LeetCode #816 — MEDIUM

Ambiguous Coordinates

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

The Problem

Problem Statement

We had some 2-dimensional coordinates, like "(1, 3)" or "(2, 0.5)". Then, we removed all commas, decimal points, and spaces and ended up with the string s.

For example, "(1, 3)" becomes s = "(13)" and "(2, 0.5)" becomes s = "(205)".

Return a list of strings representing all possibilities for what our original coordinates could have been.

Our original representation never had extraneous zeroes, so we never started with numbers like "00", "0.0", "0.00", "1.0", "001", "00.01", or any other number that can be represented with fewer digits. Also, a decimal point within a number never occurs without at least one digit occurring before it, so we never started with numbers like ".1".

The final answer list can be returned in any order. All coordinates in the final answer have exactly one space between them (occurring after the comma.)

Example 1:

Input: s = "(123)"
Output: ["(1, 2.3)","(1, 23)","(1.2, 3)","(12, 3)"]

Example 2:

Input: s = "(0123)"
Output: ["(0, 1.23)","(0, 12.3)","(0, 123)","(0.1, 2.3)","(0.1, 23)","(0.12, 3)"]
Explanation: 0.0, 00, 0001 or 00.01 are not allowed.

Example 3:

Input: s = "(00011)"
Output: ["(0, 0.011)","(0.001, 1)"]

Constraints:

4 <= s.length <= 12
s[0] == '(' and s[s.length - 1] == ')'.
The rest of s are digits.

Step 01

Brute Force Baseline

Problem summary: We had some 2-dimensional coordinates, like "(1, 3)" or "(2, 0.5)". Then, we removed all commas, decimal points, and spaces and ended up with the string s. For example, "(1, 3)" becomes s = "(13)" and "(2, 0.5)" becomes s = "(205)". Return a list of strings representing all possibilities for what our original coordinates could have been. Our original representation never had extraneous zeroes, so we never started with numbers like "00", "0.0", "0.00", "1.0", "001", "00.01", or any other number that can be represented with fewer digits. Also, a decimal point within a number never occurs without at least one digit occurring before it, so we never started with numbers like ".1". The final answer list can be returned in any order. All coordinates in the final answer have exactly one space between them (occurring after the comma.)

Baseline thinking

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

Pattern signal: Backtracking

Example 1

"(123)"

Example 2

"(0123)"

Example 3

"(00011)"

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #816: Ambiguous Coordinates
class Solution {
    public List<String> ambiguousCoordinates(String s) {
        int n = s.length();
        List<String> ans = new ArrayList<>();
        for (int i = 2; i < n - 1; ++i) {
            for (String x : f(s, 1, i)) {
                for (String y : f(s, i, n - 1)) {
                    ans.add(String.format("(%s, %s)", x, y));
                }
            }
        }
        return ans;
    }

    private List<String> f(String s, int i, int j) {
        List<String> res = new ArrayList<>();
        for (int k = 1; k <= j - i; ++k) {
            String l = s.substring(i, i + k);
            String r = s.substring(i + k, j);
            boolean ok = ("0".equals(l) || !l.startsWith("0")) && !r.endsWith("0");
            if (ok) {
                res.add(l + (k < j - i ? "." : "") + r);
            }
        }
        return res;
    }
}

// Accepted solution for LeetCode #816: Ambiguous Coordinates
func ambiguousCoordinates(s string) []string {
	f := func(i, j int) []string {
		res := []string{}
		for k := 1; k <= j-i; k++ {
			l, r := s[i:i+k], s[i+k:j]
			ok := (l == "0" || l[0] != '0') && (r == "" || r[len(r)-1] != '0')
			if ok {
				t := ""
				if k < j-i {
					t = "."
				}
				res = append(res, l+t+r)
			}
		}
		return res
	}

	n := len(s)
	ans := []string{}
	for i := 2; i < n-1; i++ {
		for _, x := range f(1, i) {
			for _, y := range f(i, n-1) {
				ans = append(ans, "("+x+", "+y+")")
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #816: Ambiguous Coordinates
class Solution:
    def ambiguousCoordinates(self, s: str) -> List[str]:
        def f(i, j):
            res = []
            for k in range(1, j - i + 1):
                l, r = s[i : i + k], s[i + k : j]
                ok = (l == '0' or not l.startswith('0')) and not r.endswith('0')
                if ok:
                    res.append(l + ('.' if k < j - i else '') + r)
            return res

        n = len(s)
        return [
            f'({x}, {y})' for i in range(2, n - 1) for x in f(1, i) for y in f(i, n - 1)
        ]

// Accepted solution for LeetCode #816: Ambiguous Coordinates
struct Solution;

trait Nums {
    fn nums(self) -> Vec<String>;
}

impl Nums for &[char] {
    fn nums(self) -> Vec<String> {
        let n = self.len();
        let mut res = vec![];
        for i in 1..=n {
            let left: String = self[..i].iter().collect();
            let right: String = self[i..].iter().collect();
            if left.starts_with('0') && left != "0" {
                continue;
            }
            if right.ends_with('0') {
                continue;
            }
            res.push(format!(
                "{}{}{}",
                left,
                if i == n { "" } else { "." },
                right
            ));
        }
        res
    }
}

impl Solution {
    fn ambiguous_coordinates(s: String) -> Vec<String> {
        let s: Vec<char> = s.chars().collect();
        let n = s.len();
        let mut res = vec![];
        for i in 2..n - 1 {
            let left = &s[1..i];
            let right = &s[i..n - 1];
            for l in left.nums() {
                for r in right.nums() {
                    res.push(format!("({}, {})", l, r));
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let s = "(123)".to_string();
    let mut res = vec_string!["(1, 23)", "(12, 3)", "(1.2, 3)", "(1, 2.3)"];
    let mut ans = Solution::ambiguous_coordinates(s);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let s = "(00011)".to_string();
    let mut res = vec_string!["(0.001, 1)", "(0, 0.011)"];
    let mut ans = Solution::ambiguous_coordinates(s);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let s = "(0123)".to_string();
    let mut res = vec_string![
        "(0, 123)",
        "(0, 12.3)",
        "(0, 1.23)",
        "(0.1, 23)",
        "(0.1, 2.3)",
        "(0.12, 3)"
    ];
    let mut ans = Solution::ambiguous_coordinates(s);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let s = "(100)".to_string();
    let mut res = vec_string!["(10, 0)"];
    let mut ans = Solution::ambiguous_coordinates(s);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #816: Ambiguous Coordinates
function ambiguousCoordinates(s: string): string[] {
    s = s.slice(1, s.length - 1);
    const n = s.length;
    const dfs = (s: string) => {
        const res: string[] = [];
        for (let i = 1; i < s.length; i++) {
            const t = `${s.slice(0, i)}.${s.slice(i)}`;
            if (`${Number(t)}` === t) {
                res.push(t);
            }
        }
        if (`${Number(s)}` === s) {
            res.push(s);
        }
        return res;
    };
    const ans: string[] = [];
    for (let i = 1; i < n; i++) {
        for (const left of dfs(s.slice(0, i))) {
            for (const right of dfs(s.slice(i))) {
                ans.push(`(${left}, ${right})`);
            }
        }
    }
    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(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.