1433. Check If a String Can Break Another String

Description

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.

 

Solutions

Solution: Greedy

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string} s1
 * @param {string} s2
 * @return {boolean}
 */
const checkIfCanBreak = function (s1, s2) {
  const permutationS1 = s1.split('').sort();
  const permutationS2 = s2.split('').sort();
  const isCanBreak = (s1, s2) => {
    for (const [index, element] of s1.entries()) {
      if (element >= s2[index]) continue;
      return false;
    }
    return true;
  };

  return isCanBreak(permutationS1, permutationS2) || isCanBreak(permutationS2, permutationS1);
};