LeetCode #1433 — MEDIUM

Check If a String Can Break Another String

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

The Problem

Problem Statement

Given two strings: s1 and s2 with the same size, check if some permutation of string s1 can break some permutation of string s2 or vice-versa. In other words s2 can break s1 or vice-versa.

A string x can break string y (both of size n) if x[i] >= y[i] (in alphabetical order) for all i between 0 and n-1.

Example 1:

Input: s1 = "abc", s2 = "xya"
Output: true
Explanation: "ayx" is a permutation of s2="xya" which can break to string "abc" which is a permutation of s1="abc".

Example 2:

Input: s1 = "abe", s2 = "acd"
Output: false 
Explanation: All permutations for s1="abe" are: "abe", "aeb", "bae", "bea", "eab" and "eba" and all permutation for s2="acd" are: "acd", "adc", "cad", "cda", "dac" and "dca". However, there is not any permutation from s1 which can break some permutation from s2 and vice-versa.

Example 3:

Input: s1 = "leetcodee", s2 = "interview"
Output: true

Constraints:

s1.length == n
s2.length == n
1 <= n <= 10^5
All strings consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given two strings: s1 and s2 with the same size, check if some permutation of string s1 can break some permutation of string s2 or vice-versa. In other words s2 can break s1 or vice-versa. A string x can break string y (both of size n) if x[i] >= y[i] (in alphabetical order) for all i between 0 and n-1.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"abc"
"xya"

Example 2

"abe"
"acd"

Example 3

"leetcodee"
"interview"

Step 02

Core Insight

What unlocks the optimal approach

Sort both strings and then check if one of them can break the other.

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 #1433: Check If a String Can Break Another String
class Solution {
    public boolean checkIfCanBreak(String s1, String s2) {
        char[] cs1 = s1.toCharArray();
        char[] cs2 = s2.toCharArray();
        Arrays.sort(cs1);
        Arrays.sort(cs2);
        return check(cs1, cs2) || check(cs2, cs1);
    }

    private boolean check(char[] cs1, char[] cs2) {
        for (int i = 0; i < cs1.length; ++i) {
            if (cs1[i] < cs2[i]) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #1433: Check If a String Can Break Another String
func checkIfCanBreak(s1 string, s2 string) bool {
	cs1 := []byte(s1)
	cs2 := []byte(s2)
	sort.Slice(cs1, func(i, j int) bool { return cs1[i] < cs1[j] })
	sort.Slice(cs2, func(i, j int) bool { return cs2[i] < cs2[j] })
	check := func(cs1, cs2 []byte) bool {
		for i := range cs1 {
			if cs1[i] < cs2[i] {
				return false
			}
		}
		return true
	}
	return check(cs1, cs2) || check(cs2, cs1)
}

# Accepted solution for LeetCode #1433: Check If a String Can Break Another String
class Solution:
    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
        cs1 = sorted(s1)
        cs2 = sorted(s2)
        return all(a >= b for a, b in zip(cs1, cs2)) or all(
            a <= b for a, b in zip(cs1, cs2)
        )

// Accepted solution for LeetCode #1433: Check If a String Can Break Another String
struct Solution;

impl Solution {
    fn check_if_can_break(s1: String, s2: String) -> bool {
        let n = s1.len();
        let mut s1: Vec<char> = s1.chars().collect();
        let mut s2: Vec<char> = s2.chars().collect();
        s1.sort_unstable();
        s2.sort_unstable();
        let mut sum1 = 0;
        let mut sum2 = 0;
        for i in 0..n {
            if s1[i] <= s2[i] {
                sum1 += 1;
            }
            if s1[i] >= s2[i] {
                sum2 += 1;
            }
        }
        sum1 == n || sum2 == n
    }
}

#[test]
fn test() {
    let s1 = "abc".to_string();
    let s2 = "xya".to_string();
    let res = true;
    assert_eq!(Solution::check_if_can_break(s1, s2), res);
    let s1 = "abe".to_string();
    let s2 = "acd".to_string();
    let res = false;
    assert_eq!(Solution::check_if_can_break(s1, s2), res);
    let s1 = "leetcodee".to_string();
    let s2 = "interview".to_string();
    let res = true;
    assert_eq!(Solution::check_if_can_break(s1, s2), res);
}

// Accepted solution for LeetCode #1433: Check If a String Can Break Another String
function checkIfCanBreak(s1: string, s2: string): boolean {
    const cs1: string[] = Array.from(s1);
    const cs2: string[] = Array.from(s2);
    cs1.sort();
    cs2.sort();
    const check = (cs1: string[], cs2: string[]) => {
        for (let i = 0; i < cs1.length; i++) {
            if (cs1[i] < cs2[i]) {
                return false;
            }
        }
        return true;
    };
    return check(cs1, cs2) || check(cs2, cs1);
}

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.