
                if c == a or c == b:
                    mx += 1
                elif cnt < k:
                    cnt += 1
                    mx += 1
                else:
                    mx -= 1
                ans = max(ans, mx)
            return ans

        a = calc("S", "E")
        b = calc("S", "W")
        c = calc("N", "E")
        d = calc("N", "W")
        return max(a, b, c, d)

// Accepted solution for LeetCode #3443: Maximum Manhattan Distance After K Changes
impl Solution {
    pub fn max_distance(s: String, k: i32) -> i32 {
        fn calc(s: &str, a: char, b: char, k: i32) -> i32 {
            let mut ans = 0;
            let mut mx = 0;
            let mut cnt = 0;
            for c in s.chars() {
                if c == a || c == b {
                    mx += 1;
                } else if cnt < k {
                    mx += 1;
                    cnt += 1;
                } else {
                    mx -= 1;
                }
                ans = ans.max(mx);
            }
            ans
        }

        let a = calc(&s, 'S', 'E', k);
        let b = calc(&s, 'S', 'W', k);
        let c = calc(&s, 'N', 'E', k);
        let d = calc(&s, 'N', 'W', k);
        a.max(b).max(c).max(d)
    }
}

// Accepted solution for LeetCode #3443: Maximum Manhattan Distance After K Changes
function maxDistance(s: string, k: number): number {
    const calc = (a: string, b: string): number => {
        let [ans, mx, cnt] = [0, 0, 0];
        for (const c of s) {
            if (c === a || c === b) {
                ++mx;
            } else if (cnt < k) {
                ++mx;
                ++cnt;
            } else {
                --mx;
            }
            ans = Math.max(ans, mx);
        }
        return ans;
    };
    const a = calc('S', 'E');
    const b = calc('S', 'W');
    const c = calc('N', 'E');
    const d = calc('N', 'W');
    return Math.max(a, b, c, d);
}

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(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

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.

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.